case study on electricity class 10

Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity, case study: 1.

Thus, V= IR

The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

3) Why tungsten is used in electric bulbs?

4) 1M ohm = 10 6 ohm

Case study: 2

Thus, total current through the circuit is the sum of current flowing through each resistor.

I = I1 + I2 + I3

Questions:

3) If n resistors of resistance R are connected in parallel then what is the equivalent resistance?

1/Re = 1/R + 1/R + 1/R +….n times 1/R

Thus, 1/Re = n/R

Case study:3

3) 1KW h = ?

2) In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

CBSE 10th Standard Science Subject Electricity Chapter Case Study Questions With Solution 2021

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Cbse 10th standard science subject electricity case study questions with solution 2021.

10th Standard CBSE

Final Semester - June 2015

The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current. The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference. $\begin{equation} V_{A B}=V_{B}-V_{A}=\frac{W_{B A}}{q} \end{equation}$ The SI unit of potential and potential difference is volt. (i) The 2 C of charge is flowing through a conductor in 100 rns, the current in the circuit is

(ii) Which of the following is true? (a) Current flows from positive terminal ofthe cell to the negative terminal of the cell outside the cell. (b) The negative charge moves from lower potential to higher potential. (c) The direction of flow of current in same as the direction of flow of positive charge. (d) All of these (iii) The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is

(iv) The number of electrons flowing per second in a conductor if 1A current is passing through it

(v) The voltage can be written as

The relationship between potential difference and current was first established by George Simon Ohm called Ohm's law. According to this law, the current through a metallic conductor is proportional to the potential difference applied between its ends, provided the temperature remain constant i.e. I $\begin{equation} \propto \end{equation}$ V or V = IR; where R is constant for the conductor and it is called resistance of the conductor. Although Ohm's law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. (i) If both the potential difference and the resistance in a circuit are doubled, then

(ii) For a conductor, the graph between V and I is there. Which one is the correct?

(iii) The slope of V - I graph (V on x-axis and I on y-axis) gives

(iv) When battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is

(v) By increasing the voltage across a conductor, the

The obstruction offered by a conductor in the path of flow of current is called resistance. The SI unit of resistance is ohm ( $\begin{equation} \Omega \end{equation}$ ). It has been found that the resistance of a conductor depends on the temperature of the conductor. As the temperature increases the resistance also increases. But the resistance of alloys like mangnin, constantan and nichrome is almost unaffected by temperature. The resistance of a conductor also depends on the length of conductor and the area of cross-section of the conductor. More be the length, more will be the resistance, more be the area of cross-section, lesser will be the resistance. (i) Which of the following is not will desired in material being used for making electrical wires?

< T	> T
= T

(iii) Two wires of same material one of length L and area of cross-section A, other is of length 2L and area A/2 . Which of the following is correct?

= R	= 4R
= 4R	= 2R

(iv) For the same conducting wire (a) resistance is higher in summer (b) resistance is higher in winter (c) resistance is same is summer or in winter (d) none of these (v) A wire of resistance 20 $\begin{equation} \Omega \end{equation}$ is cut into 5 equal pieces. The resistance of each part is

= V = V	+ V + V
+ V + V = 3V	+ V + V

(ii) When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is

(iii) There is a wire oflength 20 cm and having resistance 20 $\begin{equation} \Omega \end{equation}$ cut into 4 equal pieces and then joined in series. The equivalent resistance is

*****************************************

Cbse 10th standard science subject electricity case study questions with solution 2021 answer keys.

(i) (a): q = 2 C, t= 100 ms = 0.1 s $\begin{equation} I=\frac{q}{t}=\frac{2}{0.1}=20 \mathrm{~A} \text { . } \end{equation}$ (ii) (d) (iii) (b): W = 100 J, q = 20 C $\begin{equation} V=\frac{W}{q}=\frac{100}{20}=5 \mathrm{~V} \end{equation}$ (iv) (c): I = 1 A, t = 1 s q = It = 1 x 1 = 1C $\begin{equation} n=\frac{q}{e}=\frac{1}{1.6 \times 10^{-19}}=6.25 \times 10^{18} \end{equation}$ (v) (c): $\begin{equation} V=\frac{W}{q}=\frac{W}{I t} \end{equation}$

(i)(a): V = IR So, V' $\rightarrow$ 2 V, R' $\rightarrow$ 2R $\begin{equation} I^{\prime}=\frac{2 V}{2 R}=I \end{equation}$ (ii) (b): V $\begin{equation} \propto \end{equation}$ I. So, the graph is a straight line and passing through origin. (iii) (b): Slope of V-I graph = $\begin{equation} \frac{I}{V}=\frac{1}{R} \text { . } \end{equation}$ (iv) (c): Given: V = 9 V, 1= 0.1 A $\begin{equation} R=\frac{V}{I}=\frac{9}{0.1}=90 \Omega \end{equation}$ (v) (c): On increasing the voltage, the resistance remain same, so current will increase .

(i) (b): The electrical wire should have low resistance. (ii) (b): More is the temperature, more will be the resistance. The resistance of A is more, so temperature of A is more. (iii) (c) : $\begin{equation} R_{1}=\rho \frac{L}{A}, R_{2}=\rho \frac{2 L}{A / 2} \end{equation}$ (iv) (a): In summers, temperature is more, so resistance is more. (v) (a): $\begin{equation} R^{\prime}=\frac{R}{5}=\frac{20}{5}=4 \Omega \end{equation}$

(i) (b): In series combination, the total voltage is equal to the sum of voltage drop across each resistance. (ii) (b): R s = R 1 + R 2 + R 3 So, R s = R + R + R = 3R (iii) (a) : Resistance of each wire = 20/4 = 5 $\begin{equation} \Omega \end{equation}$ Equivalent resistance in series R s = 5 + 5 + 5 + 5 = 20 $\begin{equation} \Omega \end{equation}$ (iv) (a) : All are in series, R s = 5R= 5 x 2 = 10 $\begin{equation} \Omega \end{equation}$ (v) (a) : R s = 1 + 2 + 3 = 6 $\begin{equation} \Omega \end{equation}$ $\begin{equation} I=\frac{18}{6}=3 \mathrm{~A} \end{equation}$

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Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12 are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

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In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19 electrons (b) 2.6 × 10 18 electrons (c) 2.65 × 10 19 electrons (d) 6.25 × 10 18 electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is : (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

Here we are providing you with Class 10 Science Chapter 12 Electricity Case Study Questions, by practicing these Case Study and Passage Based Questions will help you in your Class 10th Board Exam.

Case Study Chapter 12 Electricity

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

Answer: (b) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

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Case Study Chapter 12 Electricity

Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year. Please practise these case study based Class 10 Science Questions and answers to get more marks in examinations.

Case Study Questions Chapter 12 Electricity

Case/Passage – 1

Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R 1 > R 2 .

Question: Which lamp will glow more brightly when they are connected in parallel? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: Which lamp will glow more brightly when they are connected in series? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: If the lamp of resistance R 2 now burns out and the lamp of resistance R1 alone is plugged in, will the illumination increase or decrease? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) None

Question: If the lamp of resistance R 1 now burns out, how will the illumination produced change? (a) Net illumination will increase (b) Net illumination will decrease (c) Net illumination will remain same (d) Net illumination will reduced to zero

Question: Would physically bending a supply wire cause any change in the illumination? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) It is not possible to predict from the given datas

Case/Passage – 2

The rate at which electric energy is dissipated or consumed in an electric circuit. This is termed as electric power, P = IV, According to Ohm’s law V = IR We can express the power dissipated in the alternative forms P =I 2 R=V 2 /R

If 100W – 220V is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts. The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy. The total energy used in a circuit = power of the electric circuit × time.

Question: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2 /R

Question: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in sereis and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be– (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Question: In an electrical circuit, two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J

Question: In an electrical circuit three incandescent bulbs. A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A (d) Brightness of bulb C will be less than that of B

Question: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be– (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Case/Passage – 3

Answer the following questions based on the given circuit.

Question: The equivalent resistance between points A and B is (a) 7Ω (b) 6Ω (c) 13Ω (d) 5Ω

Question: The potential drop across the 3Ω resistor is (a) 1 V (b) 1.5 V (c) 2 V (d) 3 V

Question: The current flowing through in the given circuit is (a) 0.5 A (b) 1.5 A (c) 6 A (d) 3 A

Case/Passage – 4

Answer the following questions based on the given circuit.

Question: The current through each resistor is (a) 1 A (b) 2.3 A (c) 0.5 A (d) 0.75 A

Question: The equivalent resistance between points A and B, is (a) 12 Ω (b) 36 Ω (c) 32 Ω (d) 24 Ω

Question: The potential drop across the 12Ω resistor is (a) 12 V (b) 6 V (c) 8 V (d) 0.5 V

Case/Passage – 5

Question: The equivalent resistance between points A and B (a) 6.2 Ω (b) 5.1 Ω (c) 13.33 Ω (d) 1.33 Ω

Question: The current through the 4.0 ohm resistor is (a) 5.6 A (b) 0.98 A (c) 0.35 A (d) 0.68 A

Question: The current through the battery is (a) 2.33 A (b) 3.12 A (c) 4.16 A (d) 5.19 A

Case/Passage – 6

Question: The total resistance of the circuit is (a) 2 Ω (b) 4 Ω (c) 1.5 Ω (d) 0.5 Ω

Question: The current flowing through 6Ω resistor is (a) 0.50 A (b) 0.75 A (c) 0.80 A (d) 0.25

Question: The current flowing through 0.5Ω resistor is (a) 1 A (b) 1.5 A (c) 3 A (d) 2.5 A

Case/Passage – 7

Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter. Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

Question. Which type of conductor is represented by the graph given alongside?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What is the slope of graph in (i) equal to? (a) V (b) I (c) R (d) VI

Question. Which of the following is the factor on which resistance of a conductor does not depend? (a) Length (b) Area (c) Temperature (d) Pressur

Question. What type of conductor is represented by the following graph?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What type of conductors are represented by the following graph?

Study this table related to material and their resistivity and answer the questions that follow.

Question. Which of the following is used in transmission wires? (a) Cr (b) Al (c) Zn (d) Fe

Question. Which is the best conducting metal? (a) Cu (b) Ag (c) Au (d) Hg

Question. Which of the following is used as a filament in electric bulbs? (a) Nichrome (b) Tungsten (c) Manganese (d) Silver

Question. What is the range of resistivity in metals, good conductors of electricity? (a) 10–8 to 10–6 Wm (b) 10–6 to 10–4 Wm (c) 1010 to 1014 Wm (d) 1012 to 1014 Wm

Question. Which property of the alloy makes it useful in heating devices like electric iron, toasters, immersion rods, etc.? (a) Higher resistivity (b) Do not oxidise at low temperature (c) Do not reduce at high temperature (d) Oxidise at high temperature

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Class 10 Science Chapter 11 Case Based Questions - Electricity

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Case Study - 1

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans: The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans: In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6 joule Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans: Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans: At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans: If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans: The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans: Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6 ohm

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Home > Class 10 Science Subject-wise Materials

Class 10 Science Chapter 12 Electricity

Class 10 Chapter 12 Electricity as the name suggests, covers everything about electricity in detail. The constitution of electricity, the flow of electricity in the circuit, how electricity can be regulated, and much more. The chapter also includes Ohm’s law, resistors, and heating effects of electric circuits. The questions constitute 7 marks in the CBSE Class 10 exams. The inclusion of CBSE Electricity Chapter 12 is to help students create a strong foundation especially when students want to pursue the field of science and technology.

The understanding of concepts and topics included in the NCERT Chapter 12 can be done with the help of study materials like notes of electricity class 10 CBSE, question bank, mind maps, and support materials. Preparing the right study material can help in scoring good marks in the final examination.

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CBSE Class 10 Electricity Notes

Below we have provided the links to downloadable PDFs of class 10 ch 12 science notes and get an in-depth explanation and understanding of the chapter.

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CBSE Class 10 Electricity DoE Worksheet

Below, we have provided the links to downloadable PDFs of DoE Worksheets for Electricity Class 10 to practice more questions.

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Below, we have provided the links to downloadable PDFs of Experiential Learning Activity for ch 12 class 10 Science to help students implement their acquired knowledge in the real world.

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Cbse class 10 electricity important questions.

Below, we have provided Class 10 Science Important Questions that cover all the important questions in Electricity.

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Cbse class 10 electricity mind maps.

Below, we have provided Class 10 Science Mind maps that include mind maps of the related concepts in Electricity.

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Cbse class 10 electricity question bank.

Below, we have provided Class 10 Science Question Banks that cover every typology question with detailed explanations from various resources in one place

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Below, we have provided Class 10 Science Support Materials that cover Case Study-based questions from the various concepts explained in Science NCERT chapters.

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Why download these chapter-wise pdfs.

Science Class 10 Electricity chapter can include both objective and subjective questions related to Ohm’s law, SI unit of current, and magnetic effects of currents. The study materials are exam-centric and with the help of visual study materials like mind maps can help in connecting the knowledge they have acquired. With the right preparation, working strategically can help students build a strong base and score at least 7 marks in the final exams. The chapter-wise study materials are effective for both teachers and students.

Creating a study timetable and including these chapter-wise PDFs can help students prepare the chapters in a strategic and organized manner. Just allot sufficient time for understanding and revising the concepts.
Students wouldn’t have to juggle between numerous sites to find various study material that suits their learning style. The chapter-wise study material can be accessed in one place.
Download and browse these chapter-wise PDFs on any device in the comfort of your place by using an Internet connection.
After downloading these PDFs, students can get these educational materials printed and prepare accordingly.

How Can This Chapter-wise Material Help Students?

The Science Electricity chapter-wise materials can help in completing the chapter from the 10th NCERT textbook in addition to the extra study materials. Students may efficiently prepare for the chapter by downloading chapter notes, DoE worksheets, question banks, key questions, and a plethora of additional study resources.

Students could structure and focus on a certain less strong area of the subject at hand by using science class 10 electricity notes.
When it comes to making quick and efficient preparations or refining the flow of concepts under a certain topic that you might have overlooked, mind maps are a valuable tool.
The DoE worksheets and question banks may be used to study for every category of question that will be analyzed in the tenth board examinations. Once they have mastered the material, students can make a timetable and practice answering pertinent questions.
Among the Class 10 CBSE, important questions are the recurring questions and the notion of questions you should practice for the test. You may improve your chances of receiving better exam marks by rehearsing important questions.

Educators can use the additional materials and practice questions that Educart has provided to help students practice these topics completely. To download these PDFs, the user only has to verify themselves and click the link.

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NCERT Solutions for Class 10 Science Chapter 12 Electricity

September 27, 2019 by Rama Krishna

NCERT Solutions For Class 10 Science Chapter 12 Electricity : In this article, we will provide you all the necessary information regarding NCERT solutions for class 10 science physics chapter 12 electricity. Working on CBSE class 10 physics electricity questions and answers will help candidates to score good marks in-class tests as well as in the CBSE Class 10 board exam.

Electricity class 10 NCERT solutions will not only help in board exam preparation but also helps in clearing the competitive exams like Engineering. Also, candidates can find electricity class 10 numericals with solutions which helps candidates solving their assignments. Read on to find out everything NCERT Solutions For Class 10 Science Chapter 12 Electricity .

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 12 Electricity, let’s have an overview of the list of topics and subtopics under Electricity class 10 NCERT solutions :

Electricity
Electric Current And Circuit
Electric Potential And Potential Difference
Circuit Diagram
Factors On Which The Resistance Of A Conductor Depends
Resistance Of A System Of Resistors
Heating Effect Of Electric Current
Electric Power

Free download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

विद्युत कक्षा 10 विज्ञान हिंदी में
Class 10 Electricity Important Questions
Electricity Class 10 Notes
Electricity NCERT Exemplar Solutions

Class 10 Science Electricity Mind Map

Ncert solutions for class 10 science chapter 12 intext questions.

Page Number: 200

Question 1 What does an electric circuit mean ? Answer: A continuous and closed path along which an electric current flows is called an electric circuit.

Question 2 Define the unit of current. Answer: Unit of current is ampere. If one coulomb of charge flows through any section of a conductor in one second then the current through it is said to be one ampere. I = $\frac { Q }{ t }$ or 1 A = I C s -1

Question 3 Calculate the number of electrons constituting one coulomb of charge. Answer: Charge on one electron, e = 1.6 x 10 -19 C Total charge, Q = 1 C Number of electrons, n = $\frac { Q }{ e }$ = $\frac { 1C }{ 1.6x{ 10 }^{ -19 } }$ = 6.25 x 10 18

Page Number: 202

Question 1 Name a device that helps to maintain a potential difference across a conductor. Answer: A battery.

Question 2 What is meant by saying that the potential difference between two points is IV? Answer: The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Question 3 How much energy is given to each coulomb of charge passing through a 6 V battery ? Answer: Energy given by battery = charge x potential difference or W = QV = 1C X 6V = 6J.

Page Number: 209

Question 1 On what factors does the resistance of a conductor depend ? OR List the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE2018] Answer: The resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.

Question 2 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ? Answer: The current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Therefore, smaller is the resistance of the conductor.

Question 3 Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it ? Answer: When potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm’s law i.e., V ∝ I.

Question 4 Why are coils of electric toasters and electric irons are made of an-alloy rather than a pure metal ? OR Why are alloys commonly used in electric heating devices? Given reason. [CBSE 2018] Answer: The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of a pure metal, and (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.

Question 5 Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following : (i) Which among iron and mercury is a better conductor ? (ii) Which material is the best conductor ? Answer: (i) Resistivity of iron = 10.0 x 10 -8 Ω m Resistivity of mercury = 94.0 x 10 -8 Ω m. Thus iron is a better conductor because it has lower resistivity than mercury. (ii) Because silver has the lowest resistivity (= 1.60 x 10 -8 Ω m), therefore silver is the best conductor.

Page Number: 213

NCERT Solutions for Class 10 Science Chapter 12 Electricity Intext Questions Page 213 Q1

Page Number: 216

Question 1 Judge the equivalent resistance when the following are connected in parallel : (i) 1 Ω and 106 Ω, (if) 1 Ω and 103 Ω and 106 Ω. Answer: When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance. (i) Equivalent resistance < 1 Ω. (ii) Equivalent resistance < 1 Ω.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Intext Questions Page 216 Q2

Question 3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ? Answer: Advantages of connecting electrical devices in parallel with the battery are :

In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.
In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.
In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.
In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.

Page Number: 218

Question 1 Why does the cord of an electric heater not glow while the heating element does ? Solution: Heat generated in a circuit is given by I 2 R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

Question 2 Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. Solution: Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 10 6 J

Question 3 An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s. Solution: Here, R = 20 Ω, i = 5 A, t = 3s Heat developed, H = I 2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 10 4 J

Page Number: 220

Question 1 What determines the rate at which energy is delivered by a current ? Answer: Resistance of the circuit determines the rate at which energy is delivered by a current.

Question 2 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. Answer: Here, I = 5 A, V = 220 V, t = 2h = 7,200 s Power, P = V I = 220 x 5 = 1100 W Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 10 5 J

NCERT Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions

Question 1 A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is : (a) $\frac { 1 }{ 25 }$ (b) $\frac { 1 }{ 5 }$ (c) 5 (d) 25 Answer: (d) 25

Question 2 Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) $\frac { { v }^{ 2 } }{ 2 }$ Answer: (fa) IR2

Question 3 An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be : (a) 100 W (b) 75 W (c) 50 W (d) 25 W Answer: (d) 25 W

Question 4 Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be : (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Answer: (c) 1 : 4

Question 5 How is a voltmeter connected in the circuit to measure the potential difference between two points ? Answer: A voltmeter is connected in parallel to measure the potential difference between two points.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Chapter End Questions Q6

Question 8 When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. Solution: Here, V = 12 V and I = 2.5 mA = 2.5 x 10 -3 A ∴ Resistance, R = $\frac { V }{ I }$ = $\frac { 12V }{ 2.5\times { 10 }^{ 3 }A }$ = 4,800 Ω = 4.8 x 10 -3 Ω

Question 9 A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor? Solution: Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = $\frac { V }{ R }$ = $\frac { 12V }{ 13.4\Omega }$ = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Chapter End Questions Q10

Question 11 Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω Solution: Here, R 1 = R 2 = R 3 = 6 Ω.

Question 12 Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ? Solution: Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps $ =\frac { Max.Power }{ Power\quad of\quad 1\quad lamp } \quad =\quad \frac { 1100 }{ 10 } =110$ ∴ 110 lamps can be connected in parallel.

Question 16 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes ? Solution: Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min = 1200 x $\frac { 10 }{ 60 }$ = 200 Wh 60 Thus, the TV set uses more energy than the toaster.

Question 17 An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. Solution: Here, R = 8 Ω, 1 = 15 A, t = 2 h The rate at which heat is developed in the heater is equal to the power. Therefore, P = I 2 R = (15) 2 x 8 = 1800 Js -1

Question 18 Explain the following: (i) Why is tungsten used almost exclusively for filament of electric lamps ? (ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ? (in) Why is the series arrangement not used for domestic circuits ? (iv) How does the resistance of a wire vary with its area of cross-section ? (v) Why are copper and aluminium wires usually employed for electricity transmission? Answer: (i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted. (ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature. (iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken. (iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R ∝ (1/πr 2 ). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor. (v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

Electric current, potential difference and electric current, Ohms law, Resistance, Resistivity factors on which the resistance of a conductor depends; Series combination of resistors, parallel combination of resistors; and its application on daily life; Heating effect of Electric current, electric Power, Interrelation between P, V, and R.

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	NCERT
	Class 10
	Science
	Chapter 12
	Electricity
	41

Formulae Handbook for Class 10 Maths and Science

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 200 Q1

Question 1: Name a device that helps to maintain a potential difference across a conductor. Answer: Cell or battery eliminator.

Question 2: What is meant by saying that the potential difference between two points is 1 V? Answer: As we know that V = W / q Thus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.

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Question 1: On what factors does the resistance of a conductor depend Answer: Resistance of a conductor depends upon: (i) Resistivity of the material. (ii) Length of the conductor. (iii) Cross-sectional area of the conductor.

Question 2: Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why Answer: The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R ∝ 1/A.

Download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 209 Q3

Question 4: Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal Answer: The coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons; (i) The resistivity of an alloy is higher than that of a pure metal. (ii) It has high melting point and does not oxidise.

Question 5: Use the data in Table 12.2 of NCERT book to answer the following: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor? ‘ Answer: (a) Iron because its resistivity is less than mercury. (b) Silver is the best conductor as it has least resistivity.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 213 Q1

Question 1: Judge the equivalent resistance when the following are connected in parallel. (a) 1 Ω and 10 6 Ω (b) 1 Ω , 10 3 Ω and 10 6 Ω Answer: Equivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit. Hence, in both the given cases, the equivalent resistance is less than 1 Ω.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 216 Q2

Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ? Answer: Advantages of connecting electrical devices in parallel:

When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.
Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
If one appliance fails to work, other will continue to work properly.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 216 Q4

Question 1: Why does the cord of an electric heater not glow while the heating element does? Answer: The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced ‘H’ is H = I 2 Rt Thus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Page 218 Q2

Question 3: An electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s. Answer: Given R = 20 Ω, I = 5 A, t = 30 s H = I 2 Rt = (5) 2 x 20 x 30 = 15000 J = 1.5 x 10 4 J

Question 1: What determines the rate at which energy is delivered by a current? Answer: Electric power determines the rate at which energy is delivered by a current.

Question 2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. Answer: Given I = 5 A, V = 220 V, t = 2 h Power, p = VI = 220 x 5 = 1100 W Energy consumed = Vlt = Pt = 1100 x 2 = 2200 Wh

Textbook Questions

Question 2: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2/ R Answer: (b) P = V 2/ R = I 2 R = VI Option (b) does not represent electrical power.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions Q3

Question 5: How is a voltmeter connected in the circuit to measure the potential difference between two points? Answer: A voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential of the source.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions Q6

Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? Answer: Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh. Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh. ∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions Q17

Question 18: Explain the following. (a) Why is the tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminum wires usually employed for electricity transmission? Answer: (a) It has high melting point and emits light at a high temperature. (b) It has more resistivity and less temperature coefficient of resistance. (c) (i) All appliances do not get same potential in series arrangement. (ii) All appliances cannot be individually operated. (d) R ∝ =1 / Area of cross – section. (e) They are very good conductors of electricity.

Short Answer Type Questions

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions SAQ Q1

Question 2: Should the resistance of an ammeter be low or high? Give reason. Answer: The resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.

Question 3: How does use of a fuse wire protect electrical appliances? Answer: The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions SAQ Q4

Question 6: Why is parallel arrangement used in domestic wiring? Answer: Parallel arrangement is used in domestic wiring because (i) Each appliance gets the same voltage as that of the mains supply. (ii) If one component is switched off, others can work properly. (iii) Fault in any branch of the circuit can be easily identified.

Question 7: B 1, B 2 and B 3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions SAQ Q7

Long Answer Type Questions

NCERT Solutions for Class 10 Science Chapter 12 Electricity Text Book Questions LAQ Q1

Multiple Choice Questions (MCQs) [1 Mark each]

NCERT Solutions for Class 10 Science Chapter 12 Electricity MCQs Q1

Question 4. An ammeter has 20 divisions between 0 mark and 2A mark on its scale. The least count of ammeter is (a) 0.01A (b) 0.2A (c) 0.1A (d) 1A Answer: (c) Number of divisions = 20 Maximum reading of ammeter = 2 A Least count of ammeter = 2/20 = 1/10 = 0.1 A

Question 5. A student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter. The least count of voltmeter is (a) 0.1 V (b) 0.01 V (c) 0.05 V (d) 1.0 V Answer: (c) Number of divisions = 20 Maximum reading of the voltmeter = 1 V Least count of voltmeter = 1/20 = 0.05 V

NCERT Solutions for Class 10 Science Chapter 12 Electricity MCQs Q6

Question 7. Which of the following is the correct method to connect the ammeter and voltmeter with resistance in the circuit to verify Ohm’s law? [CCE 2012] (a) Ammeter and voltmeter in series (b) Ammeter in series and voltmeter in parallel (c) Ammeter in parallel and voltmeter in series (d) Ammeter and voltmeter in parallel Answer: (b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.

Question 8. In an experiment on studying the dependence of the current I flowing through a given resistor on the potential difference V applied across it, a student has to change the value of the current. For doing this, he should change the (a) number of cells used (b) resistor itself (c) ammeter used in the circuit (d) Voltmeter used in the circuit Answer: (a) If we change the number of cells in electric circuit, the potential difference will change and as a result current flowing in the circuit changes.

Question 9. A milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions. If the pointer of the milliammeter is indicating the seventh graduation after 300 mark, the current flowing in the circuit is (a) 335 mA (b) 330 mA (c) 331mA (d) 340 mA Answer: (a) Number of divisions = 20 Least count of milliammeter = (100-0) / 20 = 5 mA Milliammeter reading = 300 + 7 x 5 = 335 mA

Question 10. If a student while studying the dependence of current on the potential difference keeps the circuit closed for a long time to measure the current and potential difference, then (a) ammeter’s zero error will change (b) ammeter will give more reading (c) voltmeter will show constantly higher readings (d) resistor will get heated up and its value will change Answer: (d) If the circuit is closed for a long time, then current flows in it for a long time which results that the resistor is heated.

NCERT Solutions for Class 10 Science Chapter 12 Electricity MCQs Q11

Question 12. When parallel resistors are of three different values, the potential difference across its terminals is [CCE 2015] (a) greatest across smallest resistance (b) greatest across largest resistance (c) equal across each resistance (d) least across the smallest resistance Answer: (c) Potential difference across each resistor is same in parallel combination of resistors.

NCERT Solutions for Class 10 Science Chapter 12 Electricity (Hindi Medium)

Electricity Study of Electric Charges at Rest and in Motion

Charge Something associated with the matter due to which it produces and experiences electric and magnetic effects. Resides on the outer surface of the conductor. Q = ne S.I. unit coulomb (C)

Electric Current (I) The time rate of flow of charge (Q) through any cross-section I = $\frac{Q}{t}$ S.I. unit ampere (A)

Types of Current Direct Current Current whose magnitude and direction does not vary with time.

Alternating Current Current whose magnitude and direction periodically changes with time.

Electric Potential Work done per unit charge V = $\frac{W}{Q}$ S.I. unit volt

Ohm’s law: If the physical conditions remain same, current I ∝ V => V = IR R-electric resistance Substances which obey ohm’s law called ohmic and that do not obey called non-ohmic substances.

Dependence of Resistance

On length (l) and area of cross-section (A) R ∝ l ∝ $\frac{1}{A}$ R = $\rho \frac{l}{A}$ ρ = resistivity Resistivity depends on the material of the conductor only.

On Temperature R t = R 0 ( 1 + αt) α = temperature coefficient of resistance

Resistance (R): Obstruction offered to flow of electrons. SI unit ohm Resistance, R ∝ $\frac{\ell^{2}}{m}$ l = length and m = mass of conducting wire

After stretching, if length increases by n times then resistance will increase by n 2 times i.e., R 2 = n 2 R 1 . Similarly if radius be reduced to $\frac{1}{n}$ times then area of cross-section decreases $\frac{1}{n^{2}}$ times so the resistance becomes n 4 times i.e.. R 2 = n 4 R 1 After stretching, if length of a conductor increases by x%, then resistance will increase by 2x% (valid only if x< 10%).

Using n conductors of equal resistance, the number of possible combinations is 2 n-1 .
If the resistances of n conductors are totally different, then the number of possible combinations will be 2 n .
If n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by $\frac{R_{s}}{R_{p}}=\frac{n^{2}}{1}$
If a wire of resistance R is cut in n equal parts and then these parts are collected to form a bundle, then equivalent resistance of combination will be $\frac{\mathrm{R}}{\mathrm{n}^{2}}$
If equivalent resistance of R 1 and R 2 in series and parallel be R s and R p respectively, then R 1 = $\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}+\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]$ and R 2 = $=\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}-\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]$

Grouping of Resistances

Series Grouping of Resistances Equivalent resistance, resistance, R s = R 1 + R 2 + … + R n In this case same current flows through each resistance but potential difference in the ratio of resistance

Parallel Grouping of Resistances $\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$ In this case same potential across each resistance but current distributes in the reverse ratio of their resistances

NCERT Solutions for Class 10 Science Chapter 12 Electricity Mind Map 1

Heating Effect of Electric Current As current flows through a conductor, the free electrons lose energy which is converted into heat. Joule’s heating law H ∝ I 2 H ∝ R H ∝ t H = I 2 Rt = VIt

Practical Applications

Electric heater, electric iron and water heater, etc. work on the principle of heating effect of current
Electric bulb glows when electric current flows through the filament of the bulb

Electric Power Rate at which electric energy is dissipated or consumed in a circuit, P = VI , or P = I 2 R = $\frac{\mathrm{V}^{2}}{\mathrm{R}}$

Watt is a smaller unit of power, its other bigger units are kilowatt (KW), Megawatt (MW) and Horsepower (HP) 1 KW = 10 3 W 1 MW = 10 6 W 1 hp =746 W The commercial unit of electrical energy is 1 Kwh. 1 Kwh = 3.6 × 10 6 J

LED It is a device which glows even if a weak electric current is allowed to flow through it

Now that you are provided all the necessary information regarding Electricity class 10 NCERT solutions and we hope this detailed article on NCERT Solutions For Class 10 Science Chapter 12 Electricity is helpful. If you have any doubt regarding this article or NCERT Solutions For Class 10 Science Chapter 12 Electricity, leave your comments in the comment section below and we will get back to you as soon as possible.

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Chapter 1 Chemical Reactions and Equations
Chapter 2 Acids, Bases and Salts
Chapter 3 Metals and Non-metals
Chapter 4 Carbon and Its Compounds
Chapter 5 Periodic Classification of Elements
Chapter 6 Life Processes
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Chapter 8 How do Organisms Reproduce?
Chapter 9 Heredity and Evolution
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NCERT Exemplar Class 10 Science Solutions for Chapter 12 - Electricity

Ncert exemplar solutions class 10 science chapter 12 – free pdf download.

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity are the study materials necessary for you to understand the questions that can be asked from the Class 10 Science Electricity chapter. It is crucial for students to get acquainted with this chapter in order to score excellent marks in their CBSE Class 10 examination. This solution provides answers to the questions provided in NCERT Class 10 Exemplar book. This page has answers to 18 MCQs, 11 short answer questions and 7 long answer questions.

To help students grasp all the concepts clearly and in-depth, we are offering free NCERT Exemplar for Class 10 Science Chapter 12 here. These exemplars will enable students to learn the correct answers to all the questions given at the end of the chapter. These NCERT Exemplars are prepared by experts and can be used by students as an effective learning tool to improve their conceptual understanding.

Take a closer look at Class 10 Science Chapter 12 NCERT Exemplar below.

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NCERT Exemplar solutions class 10 science Chapter 12 part 01

Access Answers to NCERT Exemplar Class 10 Science Chapter 12 – Electricity

Multiple choice questions.

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

ncert solutions class 10 science chapter 12 electricity fig 1

(a) maximum in (i)

(b) maximum in (ii)

(d) the same in all the cases

The answer is (d) the same in all the cases

Explanation:

There are no changes in any of the circuits, hence current will be same in all the circuits.

2. In the following circuits (Figure 12.2), the heat produced in the resistor or combination of resistors connected to a 12 V battery will be

ncert solutions class 10 science chapter 12 electricity fig 2

(a) same in all the cases

(b) minimum in case (i)

(d) maximum in case(iii)

The answer is (c) maximum in case(ii)

Explanation

Here two transistors are in series. In figure (iii) total resistance will be less than individual resistances as they are connected parallel. Higher resistance produces more heat hence option c) is the right answer.

3. Electrical resistivity of a given metallic wire depends upon

(a) its length

(b) its thickness

(d) nature of the material

The answer is (d) nature of the material

4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

Answer is (a) 10 20

n = 16 /1.6 x 10 -19

n = 10 x 10 19

n = 10 20 electrons

The number of electrons flowing is 10 20 electrons

5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

ncert solutions class 10 science chapter 12 electricity fig 3

The answer is (b) (ii)

6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

The answer is (d) 1 Ω

Maximum resistance is obtained when resistors are connected in series.

7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

Answer is (b) 1/25 Ω

Minimum resistance is obtained when resistors are connected parallel

1/R = 5 + 5 + 5 +5 +5= 25 Ω

8. The proper representation of the series combination of cells (Figure 12.4) obtaining maximum potential is

ncert solutions class 10 science chapter 12 electricity fig 5

The answer is (a) (i)

Here positive terminal of the next cell is adjacent to the negative terminal of the previous cell.

9. Which of the following represents voltage?

(a) $\begin{array}{l}\frac{Work done}{Current\times Time}\end{array} $

(b) Work done × Charge

(d) Work done × Charge × Time

10. A cylindrical conductor of length l and uniform area of crosssection A has resistance R. Another conductor of length 2l and resistance R of the same material has an area of cross-section

Answer is (c) 2A

When Length doubles

11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

ncert solutions class 10 science chapter 12 electricity fig 6

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(d) R2 > R3 > R1

The answer is (c) R3 > R2 > R1

Current flow is inversely proportional to resistance. Highest resistance will show less flow of current hence answer is c).

12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

Answer is (c) 300 %

Heat generated by a resistor is directly proportional to the square of the current. Hence, when the current becomes double, dissipation of heat will multiply by 2 =4. This means there will be an increase of 300%.

13. The resistivity does not change if

(a) the material is changed

(b) the temperature is changed

(d) both material and temperature are changed

Answer is (c) the shape of the resistor is changed

14. In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) The brightness of all the bulbs will be the same

(b) The brightness of bulb A will be the maximum

(d) The brightness of bulb C will be less than that of B

Answer is (c) Brightness of bulb B will be more than that of A

Bulbs are connected in parallel so the resistance of combination would be less than the arithmetic sum of the resistance of all the bulbs. So. there will be no negative effect on the flow of current. As a result, bulbs would glow according to their wattage.

15. In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

Answer is (c) 20 J

Equivalent resistance of the circuit is R = 4+2 = 6Ω

current, I= V/R = 6/6= 1A

the heat dissipated by 4-ohm resistor is, H = I 2 Rt = 20J

16. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

The answer is (d) 5 A

Or 1000 w = 220v x I

I = $\begin{array}{l}\frac{1000w}{220v}\end{array} $ = 4.54 A

17. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have

(a) same current flowing through them when connected in parallel

(b) same current flowing through them when connected in series

(d) different p

The answer is (b) same current flowing through them when connected in series

In series combination current does not get divided into branches because resistor receives a common current.

18. Unit of electric power may also be expressed as

(a) volt-ampere

(b) kilowatt-hour

(d) joule second

The answer is (a) volt-ampere

Volt-ampere (VA) is the unit used for the apparent power in an electrical circuit. A watt-second (also watt-second, symbol W s or W. s) is a derived unit of energy equivalent to the joule. The joule-second is the unit used for Planck’s constant.

Short Answer Questions

19. A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

ncert solutions class 10 science chapter 12 electricity fig 7

20. Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

ncert solutions class 10 science chapter 12 electricity fig 9

Current P= I 2 R

18W = I 2 x 2Ω

I 2 = 18W/ 2Ω

Maximum value of current passing through A is 3A.

Current through B = Current through C = 1/2 x Current through A

Current through B = Current through C = 1/2 x 3

Current through B = Current through C = 1.5 A

21. Should the resistance of an ammeter be low or high? Give reason.

Resistance of ammeter should be zero because ammeter should not affect the flow of current.

22. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

ncert solutions class 10 science chapter 12 electricity fig 10

Total resistance for parallel combination of 40 resistors can be calculated as follows:

Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 20 resistance will be same as potential difference across the other two resistors which are connected in parallel.

23. How does use of a fuse wire protect electrical appliances?

Fuse wire has great resistance than the main wiring. When there is significant increase in the electric current. Fuse wire melts to break the circuit. This prevents damage of electrical appliance.

24. What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Property of the conductor which resists the flow of electric current is called resistivity. Resistance for a particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow.

When length is doubled resistance becomes double and current flow reduces to half. This is the reason for the decrease in ammeter reading.

25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Commercial unit of electrical energy is kilowatt/hr

1 kw/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 10 6 J

26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

1) Let R be the resistance of the electric lamp. In series total resistance = 5 + R

1 = 10/5+R

2) V across Lamp + conductor = 10 V

V acoess Lamp = I × R = 1 * 5 = 5 Volt

27. Why is parallel arrangement used in domestic wiring?

Parallel arrangement used in domestic wiring because it provides the same potential difference across each electrical appliance.

28. B1 , B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

What happens to the glow of the other two bulbs when the bulb B1 gets fused?
What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused?
How much power is dissipated in the circuit when all the three bulbs glow together?

i) Potential difference does not get divided in parallel circuit. Hence glowing of other bulbs will not get affected when bulb one is fused.

ii) Ammeter A shows a reading of 3A. This means each of the Al. A2, and A3 show IA reading.

iii) R= V/I = 4.5V/3A= 1.5Ω

Now P= I 2 R

= (3A) 2 x 1.5 Ω

Long Answer Questions

29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

(Resistance of the bulbs in series will be three times the resistance of single bulb. Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly.

The bulbs in series combination will stop glowing as the circuit is broken and current is zero. However the bulbs in parallel combination shall continue to glow with the same brightness.

30. State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Ohm’s law states that at constant temperature potential difference (voltage) across an ideal conductor is proportional to the current through it.

ncert solutions class 10 science chapter 12 electricity fig 11

Verification of Ohm’s law

Set up a circuit as shown in Fig. consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)

First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given

Next, connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

Repeat the above steps using three cells and then four cells in the circuit separately.

ncert solutions class 10 science chapter 12 electricity fig 12

31. What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Resistivity is an inherent property of a conductor which resists the flow of electric current. Resistivity of each material is unique. SI unit of resistance is Ωm.

Experiment to study the factors on which the resistance of conducting wire depends.

Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.

Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4.

Observation:

It is observed that resistance depend on material of conductor

Length of conductor determines resistance

Resistance depends on area of cross-section.

ncert solutions class 10 science chapter 12 electricity fig 13

Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.

Now repeat the above step with the 10 W bulb in the gap XY. Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate?

You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Collect three resistors R1, R2, R3 in series to make the circuit.
Use ammeter to see the changes observed in the current flow.
Remove R1 and take the reading of potential difference of R2 and R3
Remove R2 and take the reading of potential difference of R1 and R3

ncert solutions class 10 science chapter 12 electricity fig 14

Ammeter reading was the same in each case so it can be inferred that the current remains the same in the circuit. To cross-check one can place ammeter and different places and observe the current flow.

33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Take three resistors R 1 , R 2 and R 3 , connect them in parallel to make a circuit; as shown in the figure.

Use voltmeter to take reading of potential difference of three resistors in parallel combination.

Now, remove the resistor R1 and take the reacting of the potential difference of remaining resistors combination.

Then, remove the resistor R 2 , and take the reading of potential difference of remaining resistor.

ncert solutions class 10 science chapter 12 electricity fig 15

In each case Voltmeter reading was the same which shows that the same potential difference exists across three resistors connected in a parallel arrangement.

34. What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

According to Joules heating effect heat produces in a resistor is

Directly proportional to square of current for the given resistor.
Directly proportional to resistance for a given current,
Directly proportional to the time of current flowing through the resistor.

This can be expressed as

H is heating effect, I is electric current, R is resistance and t is time.

Experiment to demonstrate Joules law of heating

Take a water heating immersion rod and connect to a socket which is connected to the regulator. It Is important to recall that a regulator controls the amount of current flowing through a device.
Keep the pointer of regulator on the minimum and count the time taken by immersion rod to heat a certain amount of water.
Increase the pointer of the regulator to the next level. Count the time taken by immersion rod to heat the same amount of water.
Repeat above step for higher levels on regulator to count the time.

It is seen that with an increased amount of electric current, less time is required o heat the same amount of water. This shows Joule’s Law of Heating.

Application:

Electric toaster, oven, electric kettle and electric heater etc. work on the basis of leafing effect of current.

35. Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination

(b) Current flowing through 4 Ω resistor

(d) Power dissipated in 4 Ω resistor (e) Difference in ammeter readings, if any

ncert solutions class 10 science chapter 12 electricity fig 16

NCERT Exemplar Class 10 Chapter 12 Electricity

Sometimes we might have wondered about what constitutes electricity or how does it flow in an electric circuit, or what controls and regulates the current through an electric circuit? In Chapter 12 Electricity, students will find answers to these questions. They will also learn about other topics like the heating effect of electric current and its applications, the circuit diagram, Ohm’s law , resistors and conductors, electrical potential and potential difference.

Topics covered in Class 10 NCERT Exemplar Solutions for Science Chapter 12 Electricity

Introduction
The potential difference – Definition of volt and voltmeter
Ohm’s law – Ohm and resistance
Factors on which the resistance of the conductor depends – Resistivity
Resistors in series – Total/resultant/overall and voltage across each resistor
Resistors in parallel
The advantage of parallel combination over the series combination
Heating effect of an electric circuit – Joule’s law of the heating effect of electric current, electric fuse and electric power.

With BYJU’S, students can excel in their studies and can score better marks in the board examination. Class 10 is an important stage of a student’s life, as it consists of topics which are necessary to understand thoroughly for future entrance exams. To help you grasp the concepts clearly, BYJU’S brings you notes , sample papers , and animation videos. For a customised learning experience, visit BYJU’S website or download BYJU’S – The Learning App.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Science Chapter 12

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Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

Class 10 Important Question
Chapter 12: Electricity

CBSE Class 10 Science Chapter-12 Electricity Important Questions with Answers - Free PDF Download

Class 10 is an essential stage in every student's career. Their performance in Class 10 acts as a base for their future studies. So it's important to score well in Class 10. The most important and challenging subject in Class 10 is Science and many students seem to get confused here and lose marks. If students want a successful career in future, then they can't afford to lose marks in this stage of their life. It's important to understand every chapter in science thoroughly to score good marks. Chapter 12 of Class 10 Science which is about electricity is one of the difficult chapters. A student who is incapable of understanding this chapter must practice Important Questions for Class 10 Science Chapter 12 . These important questions of Electricity Class 10 can make the students through on the concepts of this chapter. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Download CBSE Class 10 Science Important Questions 2024-25 PDF

Also, check CBSE Class 10 Science Important Questions for other chapters:

CBSE Class 10 Science Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11
12	Chapter 12	Electricity
13	Chapter 13
14	Chapter 14
15	Chapter 15
16	Chapter 16

Related Chapters

Study Important Questions for Class 10 Science Chapter 12 – Electricity

Very Short Answer Questions (1 Mark)

1. Which two circuit components are connected in parallel in the following circuit diagram?

(a) \[{{\text{R}}_{\text{1}}}\] and ${{\text{R}}_{\text{2}}}$ only

(b) ${{\text{R}}_{\text{1}}}{\text{,}}{{\text{R}}_{\text{2}}}$ only

(d) ${{\text{R}}_{\text{1}}}$ and ${\text{V}}$only

Ans: (a) The two circuit components that are connected parallel in the circuit diagram is and ${R_2}$ only.

2. A metallic conductor has loosely bound electrons called free electrons. The metallic conductor is

(a) negatively charged

(b) positively charged

(d) Either positively charged or negatively charged

Ans: (c) The metallic conductor is neutral.

3. Which of the following expressions does not represent the electric power in the circuit?

(a) ${\text{VI}}$.

(b) ${{\text{I}}^{\text{2}}}{\text{/R}}$

(d) ${{\text{I}}^{\text{2}}}{\text{R}}$

Ans: (b) The expression which does not represent the electric power in the circuit is ${I^2}/R$ .

4. Resistivity of a metallic wise depends on

(a) its length

(b) its shape

(d) nature of material

Ans: (d) Resistivity of a metallic wire depends on the nature of the material.

5. If the current I through a resistor is increased by \[{\mathbf{100}}\% \] the increased in power dissipation will be (assume temperature remain unchanged)

(a) \[{\mathbf{100}}\% \]

(b) \[{\mathbf{200}}\% \]

(d) \[{\mathbf{400}}\% \]

Ans: (c) The increase in power dissipation will be \[300\% \] .

6. For the circuit arrangement shown below, a student would observe.

(a) Some reading in both ammeter and voltmeter.

(b) No reading in either the ammeter or the voltmeter.

(d) Some reading in the voltmeter but no reading in the ammeter.

Ans: (c) A student will observe some reading in the ammeter but no reading in the voltmeter.

7. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected in parallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{{\text{1}}}{{\text{5}}}$

(b) ${\text{5}}$

(d) ${\text{25}}$

Ans: (d)The ratio $\dfrac{R}{{R'}}$ is $25$ .

8. The resistance of the conductor is $R$ . If its length is doubled, then its new resistance will be

Ans: (c) The new resistance is $4R$ .

9. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances ${R_1},{R_2},{R_3}$ respectively as shown in the figure. Which of the following is live?

(a) ${R_3} > {R_2} > {R_1}$

(b) ${R_2} > {R_3} > {R_1}$

(d) ${R_1} = {R_2} = {R_3}$

Ans: (a)According to the graph , ${R_3} > {R_2} > {R_1}$

10. The nature of the graph between potential difference and the electric current flowing through a conductor is

(a)parabolic

(b) circle

(d) hyperbolic

Ans: (c)The nature of the graph between potential difference and the electric current flowing through a conductor is a straight line.

11. An electric heater is salted at $1500$ w. How much heat is produced per hour?

(i) $5400$ J

(ii) $54000$ J

(iii) $5.4 \times {10^5}$

(iv) $5.4 \times {10^6}$.

Ans: (iv) The electric heater produces $5.4 \times {10^6}$ J per hour.

12. A student says that the resistance of two wires of the same length and same area of cross section is the same. This statement is correct if

(a) Both wires are of different materials

(b) Both wires are made of the same material and are at different temperatures.

(d) Both wires are made of different materials and are at the same temperature.

Ans: This statement is correct if (c)The resistance of two wires of the same length and same area of cross section is the same if both wires are made of the same material and are at the same temperature.

13. In an experiment ohm, s law a student obtained a graph as shown in the diagram. The value of resistance of the resistor is

(a) $0.1\Omega $

(b) $1.0\Omega $

(d) $100\Omega $

Ans: (d) The value of resistance of the resistor is $100\Omega $ .

14. Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is

(a) $1$ Joule

(b) $10$ Joule

(d) $100$ Joule

Ans: (c) Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is zero.

15. Three resistors are shown in the figure. The resistance of the combination is

(a)$3\Omega $

(b) $6\Omega $

(d) $7\Omega $

Ans: (c) The resistance of the combination is $9\Omega $ .

16. Name a device that helps to maintain a potential difference between across a conductor.

Ans: A device that helps to maintain a potential difference between conductors is the battery.

17. What determines the rate at which energy is delivered by a current?

Ans: The rate at which energy is delivered by a current is determined by electric power.

18. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{1}{5}$

Ans: (d)A wire of resistance ${\text{R}}$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are ${\text{R'}}$. In this cases, the ratio$\dfrac{R}{{R'}}$ is $25$ .

19. Which of the following terms does not represent electrical power in a circuit?

(a) ${I^2}R$

(b) $I{R^2}$

(d) ${V^2}/R$

Ans: (b) The term that does not represent electrical power in a circuit is $I{R^2}$ .

20. An electric bulb is rated $220$ V and $100$ W. When it is operated on $110$ V, the power consumed will be:

(a) $220$ W

(b) $75$ W

(d) $25$ W

Ans: (d)The power consumed will be $25$ W.

21. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be:

Ans: (c) The ratio of heat produced in series and parallel combination will be $1:4$ .

22. A wire of resistance $R$ is bent in the form of a closed circle, what is the resistance across a diameter of the circle?

Ans: The resistance across a diameter of the circle

23. A charge of $6$ C is moved between two points $P$ and $Q$ having, potential $10$ V and $5$ V respectively. Find the amount of work done.

Ans: The amount of work done, $W = q({V_2} - {V_1})\;$

$\; = {\text{ }}6\left( {10 - 5} \right)$

\[ = {\text{ }}30\] joule

24. Name the physical quantity whose SI unit is JC

Ans: The physical quantity whose SI unit is JC is Potential.

25. Why are copper wires used as connecting wires?

Ans: Copper wires are used as connecting wires because in case of copper the electrical resistivity for it is low.

26. A wire of resistivity $p$ is stretched to double its length. What is its new resistivity?

Ans: When a wire of resistivity ${\text{p}}$ is stretched to double its length, then the new resistivity remains the same because resistivity depends on the nature of material.

27. What is the resistance of the connecting wire?

Ans: The resistance of a connecting wire made of a good conductor is extremely low.

28. What is the resistance of an ammeter?

Ans: An ammeter's resistance is very minimal, and in an ideal ammeter, it is zero.

29. What is the resistance of a Voltmeter?

Ans: An ideal voltmeter's internal resistance is infinite.

Short Answer Questions (2 Marks)

1. How does use of fuse wire protect electrical appliances?

Ans: When a large quantity of current passes through the circuit, the temperature of the wire rises and the fuse wire melts. This prevents current from flowing into the house's other circuits, saving electrical appliances.

2. Calculate the resistance of an electric bulb which allows a $10$A current when connected to a $220$ V power source?

Ans: It is given from the question that an electric bulb allows a $10$A current when connected to a $220$ V power source.

\[I = 10\] A ,

\[V = 220\]V

\[R = \dfrac{V}{I}\]

\[ = \dfrac{{220}}{{10}}\]

\[ = 22\] ohm

3. (i) Identify the $V - I$ graphs for ohmic and non-ohmic materials.

Ans: The $V - I$ graphs for ohmic and non-ohmic materials respectively can be represented as shown below:

(ii) Give one example of each.

Ans: (ii) Some examples of ohmic material are Copper, Nichrome and some examples of Non-ohmic material are Diode, Transistor.

4. What do the following symbols represent in a circuit? Write the name and one function of each?

Ans: (i) It symbolises a battery that maintains a potential difference across the circuit element to allow current to flow.

Ans: It's an ammeter that measures how much current is flowing across a circuit.

5. Define the term “volt”?

Ans: If $1$joule of energy is transferred between two points A and $B$, the potential difference between them is one volt. In an electric circuit, work is done to move one coulomb of charge from one point to another field.

6. Why does the connecting rod of an electric heater not glow while the heating element does?

Ans: As its resistance is lower than that of the heating element, the connecting cord of an electric heater does not glow. As a result, the heating element produces more heat than the connecting cord, and it glows

7. A number of \[n\] resistors each of resistance \[R\] are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit is series combination and parallel combination?

Ans: Total effective resistance of the circuit when in series combination \[{R_s} = nR\]

And for parallel combination is \[{R_p} = \dfrac{R}{n}\] and

$\dfrac{{{R_s}}}{{{R_p}}} = \dfrac{{nR}}{{\dfrac{R}{n}}}$

The ratio will be ${n^2}$ .

8. Draw a schematic diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \] and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series.

Ans: The circuit diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \] and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series can be represented as show below,

9. A copper wire has diameter \[0.5\] mm and Resistivity of \[1.6 \times {10^{ - 8}}\Omega m\] What is the length of this wire to make its resistance ? How much does the resistance change if diameter is doubled?

Ans: Diameter of the copper wire, $D = 0.5 \times {10^{ - 3}}$ m

$P = 1.6 \times {10^{ - 8}}$

$R = \dfrac{{\rho l}}{A}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}}$

$ = \dfrac{{4\rho l}}{{\pi {D^2}}}$

$ \Rightarrow l = \dfrac{{\pi R{D^2}}}{{4\rho }}$

$l = \dfrac{{3.14 \times 10 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{4 \times 1.62 \times {{10}^{ - 8}}}}$

$ = 121.14$

Length of the wire, $l = 121.14$ m

New $R' = \dfrac{{4\rho l}}{{\pi {{(D')}^2}}}$

$ = \dfrac{1}{4}\dfrac{{4\rho l}}{{\pi {D^2}}}$

$ = \dfrac{1}{4}R$

Length of the wire to make its resistance $10\Omega $ is $121.14$m and when the diameter is doubled the new resistance will be one fourth that of the old one.

10. Alloys are used in electrical heating devices rather than pure metals. Give a reason.

Ans: Alloys are utilised in electricity heating devices rather than pure metals because alloys have a higher resistivity and hence produce more heat. Furthermore, alloy is non-combustible (or oxidises easily at higher temperature).

11. On what factor does the resistance of a conductor depend?

Ans: The factors on which Resistance depends are:-

(a) Length of the conductor

(b) Area of cross - section

(d) Nature of material

12. Calculate the number of electrons consisting of one coulomb of charge?

Ans: Let $x = $ no. of electrons

Charge on $1$ electron $ = 1.6 \times {10^{ - 19}}$ C , that is

$x = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$x = 6.25 \times {10^{18}}$

The number of electrons consisting of one coulomb of charge is $6.25 \times {10^{18}}$ .

13. What does an electric circuit mean?

Ans: An electric circuit is a current route that is both continuous and closed. Current can flow through an electric circuit if it is complete.

14. Define the unit of current.

Ans: The ampere is the SI unit for electric current. If $1$ coulomb charge flows per second across a conductor cross-section, the current is said to be $1$ ampere.

15. Calculate the number of electrons constituting one coulomb of charge.

Ans: The charge on one electron $ = 1.6 \times {10^{ - 19}}$ coulomb.

Number of electrons in one coulomb of charge$ = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$ = 6.25 \times {10^{18}}$

16. What is meant by saying that the potential difference between two points is \[1\] v?

Ans: If $1$ joule of labour is required to move a charge of $1$ coulomb from one location to another, the potential difference between the two points is said to be $1$ volt.

17. Ammeter burns out when connected in parallel. Give reasons.

Ans: When a low-resistance wire is connected in series, a huge quantity of current travels through it, causing it to be burned, or short-circuited.

18. Judge the equivalent resistance when the following are connected in parallel:

(a) Equivalent resistance of \[1\Omega \] and \[{10^6}\Omega \]

Ans: When the resistances are connected in a parallel arrangement, the resultant resistance is given by:

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \cdots $

$\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 6}}$

$R = 1\Omega $

The equivalent resistance is $1\Omega $ .

(b) Equivalent resistance of \[1\Omega \] , \[{10^3}\Omega \] and \[{10^6}\Omega \]

Ans: $\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 3}} + {10^{ - 6}}$

19. An electric iron of resistance \[20\] takes a current of \[5\] A. Calculate the heat developed in \[30\] s.

Ans: Resistance of electric iron, $R = 20\Omega $ , current, $I = 5$ A and time $ = 30$ s.

Heat generated $H = {I^2}Rt$

$ = {5^2} \times 20 \times 30$

$ = 15000$ j

The heat developed in $30$ second is $15000$ j .

20. Compute the heat generated while transferring \[96000\] coulomb of charge in one hour through a potential difference of \[50\] V

Ans: The given information is as shown below,

Charge transferred, $Q = 96000$

Potential Difference, $V = 50$ V.

Heat generated, $H = VQ$

$ = 50 \times 96000$

$ = 4800000$ j

$ = 4.8 \times {10^6}$ j

The heat generated while transferring $96000$ coulomb of charge in one hour

through a potential difference of $50$ V is $4.8 \times {10^6}$ j

21. An electric motor takes \[5\] A from a \[220\] V line. Determine the power of the motor and energy consumed in \[2\] h.

Ans: Given that current drawn by electric motor $I = 5$ A.

The line voltage $V = 220$ V

Time, $t = 2$ h

Power of motor , $P = VI$

$ = 220 \times 5$

$ = 1100$ W and

the energy consumed $E = Pt$

$ = 2 \times 1100$

$ = 2.2$ KWh

The power of the motor and energy consumed in $2$ h are $1100$ W and $2.2$ kWh respectively.

22. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans: A voltmeter is connected in parallel to the resistance across the place where the potential difference is to be determined.

23. When a \[12\] v battery is connected across an unknown resistor, there is a current of \[2.5\] mA in the circuit. Find the value of the resistance of the resistor.

Ans: Given that Voltage of battery, $V = 12$ V,

Current, $I = 2.5$ mA

$ = 2.5 \times {10^{ - 3}}$ A

Resistance, $R = V/I$

$ = \dfrac{{12}}{{2.5 \times {{10}^{ - 3}}}}$

$ = 4800\Omega $

The value of the resistance of the resistor is $4800\Omega $ .

24. Several electric bulbs designed to be used on a \[220\] V electric supply line, are rated \[10\] W. How many lamps can be connected in parallel with each other across the two wires of \[220\] V line if the maximum allowable current is \[5\] A?

Ans: The given information is as shown below

Each bulb is rated as $10$ W, $220$ V,

It draws current, $I = P/V$

$ = \dfrac{{10}}{{220}}$ V

$ = 1/22$ A.

The maximum allowable current is $5$ A and all lamps are connected in parallel. Therefore the maximum number of bulbs joined in parallel with each other $ = 5 \times 22$whichis$110$.

25. Two lamps, one rated \[100\] W at \[220\] V, and the other \[60\] W at \[220\] V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is \[220\] V?

Ans: Current drawn by ${1^{st}}$lamp rated $100$ W at $220$ , $V = P/V$

\[ = 100/220\]

\[ = 5/11\] A.

Current drawn by ${2^{nd}}$lamp rated $60$ W at \[220\] , \[V = 60/220\]\[ = 3/11\;\]A.

In parallel arrangement the total current \[ = {\text{ }}3/11 + {\text{ }}5/11\]

\[ = {\text{ }}8/11\]

\[ = {\text{ }}0.73\] A.

Current drawn from the line if the supply voltage is $220$ V is $0.73$ A .

26. Which uses more energy, a \[250\] W TV set in \[1\] hour, or a \[1200\] W toaster in \[10\] minutes?

Ans: Energy used by a TV set of power $250$ W in $1$ hour $ = Pt$

Energy used by toaster of power \[1200\] W in \[10\] minute \[\left( {10/60h} \right) = {\text{ }}200\] Wh.

A $250$ W TV set in $1$ hour uses more energy.

27. An electric heater of resistance \[8\] draws \[15\] A from the service mains for \[2\] hours. Calculate the rate at which heat is developed in the heater.

Ans: Resistance of electric heater, \[R{\text{ }} = {\text{ }}8\Omega \],

current, \[I{\text{ }} = {\text{ }}15\] A.

Rate at which heat developed in the heater $ = \dfrac{{{I^2}Rt}}{t}$

$ = 1800$ W.

The rate at which heat is developed in the heater is $1800$ W.

28. In the given figure what is the ratio of current in \[A\]

Ans: Observe that it I clearly known to us that $V = IR$

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{R}{{2R}}$

$ = \dfrac{1}{2}$

The ratio of current is$1/2$.

29. Two wires of equal cross sectional area, one of copper and other of managing have the same resistance. Which one will be longer?

Ans: Using the equation $\rho = \dfrac{{RA}}{l}$ , where $\rho $ is the resistivity, $R$ is the resistance and $A$ the area

We can see that copper wire has a lower resistance than manganin, hence copper will last longer.

30. A Rectangular block of iron has dimensions $L \times L \times b$ . What is the resistance of the block measured between the two square ends? Given $p$ resistivity.

Ans: $R = \dfrac{{pb}}{{{L^2}}}$is the resistance of the block measured between the two square ends

31. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances?

Ans: ${R_{series}} = 3R$

${R_{parallel}} = R/3$

The ratio of Resistances is $9$ .

32. Justify for any pair of resistance the equivalent resistance equivalent resistance in parallel.

Ans: As $R = V/I$

From the graph for any pair of resistance the equivalent resistance in series is greater than equivalent resistance in parallel.

$A = $ Series, $B = $ Parallel

33. How many bulbs of \[81\] should be joined in parallel to draw a current of \[2\] A from a battery of \[4\] V?

Ans: $R = V/I$

$ = 2\Omega $

let $n$ be the number of bulbs.

$1/R = 1/{R_1} + 1/{R_2} + \cdots + 1/{R_n} = \dfrac{n}{8}$

$ \Rightarrow \dfrac{1}{2} = \dfrac{n}{8}$

$ \Rightarrow n = 4$

Number of bulbs are $4$ .

34. Two cubes \[A\] and \[B\] are of the same material. The side of \[B\] is thrice as that of \[A\] . Find the ratio \[{R_A}/{R_B}\] .

Ans: The value of ${R_A} = \dfrac{{pL}}{A}$ and

${R_B} = \dfrac{{p3L}}{{9A}}$

${R_A}:{R_B} = 3:1$

35. If there are $3 \times {10^{11}}$ electrons are flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron $1.6 \times {10^{19}}$

Ans: Observe as shwn below,

Using the equation

$ = 3 \times {10^{11}} \times 1.6 \times {10^{19}}$C

$ = 4.8 \times {10^8}$C

$ = \dfrac{{4.8 \times {{10}^8}}}{{2 \times 60}}$

$ = 4 \times {10^7}$ A

The current flowing through the circuit is $4 \times {10^7}$ A.

36. A nichrome wire of resistivity $100$ W m and copper wire of resistivity $1.62$ M ohm-m of the same length and same area of cross section are connected in series , current is passed through them, why does the nichrome wire get heated first?

Ans: Looking at the equation

$Q = {I^2}RT$

$Q = {I^2}(pL/A)t$

Henceforth, because nichrome wire has a higher resistance than copper wire, it must be heated first.

37. What is represented by joule/coulomb?

Ans: The potential difference is represented by the joule/coulomb.

38. A charge of $2$ C moves between two plates, maintained at a p.d of $IV$ . What is the energy acquired by the charge?

Ans: The energy acquired by the charge, $W = QV$

The energy acquired is $2$ J

39. Which has more resistance: $100$ W bulb or $60$ W bulb?

Ans: As, it is clearly known that$R \propto \dfrac{1}{P}$, thus the resistance of $60$ W bulb is more.

40. What happens to the current in a circuit if its resistance is doubled?

Ans: As current and resistance are inversely proportional, the current is reduced to half of its previous value.

41. What happens to the resistance of a circuit if the current through it is doubled?

Ans: Resistance is unaffected since the circuit's resistance is independent of the current flowing through it.

42. How does the resistance of a wire depend upon its radius?

Ans: As $R \propto \dfrac{1}{A}$

$ \Rightarrow R \propto V$

Resistance of a wire is directly proportional to its radius.

43. Two wires are of the same length, same radius, but one of them is of copper and the other is of iron. Which will have more resistance.

Ans: Since \[R{\text{ }} = {\text{ }}p1/A\] ,

but A and I are the same. It is solely determined by resistivity, hence iron has a higher resistance.

44. Two wires of same material and same length have radii $R$ and $r$ Compare their resistances.

Ans: Suppose $R$ and $r$ are resistances, then $R = r$ as $p$ and $I$ are the same.

Short Answer Questions (3 Marks)

1. Two metallic wires A and B are connected, wire A has length I and radius $r$ , while wire B has length $2l$ and radius $2r$ . Find the ratio of total resistance of series combination and the resistance of wire $A$ , if both the wires are of the same material?

Ans: Observe as shown below,

Resistance of metallic wire $A$ , ${R_1} = \dfrac{{\rho l}}{A}$

Resistance of metallic wire $B$ , ${R_2} = \dfrac{{\rho 2l}}{{4\pi {r^2}}}$

Total resistance in series is $R = {R_1} + {R_2}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}} + \dfrac{{2\rho l}}{{4\pi {r^2}}}$

$ = \dfrac{{3\rho l}}{{2\pi {r^2}}}$

The ratio of the total resistance is series to the resistance of $A$ is

$\dfrac{R}{{{R_1}}} = \dfrac{{\dfrac{{\rho l}}{{\pi {r^2}}}}}{{\dfrac{{3\rho l}}{{2\pi {r^2}}}}}$

The ratio of the total resistance is series to the resistance of $A$ is $2/3$ .

2. Should the heating element of an electric iron be made of iron, silver or nichrome wire? Justify giving three reasons?

Ans: The following reasons can be found, why the heating element of an electric iron is composed of nichrome wire.

(1) Due to the high resistance, the passage of current generates additional heat.

(2) High melting point.

(3) At high temperatures, it does not easily oxidised (or burn).

3. (a) Define electric resistance of a conductor?

Ans: A conductor's electric resistance is defined as the resistance it provides to the flow of current.

That is $R = V/I$ and its S.I. unit is ohm , $\Omega $ .

(b) A wire of length $L$ and resistance $R$ is stretched so that its length is doubled and the area of the cross section is halved. How will its

(i) resistance change

Ans: It is clearly known that resistance, $R = \dfrac{{\rho l}}{A}$

New length $L' = 2L$ and

$A' = \dfrac{A}{2}$

Therefore ${R^1} = \dfrac{{\rho L'}}{{A'}}$

Therefore, the resistance of a wire becomes 4 times its original resistance.

(ii) resistivity change?

Ans: The size of a wire has no bearing on its resistance. As a result, resistance does not vary.

4. Two resistors of resistance $R$ and $2R$ are connected in parallel in an electric circuit. Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: Power consumed by $R$ , ${\rho _1} = \dfrac{{{V^2}}}{R}$

Power consumed by$2R$ , ${\rho _2} = \dfrac{{{V^2}}}{{2R}}$

Ratio $\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{{\dfrac{{{V^2}}}{R}}}{{\dfrac{{{V^2}}}{{2R}}}}$

The ratio of the electric power consumed by $R$ and $2R$ is $2:1$ .

5. The length of different metallic wires but of same area of cross section and made of the same material are given below

Wire	Length
$A$	$1\,m$
$B$	$1.5\,m$
$C$	$2.0\,m$

(i) Out of these two wires which wire has higher resistance.

Ans: As $R \propto l$ (length of the conductor) and since length of wire $C$ is more than $A$ and $B$ , wire $C$ has higher resistance.

(ii) Which wire has higher electrical resistance? Justify your answer.

Ans: The electrical resistivity of a wire is determined by the nature of the material, not by its dimensions. As a result, the resistivity of all wires is the same as the substance of all wires.

6. Two resistors of resistances $R$ and $2R$ are connected in series with an electrical circuit? Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: It is clearly known thatElectric power consumed by $R$ , ${P_1} = {I^2}R$

Also, electric power consumed by $2R$ , ${P_2} = {I^2}2R$

$\dfrac{{{P_1}}}{{{P_2}}} = 1/2$

The ratio of the electric power consumed by $R$ and $2R$ is$1:2$.

7. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. the ratio of heat produced in series and parallel combinations would be

Ans: (c)Let resistance of each wire is $R$

In series, resistance is $ = 2R$

Heat produced, ${H_1} = \dfrac{{{V^2}}}{{2R}}t$

In parallel total resistance $ = R/2$

Heat produced, ${H_2} = \left( {\dfrac{{{V^2}}}{{\dfrac{R}{2}}}} \right)t$

$ = \dfrac{{2{V^2}}}{R}t$

${H_2} = 4{H_1}$

$\dfrac{{{H_1}}}{{{H_2}}} = 1/4$

The ratio of heat produced in series and parallel combinations is $1:4$ .

8. Calculate the following of a circuit shown in the figure.

(i) effective resistance

Ans: Effective resistance, $R = {R_1} + {R_2}$

$ = 5 + 10$

The effective resistance is $15\Omega $ .

(ii) current

Ans: Current, $I = V/R$

The current is $0.133$ A.

(iii) Potential difference across $10\Omega $ resistor

Ans: Potential difference across $10\Omega $

$ = \dfrac{2}{{15}} \times 10$

The potential difference across $10\Omega $ is $1.33$ volt.

9. A Piece of wire of resistance $20\Omega $ is drawn out so that its length is increased to twice its original length to calculate the resistance of the wire is the new situation?

Ans: Resistance of wire$ = 20\Omega $

As $R = \dfrac{{\rho l}}{A}$

And as length of a wire is increased, its area of cross- section decreases, and volume of the wire remains constant.

$R' = \dfrac{{\rho l'}}{{A'}}$

$ = \dfrac{{4\rho l}}{{A'}}$

$\dfrac{{R'}}{R} = \dfrac{4}{1}$

$ \Rightarrow R' = 80$

The resistance of the wire is the new situation is $80\Omega $ .

10. A battery made of $5$ cells each of $2$ V and have internal resistance $0.1\Omega ,0.2\Omega ,$$0.3\Omega ,0.4\Omega $ and \[0.5\Omega \] is connected across \[10\Omega \] resistance. Draw a circuit diagram and calculate the current flowing through$10\Omega $resistance?

Ans: The internal resistance , \[0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5\Omega \]

Total resistance \[ = 1.5 + 10\]

\[ = 11.5\]

\[I = V/R\]

\[ = 10/11.5\]

\[I = {\text{ }}0.869\] A

Current flowing through $10\Omega $ resistance is $0.869$ A.

11. In the circuit diagram given here Calculate-

(a) The total effective resistance

Ans: As resistances are in parallel

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$

$ = \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{10}}$

$ = \dfrac{8}{{10}}$

$R = \dfrac{{10}}{8}$

Total effective resistance is $\dfrac{{10}}{8}\Omega $

(b) The total current

Ans: Total current, $I = V/R$

$ = \dfrac{6}{{10/8}}$

The total current is $4.8$ A.

Ans: From the circuit diagram, ${I_1},{I_2}$ and ${I_3}$ be the current through $2\Omega ,5\Omega $ and $10\Omega $ respectively

Therefore, ${I_1} = \dfrac{V}{{{R_1}}}$

${I_2} = \dfrac{V}{{{R_2}}}$

${I_3} = \dfrac{V}{{{R_3}}}$

Current through each$2\Omega ,5\Omega $ and $10\Omega $ is $3$ A ,$1.2$ A, and $0.6$ A respectively.

12. You have two circuits Compare the power used in \[2\Omega \] resistor in each case.

(i) a \[6V\] battery is series with \[1\Omega \] and \[2\Omega \] resistor

Ans: Potential difference, $V = 6V$

${R_1} = 1\Omega $

${R_2} = 2\Omega $

Total Resistor $ = {R_1} + {R_2}$

$I = \dfrac{V}{R}$

$ = \dfrac{6}{3}$

${P_1} = \dfrac{{{V^2}}}{R}$

The power used in $2\Omega $resistor is $8$ W.

(ii) a \[4V\] battery in parallel with \[12\Omega \] and \[2\Omega \] resistor

Ans: Potential difference, $V = 4V$

${R_1} = 12\Omega $

${P_2} = \dfrac{{{V^2}}}{R}$

The ratio of both power is $1:1$ .

13. How much energy is given to each coulomb of charge passing through a \[6\] volt battery?

Charge, $Q = 1C$

Energy = total work done

\[ = {\text{ }}Q{\text{ }}x{\text{ }}V\]

\[ = {\text{ }}1x6\]

\[ = {\text{ }}6\] joule.

Energy given to each coulomb of charge passing through a $6$ volt battery is $6$ joules.

14. On what factor does the resistance of a conductor depend?

Ans: A conductor's resistance is determined by the following factors:

(i) length of conductor

(ii) Area of cross-section

(iii) Temperature

(iv) Conductors are made from a variety of materials.

15. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Ans: When linked to the same source, current flows more freely through a thick wire than via a small wire of the same material. It's because resistance rises as thickness decreases.

16. Let the resistance of an electric component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Ans: T he electrical component's resistance R remains unchanged, but the potential difference across its ends falls to half of its original value. As a result of Ohm's law, new current is reduced to half of its initial value.

17. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans: The following reasons are why the coils of electric toasters and electric irons are built of an alloy rather than a pure metal:

(i) An alloy's resistivity is higher than that of pure metal.

(ii) An alloy does not rust quickly at high temperatures.

18. Draw a schematic diagram of a circuit consisting of a battery of three cells of \[2V\] , each, a \[5\Omega \] resistor, \[8\Omega \] resistors and a \[12\Omega \] and a plug key, all connected in series.

Ans: The diagram of circuit is as follows

19. An electric lamp of \[100\Omega \] , a toaster of resistance \[50\Omega \] and a water filter of resistance \[500\Omega \] are connected in parallel to a \[220V\] source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Ans: It can be found from the question, voltage, $V = 220V$

${R_1} = 100\Omega $

${R_2} = 50\Omega $

${R_3} = 500\Omega $

\[1/R = 1/100 + 1/50 + 1/500\]

\[R{\text{ }} = {\text{ }}500/16\]

\[ = {\text{ }}31.25\Omega \]

The resistance of electric iron, which draws as much current as all three appliances take together is $31.25\Omega $ .

Current passing through electric iron, \[I = V/R\]

\[ = 220/31.25\]

\[ = 7.04\] A.

That is the current passing is $7.04$ A.

20. What is (a) the highest total resistance that can be secured by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: When all four resistances must be connected in series, the highest resistance is achieved. In that instance, the outcome

$R = {R_1} + {R_2} + {R_3} + {R_4}$

\[ = {\text{ }}4 + 8 + 12 + 24\]

\[ = {\text{ }}48\Omega \]

The highest resistance is \[48\Omega \] .

(b) The lowest total resistance that can be secured by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: All four resistances must be connected in parallel to produce the lowest resistance.

$1/R = 1/{R_1} + 1/{R_2} + 1/{R_3} + 1/{R_4}$

\[ = {\text{ }}1/4{\text{ }} + 1/8{\text{ }} + 1/12{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}12/24\]

The lowest resistance is $2\Omega $ .

21. Why does the cord of an electric heater not glow while the heating element does?

Ans: When connected to the voltage source, the cord of a heater and the cord of an electric heater are connected in series and carry the same current.

Because the resistance of the cord is so low in comparison to the resistance of the heater element.

As a result, the amount of heat created in the cord is extremely low, but significantly higher in the heater element. As a result, the heating element glows, but the cord does not.

22. A copper wire has diameter \[0.5\] mm and resistivity of \[1.6 \times {10^{ - 8}}\] m. What will be the length of this wire to make its resistance \[10\] ? How much does the resistance change if the diameter is doubled?

Ans: The diameter of wire, d = 0.5 mm,

Resistivity, $\rho = 1.6 \times {10^{ - 8}}$

resistance $R = {\text{ }}10{\text{ }}\Omega $

\[R = \rho L/A\]

$L = \dfrac{{\pi {D^2}R}}{{4\rho }}$

\[ = \dfrac{{22 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{7 \times 4 \times 1.6 \times {{10}^{ - 8}}}}\]

\[ = {\text{ }}122.5\] m

As resistance is inversely proportional to the cross-section area of wire, when the diameter is doubled for a given length of material, the resistance reduces.

23. A battery of \[9V\] is connected in series with resistance of \[0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega \] and \[12\Omega \]respectively. How much current would flow through the \[12\] resistor?

Ans: Potential difference $V = 9V$ .

Total resistance \[ = {\text{ }}0.2{\text{ }} + 0.3{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}12{\text{ }}\]

\[ = {\text{ }}13.4{\text{ }}\Omega \]

Current in the circuit \[I{\text{ }} = {\text{ }}V/R{\text{ }}\]

\[ = {\text{ }}9{\text{ }}V{\text{ }}/{\text{ }}13.{\text{ }}4{\text{ }}\] A.

\[ = {\text{ }}0.67\]

In a series circuit the same current flows through all the resistance, hence current of \[0.67\] A will flow through \[12{\text{ }}\Omega \] resistor.

24. How many \[176\Omega \] resistors (in parallel) are required to carry \[5\] A on a \[220\] V line?

Ans: Let the resistors of \[176\Omega \] be joined in parallel.

Their combined resistance,

\[1/R = 1/176 + 1/176 \ldots \ldots \] times

\[ = n/176\;\] or

$ \Rightarrow R = 176/n\Omega $

Given that \[V = 220V\] and

\[I = 5\] A

\[R = V/I\]

\[ = 176/n\]

\[ = 220/5\]

\[ = 44\Omega \;\;\;\;\;\;\]

\[n{\text{ }} = {\text{ }}176/44{\text{ }}\]

\[ = {\text{ }}4\] ,

A number of $4$ resistors should be joined in parallel.

25. Show how you would connect three resistors, each of resistance \[6\Omega \] so that the combination has resistance of

(i) \[9\Omega \]

Ans: From the question, ${R_1} = {R_2}$

Join three resistors as below to get net resistance of $9\Omega $ :

(ii) \[4\Omega \]

Ans: Join three resistors as below to get \[4\Omega \] net resistance :

26. A hot plate of an electric oven connected to a \[220V\] line has two resistance coils \[A\] and \[B\] . Each of \[24\Omega \] resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?

Ans: From the question, potential difference\[V = {\text{ }}220{\text{ }}V\] .

Resistance of coil $A = $ Resistance of coil \[B\]

\[ = 24{\text{ }}\Omega \]

When coil is used separately, the circuit \[I = V/R\]

\[ = 220V/24\Omega \]

\[ = {\text{ }}9.2{\text{ }}A\]

The current is\[9.2{\text{ }}A\].

When coils are used in series total resistance $R = {R_1} + {R_2}$

\[ = 24 + 24\]

\[ = 48\Omega \]

The current flowing, \[I = V/R\]

\[ = 220V/48\Omega \]

\[ = 4.6\] A

The current is $4.6$ A.

Two coils are joined in parallel.

Total resistance \[R = {\text{ }}1/24{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}2/24\]

\[R = 12{\text{ }}\Omega \] .

Current \[I = V/R\]

\[ = 220V/12\Omega \]

\[ = 18.3\] A.

The current is $18.3$ A.

27. Compare the power used in the \[2\Omega \] resistor in each of the following circuits:

(i) a \[6\] volt battery in series with \[1\Omega \] and \[2\Omega \] resistors and,

Ans: Suppose a \[2\Omega \] resistor is joined to a $6$ V battery in series with \[1\Omega \] and \[2\Omega \] resistors.

Total resistance \[R = 2 + 1 + 2\]

\[ = 5\Omega \]

Current \[I = 6V/5\Omega \]

\[ = 1.2\] A

Power used in $2A$ resistor \[ = {I^2}R\]

\[ = 2.88\] W

Power used is $2.88$ W.

(ii) a \[4V\] battery in parallel with \[12\Omega \] and \[2\Omega \] resistors.

Ans: Suppose \[2\Omega \] resistor is joined to a $4V$ battery in parallel with \[12\Omega \] resistor and \[2\Omega \] resistors,

the current flowing in \[2\Omega = 4V/2\Omega \]

Power used in \[2\Omega \] resistor \[ = {I^2}R\]

\[ = 8\] W

Ratio \[ = 2.88/8\]

\[ = 0.36:1\;\]

The ratio of power used is \[0.36:1\] .

28. In the given figure what is ratio of ammeter reading when \[J\] is connected to \[A\] and then to \[B\]

Ans: Connect $J$ to $A$ , then

\[ = 0.6A\]

When $J$ is connected to $B$\[V = 1 + 2 + 3 + 4\]

Ratio of ammeter reading when J is connected to $A$ and then to $B$ is $3:10$ .

29. Given a resistor each of resistors \[R\] . How will you combine them to get the

(i) maximum effective resistance?

Ans: For maximum resistance $R = nr$ , this is the same as combining a series of numbers.

(ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

Ans: For minimum resistance $R' = r/n$ , this is the same as combining a series of numbers.The ratio of the maximum to minimum resistance is $R/R' = {n^2}$.

30. A wire of length \[L\] and resistance \[R\] is stretched so that its length is doubled. How will its

(a) Resistance change

Ans: The resistance of a wire is determined by its length, cross-sectional area, and resistivity as\[{\text{R = }}\dfrac{{{{\rho l}}}}{{\text{A}}}\]

Hence, if the length is doubled and area is halved, then we have

$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_2}}}{{{A_2}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}}$

$ = \dfrac{{{l_2}{A_1}}}{{{l_1}{A_2}}}$

Therefore, ${R_2} = 4{R_1}$

Hence, resistance of the wire becomes four times the original value.

(b) Resistively change?

Ans: The substance from which wire is formed has a property called wire resistivity. As a result, changing the wire's size has no effect on its resistivity.

31. Two students perform the experiments on series and parallel combinations of two given resistors \[{R_1}\] and \[{R_2}\] and plot the following \[V - I\] graphs.

Ans: Both students are correct because

\[AV/A1 = \;\] resistance $R$ and

\[A1/AV = l/R\;\]

The term "series" refers to high resistance, while "parallel" refers to low resistance.

32. A household uses the following electric appliances. Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\]

(i) Refrigerator of rating \[4\] for ten hours each day.

Ans: Month of June has \[30\] days.

Refrigerator of $400$ W is running $2$ hours each day.

Total hours it is run in $30$ days \[ = 2 \times 30\]

\[ = 60\] h

Energy consumed in kWh is \[ = 400{\text{ }} \times {\text{ }}60/1000{\text{ }}\]

\[ = {\text{ }}24\] kWh

(ii) Two electric fans of rating \[80\]each for twelve hours each day.

Ans: Two electric fans of \[80\] W are run \[12\] hours each day.

Total hours they are run in \[30\] days \[ = 12 \times 30\]

Energy consumed in kWh is \[ = 2 \times 80 \times 360/1000\]

$ = 57.6$ kWh

(iii) Six electric tubes of rating \[18\] W each for 6 hours each day.

Ans: Six electric tubes each of \[18\] W are run $6$ hours daily.

Total hours it is run in $30$ days \[ = 6 \times 30\]

\[ = 180\] h

Energy consumed in kWh is \[ = 6 \times 18 \times 180/1000\]

\[ = 19.44\] kWh

Net energy consumed in the month of June is \[ = 24 + 57.6 + 19.44\]

\[ = 101.04\] kWh

Thus, the electric bill is \[ = 3 \times 101.04\]

\[ = Rs303.12\]

Long Answer Questions (5 Marks)

1. Two wires \[A\] and \[B\] are of equal length, different cross sectional areas and made of the same metal.

(a) (i) Name the property which is same for both the wires,

Ans: Resistivity - As resistivity is a property of a substance, it is constant for both wires.

(ii) Name the property which is different for both the wires.

Ans: Resistances - As the cross sectional areas of each wires are different, they are treated as separate objects.

(b) If the resistance of wire \[A\] is four times the resistance of wire \[B\] , calculate

(i) the ratio of the cross sectional areas of the wires and

Ans: Since \[R = \dfrac{{\rho l}}{A}\]

For wire $A$ , ${R_1} = \dfrac{{\rho l}}{{{A_1}}}$

For wire $B$ , ${R_2} = \dfrac{{\rho l}}{{{A_2}}}$

\[ \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{A_1}}}{{{A_2}}}\]

Since ${R_1} = 4{R_2}$

$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = 1:4$

\[\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi {r_1}^2}}{{\pi {r_2}^2}}\]

\[ = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\]

Ratio is \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\] .

(ii) The ratio of the radii of the wire.

Ans: \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} = 1/4\]

Ratio is $1:2$ .

2. (a) State ohm’s law?

Ans: If the temperature and other physical conditions of the conductor stay constant, the electric current flowing through the conductor is precisely proportional to the potential difference across the conductor's end.

(b) The value of \[(I)\] current flowing through a conductor for the corresponding values of \[(V)\] potential difference are given below

${\text{I}}$ (Amperes)	$0.5$	$1.0$	$1.5$	$2.5$	$3$
${\text{V}}$ (Volts)	$1$	$2$	$3$	$4.5$	$5$

Plot a graph between \[V\] and \[I\] and also calculate resistance.

Ans: Along $x$ -axis \[IV = 1\] cm

The resistance is $1.67\Omega $ .

3. (a) Define electrical energy with S.I. unit?

Ans: The effort done by a source of electricity to sustain current in a circuit is known as electrical energy. The joule is its SI unit.

(b) A household uses the following electric appliance; Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\] .

(i) Refrigerator of rating \[400\] w for ten hour each day.

Ans: Electricity consumed by refrigerator in one day $ = $ power $ \times $ time

\[ = 400 \times \;10\]

\[ = {\text{ }}4000\] Wh

\[ = 4\] kwh

Therefore the electricity consumed is $4$ KWh.

(ii) Two electric fans of rating \[80\] w each for twelve hours each day.

Ans: Electricity consumed by $2$ electric fans in $1$ day, power $ \times $ time

\[ = 2 \times 80 \times 12\]

\[ = {\text{ }}1.92\] kwh

Therefore the electricity consumed is $1.92$ KWh.

(iii) Six electric tubes of rating \[18\] w each for 6hours each day.

Ans : Electricity consumed by $6$ electric tubes in $1$ day \[ = 6 \times 18 \times \;6\]

\[ = {\text{ }}0.648\] kwh

Therefore the electricity consumed is $0.648$ KWh.

Total energy consumed in one day \[ = 4 + 1.92 + 0.648\]

\[ = 6.548\] kwh

Total energy consumed in one month \[ = 6.568 \times 30\]

\[ = 197.04\] kwh

Cost of $1$ unit (kwh) $ = $ Rs \[3.00\]

Cost of \[197.04\] kwh \[ = 197.04 \times 3\]

Electricity bill \[ = Rs591.12\]

The electricity bill of the household for the month of June is Rs. $591.12$.

4. Redraw the circuit of question \[1\] , putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the \[12\Omega \] resistors. What would be the reading in the ammeter and voltmeter?

Ammeter $A$ has been joined in series with circuit and voltmeter $V$ is joined in parallel to $12$ ohms resistor.

Total voltage of battery \[V = 3x2\]

\[ = 6\] V.

Total resistance \[R = {R_1} + {R_2} + {R_3}\]

\[ = 5\Omega + 8\Omega + 12\Omega \]

\[ = 25\Omega \]

Ammeter reading \[ = I\]

\[ = 6/25\]

\[ = 0.24\] A.

Voltmeter reading \[ = {\text{ }}IR\]

\[ = 0.24{\text{ }}x{\text{ }}12\]

\[ = 2.88\] V.

The reading in the ammeter and voltmeter is $0.24$ A and $2.88$ V respectively.

5. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Ans: The following are the benefits of connecting electrical equipment in parallel with the battery rather than in series:

(i) Each connecting electrical device will have the same voltage, and the device will take current according to its resistance.

(ii) It is possible to use separate on/off switches.

(iii) As the total resistance in the parallel circuit falls, a large current can be pulled from the cell.

(iv) Even if one electrical gadget is broken, other devices continue to function normally.

6. How can three resistors of resistance \[2\Omega ,3\Omega \]and \[6\Omega \] be connected to give a total resistance of

(a) \[4\Omega \]

Ans: If we connect resistance of $3\Omega $ and $6\Omega $ in parallel and resistance of $2\Omega $ is connected in series with the combination, then total resistance of combination is $4\Omega $ .

(b) \[9\Omega \]

Ans: If all the three resistance are joined in parallel the resulting resistance will be $3\Omega $ .

7. The value of current \[I\] flowing in a given resistor for the corresponding values of potential difference \[V\] across the resistor are given below:

${\text{I}}$ (amperes)	$0.5$	$1.0$	$2.0$	$3.0$	$4.0$
${\text{V}}$ (volts)	$1.6$	$3.4$	$6.7$	$10.2$	$13.2$

Plot a graph between \[V\] and \[I\] and calculate the resistance of that resistor.

Ans: From the given data the \[I - V\] graph is a straight line as shown below:

Resistance of resistor \[R{\text{ }} = {\text{ }}{V_A} - {V_B}/{1_A} - {1_B}\]

\[ = {\text{ }}12{\text{ }}V{\text{ }}--{\text{ }}6{\text{ }}V/{\text{ }}3.6{\text{ }}A{\text{ }}--{\text{ }}1.8{\text{ }}A\]

\[ = {\text{ }}6V/{\text{ }}1.8{\text{ }}A{\text{ }}\]

\[ = {\text{ }}3.3{\text{ }}\Omega \]

8. Explain the following:

(a) Why is tungsten used almost exclusively for filament of electric lamps?

Ans: For the filament of electric lamps, we need a robust metal with a high melting point. Because of its high melting point, tungsten is utilised only for electric lamp filament.

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Ans: Electric heating device conductors are composed of an alloy because it has a higher resistance than pure metal and a higher melting point, which prevents it from oxidising at high temperatures.

Ans: As the current to all appliances remains constant despite varying resistance, each appliance cannot be turned on or off independently.

(d) How does the resistance of wire vary with its area of cross-section?

Ans: As resistance of a wire is inversely proportional to its cross-section area, the resistance will decrease when the area of cross section increases.

(e) Why are copper and aluminum wires usually employed for electric transmission?

Ans: As copper and aluminium wires are good conductors with low resistance, they are commonly utilised for electrical transmission. They can also be drawn into thin wires since they are ductil

Important Questions of Chapter 12 Class 10 Science - Free PDF Download

Students who are weak in Science and do not have a strong core knowledge might find the chapter electricity quite confusing. This chapter is full of theories and diagrams which confuse students and act as a barrier in their way of achieving good marks. Students must plan to avoid this situation so that they can score the highest possible marks in the final exams. The best way to overcome this problem is a continuous practice. Students can solve some of Class 10 Science Chapter 12 Important Questions regularly. They must induct these practice hours in their preparation schedule. So that every day some time is given to revise and practice. This will make the students efficient and thorough.

Students who are unable to solve the questions must take the help of Important Questions of Electricity Class 10 with Solutions which is available on Vedantu for free. Students can download these questions in PDF format. The downloaded CBSE Class 10 Science Chapter 12 Important Questions act as a guide in preparation for the final exams.

Class 10 Science Ch 12 Important Questions

Students will learn about the concepts and theories of electricity when they will study Chapter 12 of Class 10. They will gain a detailed knowledge of this chapter after practising Important Questions for Class 10 Science Chapter 12. Some of the knowledge that the students will learn are as follows:

Electricity

Electricity is considered as a set of physical phenomena which are associated with the presence and motion of electric charge. It is also believed that electricity is somehow related to magnetism which is why both are part of a phenomenon which is called the phenomenon of electromagnetism. This was described in Maxwell's equations. There are many other phenomena which are related to electricity such as lightning, static electricity, electric heating, electric discharges and many others.

An electric field gets produced, when there is some kind of presence of positive or negative charge. The positive and negative charges are considered as electric charges. When the electric charge moves it is known as Electric current, and it produces a magnetic field nearby.

Attributes of Electricity

There are two primary attributes of electricity: voltage and current. They both have different properties and are quite different from each other, but only in electronic circuits, they are interrelated. Absence of any of these attributes can harm the circuit operation and ruin it. Let's discuss both these attributes:

Voltage is considered as a force which is responsible for making the current flow in a circuit. Voltage gets measured in volts. In order to get a better understanding of voltage, try to think of a water faucet, in which voltage can get compared to the pressure at which the water is coming out of the faucet. A slow-flowing stream from a faucet resembles a low voltage circuit. In contrast, a high flowing stream resembles a high voltage circuit. A voltage is a mandatory requirement for a current to flow.

The movement of electrical charges is termed as current. When electrons flow through an electronic circuit, it generates a current. Current is measured in terms of amperes (amp). Here also we can take the example of a water faucet to understand the concept of current. It is assumed that the current is at a high level when more water flows in an hour through the faucet, whereas when the flowing water level is low, the current is also low. After the rain, you must have seen that river flows faster than the usual speed because the current remains high at that point of time as more amount of water passes during rains.

Conductors and Insulators

Another thing that students are going to learn while practising the Electricity Class 10 Important Questions is about the significant classification of elements which is done on the basis of their conductivity of electric charge, i.e. conductors and insulators.

Conductors:

In simple terms, conductors is the name given to those materials which allow electricity to flow through them easily without any difficulties. The conductors' property that makes them capable of allowing electricity to flow through them is termed as conductivity. When electrons start flowing in the conductors, they start to produce an electric current. The force which is required for the current to move through the conductor is known as the voltage. Some examples of conductors are copper, gold and iron. Let's discuss some properties of a Conductor.

Materials which are considered as conductors have a minimal resistance because electricity flows through them.

The inductance of the electric conductor occurs when there is a high voltage drop in the conductor.

Inside a perfect conductor, the electric field is considered to be zero. It is zero because it helps to keep the electrons calm so that they don’t accelerate.

There is no electric charge inside the material which is considered as a conductor.

Insulators are considered to be materials within which free flow of electrons from one particle of the element to the other particle is interrupted. If a certain amount of charge is transferred to such an element at a given point of time, then the charge does not get distributed in the surface and remains at the same position. The most common process to charge these elements is by rubbing or charging it through induction. Some examples of insulators are wood, plastic and glass. Let's discuss some properties of insulators:

There are no free electrons in such material because all the electrons are tightly held with each other.

The ability of these materials to stop the electric current from passing through them is known as resistance.

Dielectric length of insulators is vast. Dielectric strength is considered as the maximum electric field that an insulator can handle without suffering an electrical breakdown.

High air permeability is a feature of good insulators as they allow air to pass through their pores.

Important Questions for Electricity Class 10

To give the students an overview of the Important Questions of Chapter 12 Class 10 Science , we are listing here some of the questions which are most likely to come in the exams:

What is Electricity?

What are the attributes of electricity?

Write the SI unit of resistivity.

What do you mean by electric current? Mention and define the SI unit of electric current.

What do you understand by electrical resistivity of a material? Describe an experiment which will show the factors on which the resistance of a conducting wire depends.

What do you understand by a conductor? State its properties.

What do you understand by an insulator? State its properties.

Differentiate between indicator and conductor.

State the difference between electric energy and power.

Mention the commercial unit of electric energy.

Convert the unit of electric energy into joules.

Benefits of Important Questions for Class 10 Science Chapter 12

Class 10 Science Chapter 12 Important Questions offer significant advantages for board exam preparation.

Vedantu's team conducts thorough research to compile a list with a high probability of exam inclusion.

Expert reviews from seasoned professionals with extensive experience ensure accuracy.

Questions adhere to the CBSE board format, providing students insight into exam patterns.

Detailed and explanatory solutions accompany these important questions for comprehensive understanding.

Important Related Links for CBSE Class 10 Science

CBSE Class 10 Science Study Materials

These important questions serve as a valuable reference, elucidating key concepts and aiding comprehensive exam preparation. Whether unravelling Ohm's Law or comprehending circuit configurations, these important questions offer understanding. For students grappling with scientific complexities, relying on "Electricity Class 10 Important Questions" proves beneficial, promising improved comprehension and heightened confidence. This resource is not just a tool for academic success but a key to unlocking one's potential in understanding and excelling in the captivating subject of electricity.

FAQs on Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

1. How will Vedantu’s important questions for Class 10 Science Chapter 12 will help you to score well?

Ans: Vedantu’s important questions for Class 10 Science Chapter 12 helps the students by providing an excellent exam strategy to prepare for the board examinations. Science is an essential subject for students who are aspiring for various competitive exams like JEE, NEET, etc in future. Every student aspiring to excel in the board examinations requires deep knowledge and thorough understanding of this subject. Therefore, the important questions for CBSE Class 10 Science help the students in preparing for their exams. All the questions present in these materials are framed under the latest CBSE curriculum and guidelines.

2. Can I download the important questions for Class 10 Science Chapter 12 from Vedantu website for free?

Ans: Yes, the download option is available on Vedantu website and mobile application for Class 10 Science Chapter 12. Any student can download these study materials from Vedantu website in PDF format at absolutely free of cost. These are high-quality study materials created by our in-house subject matter experts as per the latest CBSE curriculum and guidelines.

3. Define electricity.

Ans: Electricity is one of the most important aspects of our society. Electricity is a set of physical phenomena that has been shaping up our civilization ever since the wake of the industrial revolution all over the entire industries and businesses. Today, life without electricity would be unimaginable and would lead to total chaos if we somehow lose this important source of energy.

4. What are the topics and subtopics covered under this chapter of Class 10 Science?

Resistivity and Resistance

Factors that affect the Resistance of a Conductor

Parallel and Series Combination of Resistors and their applications

Heating Effect of Electric Current and its Applications

Electric Power

The interrelation between P, V, I and R

5. Define the following terms-

Potential difference

Ans: One volt- One volt is defined as the potential difference between two points in a conductor that carries current. Here, one joule of work is done to locomote a charge from one place to the other. It is the SI unit of potential difference.

Potential Difference- Potential difference is defined as the work done to move a charge from one point to another in a current-carrying conductor. The formula of potential difference is 1 volt- 1 joule/ 1coulomb. A voltmeter is used to measure the potential difference. A voltmeter is connected parallelly in a circuit.

6. Define resistance.

Ans: Resistance is defined as the internal property of a substance that offers obstruction to the current flowing through it. The electric resistance of a substance is inversely proportional to its area of cross-section (A) and directly proportional to its length (l). This means when the area will increase the resistance will decrease and vice-versa. And if the length increases, the resistance will also increase.

The formula is: R=⍴*L/A

Here R is the resistance, L is the length, A is the Area and ⍴ is a constant. The SI unit of resistance is the ohm.

7. Define resistivity.

Ans: The resistivity can be defined as the resistance offered by a current-carrying conductor that has an area of cross-section equal to one centimetre square with a length equal to one centimetre. This means it is the resistance of a one-centimetre cube of the current-carrying conductor. The formula for resistivity is ⍴=R*L/A

Here R is the resistance, L is the length and A is the area. The SI unit of resistivity is ohm*metre. To study more about resistivity, students can download the Important Questions free of cost from the Vedantu website or mobile app.

8. Define ohm’s law.

Ans: Ohm’s law states that the potential difference (V) across a current-carrying conductor is directly proportional to the current (I) flowing through it. Mathematically, V ∝ I or V= IR.

Here, V is the potential difference, I is the current and R is a constant called resistance of the conductor. Ohm’s law is followed by both electrolytic conductors as well as metallic objects.

9. Define Joule’s law of heating.

Ans: Joule’s law of heating states that the heat generated in a current-carrying resistor is directly proportional to the square of the current, resisting capability of the resistor and time till which the current flows through the resistor. The formula of Joule’s Law of Heating is given as

H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule.

CBSE Class 10 Science Important Questions

Cbse study materials.

NCERT Solutions for Class 10 Science Chapter 12 Electricity

Study Reference for Class 10 Chapter 12 Electricity


Conductors	Silver	1.60 × 10
	Copper	1.62 × 10
	Aluminium	2.63 × 10
	Tungsten	5.20 × 10
	Nickel	6.84 × 10
	Iron	10.0 × 10
	Chromium	12.9 × 10
	Mercury	94.0 × 10
	Manganese	1.84 × 10
	Constantan (alloy of Cu and Ni)	49 × 10
Alloys	Manganin (alloy of Cu, Mn and Ni)	44 × 10
	Nichrome (alloy of Ni, Cr, Mn and Fe)	100 × 10
	Glass	10 − 10
Insulators	Hard rubber	10 − 10
	Ebonite	10 − 10
	Diamond	10 − 10
	Paper (dry)	10

(amperes )
(volts)

(volts)
(amperes )

Chapter 12 Electricity Class 10 NCERT Solutions

Ncert solutions for class 10 science chapters:.

What is the resistance of a Voltmeter?

What does an electric circuit mean, what is meant by saying that the potential difference between two points is 1v, how is a voltmeter connected in the circuit to measure the potential difference between two points, why are coils of electric toasters and electric irons made of alloy rather than pure metals, contact form.

NCERT Solutions for Class 10 Science Chapter 12 Electricity

The Class 10 NCERT Solutions for Science Chapter 12 Electricity includes all the intext and exercise questions. Class 10 Science Chapter 12 Electricity NCERT questions and answers help students to clear their doubts and to obtain good marks in Class 10 board exam. All the solutions provided in this article are strictly based on the CBSE syllabus and curriculum.

Class 10 Science Chapter 12 NCERT Questions and Answers

Class 10 Science Chapter 12 Electricity NCERT Questions and Answers are prepared by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 10 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 10 Science Chapter 12 Intext Questions

Intext Question (Page No. 200)

Question 1: What does an electric circuit mean?

Answer: A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric power source, wires, switches and electric devices like resistors bulbs etc.

Question 2: Define the unit of current.

Answer: When one-coulomb charge flows through an electric device in a circuit in one second, then the current flowing through the device is said to be one ampere.

Question 3: Calculate the number of electrons constituting one coulomb of charge.

Answer: Charge on one electron, 𝑒 = 1.6 × 10 − 19 C

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 1

Intext Question (Page No. 202)

Question 1: Name a device that helps to maintain a potential difference across a conductor.

Answer: Cell or battery maintain potential difference across a conductor.

Question 2: What is meant by saying that the potential difference between two points is 1 V?

Answer: When 1 J of work is done to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.

Question 3: How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer: Potential difference created by battery Δ𝑉 = 6 V

Potential difference between two points in a circuit is defined as energy required (or work done) in moving one coulomb of charge from one point to the other.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 2

Intext Question (Page No. 209)

Question 1: On what factors does the resistance of a conductor depend?

Answer: The resistance of a resistor depends on

Properties of the material of conductor, or resistivity of a material
Length of conductor
Area of cross-section of the conductor
The resistivity of a material depends on temperature. Therefore, the resistance of a resistor also depends on the temperature of the conductor.

Question 2: Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer: Resistance of a conductor depends on resistivity 𝜌, length 𝑙 and on the area of cross-section 𝐴 as

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 3

∴ when connected to the same power source, current through thick wire is more compared to thin wire of same material.

Question 3: Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer: The change in the current flowing through the electrical component can be determined by Ohm’s Law. According to Ohm’s Law, the current is given by I = V/R Now, the potential difference is reduced to half keeping the resistance constant, Let the new voltage be V’ = V/2

Let the new resistance be R’ = R and the new amount of current be I’.

The change in the current can be determined using Ohm’s law as follows:

Therefore, the current flowing the electrical component is reduced by half.

Question 4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc.

Question 5: Use the data in Table 12.2 to answer the following – Table 12.2 Electrical resistivity of some substances at 20°C

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 5

(a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?

Answer: (a) Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.

(b) Among all the materials listed in the table, silver is the best conductor because the resistivity of silver is lowest among all, i.e., 1.60 × 10 –8 .

Intext Question (Page No. 213)

Question 1: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 6

Question 2: Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 7

Resisters are connected in series. So, the net resistance in the circuit = 5 Ω + 8 Ω + 12 Ω = 25 Ω Net potential = 6 V

Now for the 12 Ω resistor, current = 0.24 A

So, using Ohm’s law V = 0.24 × 12 V = 2.88 V

Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.

Intext Question (Page No. 216)

Question 1: Judge the equivalent resistance when the following are connected in parallel –

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 9

From the above two problems, when very low resistance is connected to very high resistance, the resistance of the combination will be close to low resistance.

Question 2: An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 10

To draw the same current resistance of electric iron required is 31.25 Ω and current through it is 7.04 A

Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer: To operate properly, different electric devices need a different amount of current. In series combination all the devices get the same current, whereas in parallel combination potential difference across all the resistors is same and current will be distributed according to resistance.

In series combination, if one device fails circuit becomes broken and other devices stop working. Whereas in parallel combination all devices are independently connected to mains, even if one device fails other devices continue to work.

Question 4: How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Answer: Case 1: The circuit diagram below shows the connection of three resistors

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 12

From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 13

The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows: R eq = 2 Ω +2 Ω = 4 Ω Hence, the total resistance of the circuit is 4 Ω.

Case 2: When all resistors are connected in parallel, the equivalent resistance is,

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 14

Question 5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer: Highest resistance is possible when all resistors are connected in series.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 16

R = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω ∴ highest resistance possible is 48 Ω

Lowest resistance is possible when all resistors are connected in parallel

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 17

∴ Lowest resistance possible is 2 Ω

Intext Question (Page No. 218)

Question 1: Why does the cord of an electric heater not glow while the heating element does?

Answer: When the same current flows through the conducting wire and heating element, the heat generated (I 2 Rt) is very high in heating element compared to conducting wire. This is due to the resistance of the heating element is very high compared to the resistance of conducting wire. Thus, when same current flows through the conducting wire and heating element, heating element gets hot and glows.

Question 2: Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

Answer: From Joule’s law of heating, the heat generated can be written as H = V × I × t where, V is the voltage, V = 50 V I is the current t is the time in seconds, The amount of current can be calculated as follows:

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 19

Therefore, the heat generated when 96000 coulomb of charge flows through a potential difference of 50 V is 4.8 × 106 J

Question 3: An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer: From Joule’s law of heating, the heat generated can be written as

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 20

Therefore, the heat generated in 30 s is 15 × 10 3 J

Intext Question (Page No. 220)

Question 1: What determines the rate at which energy is delivered by a current?

Answer: Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Question 2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 21

NCERT Solutions for Class 10 Science Chapter 12 Exercise Questions

Question 1: A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 22

Question 2: Which of the following terms does not represent electrical power in a circuit?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 23

Question 3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 𝑊 (b) 75 𝑊 (c) 50 𝑊 (d) 25 𝑊

Answer: (d) 25 𝑊

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 24

The resistance of the bulb remains constant if the supply voltage is reduced to110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power

Hence, the option (d) is correct.

Question 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer: (c) 1:4

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 26

Therefore, the ratio of heat produced in series and parallel combinations is 1:4. Hence, the option (c) is correct.

Question 5: How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: Voltmeter is always connected in parallel with circuit element to measure the potential difference across it.

Question 6: A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10 – 8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 27

Question 7: The values of current flowing in a given resistor for the corresponding values of potential difference 𝑉 across the resistor are given below –

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between 𝑉 and 𝐼 and calculate the resistance of that resistor.

Answer: The plot between voltage and current is called VI characteristic. The voltage is plotted on x-axis and current is plotted on y-axis.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 28

Question 8: When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 29

Question 9: A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer: In series combination current through all the resistors will be the same. Resistance of resistors when connected in series is given by

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 30

So, the current through the 12 Ω resistor will be same as 0.67 A.

Question 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer: Let the required number of resistors be 𝑛. Given, Current I = 5A and Potential difference V = 220V Now, from Ohm’s law, V = IR

Now for 𝑛 number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 32

Therefore 4 resistors of 176 Ω are required to draw 5 A current on 220 V line.

Question 11: Show how you would connect three resistors, each of resistance 6 Ω so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer : If we connect all the three resistors in series, their equivalent resistor would 6 Ω + 6 Ω + 6 Ω =18 Ω, which is not the desired value. Similarly, if we connect all the three resistors in parallel, their equivalent resistor would be

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 33

which is again not the desired value. We can obtain the desired value by connecting any two of the resistors in either series or parallel.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 34

If two resistors are connected in parallel, then their equivalent resistance is

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 35

The third resistor is in series, hence the equivalent resistance is calculated as follows: R = 6 Ω + 3 Ω = 9 Ω

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 36

When two resistors are connected in series, their equivalent resistance is given by

R = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is calculated as follows:

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 37

Question 12: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 38

Question 13: A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 39

Question 14: Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 40

Question 15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 41

Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: Energy consumed by an electrical appliance is given by H = Pt Power rating of TV set is 250 W Energy consumed by TV set 1 hour = 250 × 60 × 60 = 900000 J Power rating of toaster is 1200 W Energy consumed by toaster in 10 minutes = 1200 × 10 × 60 =720000 J Hence, TV set uses more energy than toaster.

Question 17: An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

NCERT Solutions for Class 10 Science Chapter 12 Electricity image 42

Question 18: Explain the following.

(i) Why is tungsten used almost exclusively for filament of electric lamps?

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

(in) Why is the series arrangement not used for domestic circuits?

(iv) How does the resistance of a wire vary with its area of cross-section?

(v) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: (i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted.

(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.

(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working because the whole circuit is broken.

(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R ∝ (1/πr 2 ). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor.

(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

Topics covered under Class 10 Science Chapter 12 Electricity

Below we have listed the topics discussed in NCERT Solutions for Class 10 Science Chapter 12. The list gives you a quick look at the different topics and subtopics of this chapter.

Section in NCERT Book	Topics Discussed
12.1	Electric Current and Circuit
12.2	Electric Potential and Potential Difference
12.3	Circuit Diagram
12.4	Ohm’s Law
12.5	Factors on Which the Resistance of a Conductor Depends
12.6	Resistance of a System of Resistors
12.7	Heating Effects of Electric Current
12.8	Electric Power

NCERT Solutions for Class 10 Science Chapter 12 – A Brief Discussion

Chapter Overview: Electricity has an important place in modern society. It is a controllable and convenient form of energy for a variety of uses in homes, schools, hospitals, industries and so on. In this chapter students will learn about electric current, electric circuit, circuit diagram. This chapter also covers Ohm’s law, resistor etc.

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Examining the evolution of energy storing in the ecuadorian electricity system: a case study (2006–2023).

1. Introduction

2. methodology and problem statement.

Data Collection: Data were gathered from authoritative sources, including MEM, ARCERNNR, and ISO-CENACE. The dataset spans from 2006 to 2023, providing a robust temporal framework for analysis.
System Structure Analysis: An in-depth examination of the Ecuadorian electricity system was conducted, focusing on its composition of 58 generation companies, 19 distribution companies, and 1 transmission company. This analysis included evaluating design considerations essential for ensuring the reliable delivery of electricity with required levels of quality and safety.
Operational Dynamics and Challenges: The study analyzed the dynamics of energy demand and supply within the system, considering the integration of various energy generation technologies such as hydroelectric, thermal, and non-conventional sources. Special attention was given to operational characteristics presented by run-of-river hydroelectric plants.
Hydropower Capacity and Storage Analysis: The study specifically focused on hydroelectric power’s participation and capacity in the Ecuadorian electricity system. This included evaluating reservoir storage capacity and the productivity of hydropower facilities in relation to hydropower run-of-river plants, using dam parameters to estimate maximum energy storing, and analyzing implications for energy security.
Statistical Analysis: Historical data underwent statistical analysis to identify key factors influencing the energy-storing availability of hydropower plants and, with this, the study infers the system’s energy dependency with respect to the stochasticity of inflows. This involved examining trends over maximum power demand and energy consumption due to fluctuating energy supply.
Impact Assessment: Lastly, the study evaluated the overall impact of run-of-river hydroelectric generation on the energy system, considering current operational conditions and potential future scenarios. This assessment aimed to identify vulnerabilities and propose strategies to optimize energy storing and enhance system resilience.

3. Overview of the Ecuadorian Electricity System and Its Operational Challenges

4. results and discussions.

Increased Vulnerability to Climate Variability: Run-of-river hydropower plants are highly dependent on consistent water flow, making the energy supply susceptible to seasonal and annual variations in rainfall. During dry periods or droughts, the energy output from these plants significantly decreases, leading to potential power shortages.
Economic Impact: To compensate for the variability and unreliability of run-of-river hydroelectric plants, Ecuador may need to invest in alternative energy sources, such as thermal power plants, which are generally more expensive to operate. This reliance on costly backup generation can lead to higher electricity prices for consumers and strain the national budget.
Operational Challenges: Run-of-river hydropower plants have to work in accordance with intermittent nature inflows, which requires advanced power grid management techniques to balance supply and demand. The electric system and the power grid, without sufficient energy storage or flexible backup generation, may face stability issues, increasing the risk of blackouts and reducing overall system reliability.
Environmental Consequences: Although run-of-river hydroelectric plants are a clean energy source, an increased reliance on thermal power plants during low water flow periods can lead to higher carbon emissions, counteracting environmental benefits. Additionally, the construction of new thermal plants and associated infrastructure can have negative environmental impacts. Furthermore, each hydroelectric plant should maintain an ecological inflow. In other words, downstream of dams, an inflow should be ensured throughout the year, especially during dry periods, to support flora and fauna. Unfortunately, this is not always implemented.
Need for Infrastructure Investment: Ensuring energy security with predominantly run-of-river hydropower plants necessitates significant investments in infrastructure, including the development of modern power grids capable of accommodating distributed generation, energy storage solutions and enhanced demand management systems.
Policy and Planning Implications: An effective energy policy and strategic planning are crucial to mitigate risks associated with the reliance on run-of-river hydropower plants. This includes diversifying the energy mix, investing in renewable energy sources like wind and solar, improving the water management system, and enhancing regional energy interconnections to stabilize supply during periods of low hydroelectric production. In addition, one issue with Ecuador’s electrical system is not a lack of regulatory and planning instruments but rather challenges in financing and the billing and payment structure. While users do pay for electricity services, these funds remain with distribution companies. Due to conflicting regulations, some of these companies do not pay for generators and transmission services. As a result, there is insufficient funding for maintenance and the construction of new electric facilities.
Energy Security Risks: Prolonged reliance on running river plants without addressing their limitations is leading to chronic energy insecurity in all systems. This may undermine public confidence in the energy system, disrupt economic activities, and impede national development goals.

5. Conclusions

Author contributions, data availability statement, conflicts of interest.

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Oscullo Lala, J.; Carvajal Mora, H.; Orozco Garzón, N.; Vega, J.; Ohishi, T. Examining the Evolution of Energy Storing in the Ecuadorian Electricity System: A Case Study (2006–2023). Energies 2024 , 17 , 3500. https://doi.org/10.3390/en17143500

Oscullo Lala J, Carvajal Mora H, Orozco Garzón N, Vega J, Ohishi T. Examining the Evolution of Energy Storing in the Ecuadorian Electricity System: A Case Study (2006–2023). Energies . 2024; 17(14):3500. https://doi.org/10.3390/en17143500

Oscullo Lala, José, Henry Carvajal Mora, Nathaly Orozco Garzón, José Vega, and Takaaki Ohishi. 2024. "Examining the Evolution of Energy Storing in the Ecuadorian Electricity System: A Case Study (2006–2023)" Energies 17, no. 14: 3500. https://doi.org/10.3390/en17143500

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