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NCERT Solution for Class 10 Physics Chapter 12 - Electricity

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NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 10 Physics board exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 12 - Electricity.

All our solutions for Chapter 12 - Electricity are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 10 Physics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

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NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 200

Solution 1

An electric circuit is a continuous conducting path that consists of electric devices, switching devices, source of electricity, etc. connected by conducting wires.

Concept Insight:- The figure given below is an example of an electric circuit.

Solution 2

The unit of electric current is ampere (A).

1 ampere is defined as the flow of 1 coulomb of charge through a wire in 1 second.

Solution 3

One electron possesses a charge of 1.6 × 10 -19 C, i.e., 1.6 × 10 -19 C of charge is contained in 1 electron.

1 C of charge is contained in   electrons

Therefore, 6.25 x 1018 electrons constitute one coulomb of charge.  

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 202

Solution 1

A source of electricity such as cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

Solution 2

If 1 J of work is done to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.

Solution 3

Concept Insight:- The amount of work done to move to each coulomb of charge is equal to the energy given it.

The amount of work is given by the expression,

Potential difference = 

Work Done = Potential Difference x Charge

Here,

Charge = 1 C

Potential difference = 6 V

Work Done = 6 x 1  = 6J

Therefore, 6 J of energy is given to each coulomb of charge passing through a battery of 6 V.

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 209

Solution 1

The resistance of a conductor depends upon the following factors:

(a) Material of the conductor, i.e., resistivity of the material.

(b) Temperature of the conductor

(c) Length of the conductor

(d) Cross-sectional area of the conductor

Solution 2

Concept Insight:- Resistance of a wire is given by the relation, begin mathsize 14px style straight R equals straight rho l over straight A end style 

where,

begin mathsize 14px style straight rho end style = Resistivity of the material of the wire

l = Length of the wire

A = Area of cross-section of the wire

Resistance is inversely proportional to the area of cross-section of the wire.

Thicker the wire, larger is its area of cross-section and hence lower is the resistance of the wire. Therefore, current can flow more easily through a thick wire than a thin wire.

Solution 3

Concept Insight:- The current flowing through the component is given by Ohm's law.

V = IR

or, I =

where,

Resistance of the electrical component = R

Potential difference = V

Current = I

The potential difference is reduced to half, keeping the resistance constant.

Let the new amount of current be I '.

Therefore, from Ohm's law, we obtain the amount of new current.

begin mathsize 14px style straight I apostrophe equals fraction numerator straight V apostrophe over denominator straight R apostrophe end fraction equals fraction numerator begin display style straight V over 2 end style over denominator straight R end fraction equals 1 half open parentheses straight V over straight R close parentheses equals 1 half cross times straight I equals straight I over 2 end style

Therefore, the amount of current flowing through the electrical component is reduced by half.

Solution 4

Concept Insight:-  Resistivity and melting point are two important factors here.

The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

Solution 5

(a) Resistivity of iron = 10.0 x 10-8 ohm m

Resistivity of mercury = 94.0 x 10-8 ohm m

Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

 (b) It can be observed from the table that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Concept Insight:- Less resistivity means better conductance.

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 213

Solution 1

Three cells of potential 2 V each connected in series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 ohm, 8 ohm and 12 ohm respectively connected in series with a battery of potential 6 V and a plug key.

Concept Insight:- The voltages of cells in series add up to give the final voltage. 






Concept Insight:- The voltages of cells in series add up to give the final voltage.

Solution 2

To measure the current flowing through the resistors, an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the 12 ohm resistor, a voltmeter should be connected in parallel to this resistor, as shown in the following figure.

The resistances are connected in series.

Concept Insight:- Ohm's law can be used to obtain the readings of ammeter and voltmeter.

According to Ohm's law,

V = IR,

where,

Potential difference, V = 6 V

Current flowing through the circuit/resistors = I

Resistance of the circuit, R = 5 + 8 + 12 = 25 ohm

I = = 0.24 A

Potential difference across 12 ohm resistor = VI

Current flowing through the 12 ohm resistor, I = 0.24 A

Therefore, using Ohm's law, we obtain

V1 = IR = 0.24 x 12 =  2.88 V

Therefore, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter will be 2.88 V.

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 216

Solution 1

(a) When 1 ohm and 106 ohm are connected in parallel:

Concept Insight:-  For parallel combination, equivalent resistance R is given by

(1/R) = (1/R1)+ (1/R2)


Therefore, equivalent resistance ≈1 ohm

(b) When 1 ohm, 103 ohm and 106 ohm are connected in parallel:

Concept Insight:- For parallel combination, equivalent resistance R is given by

(1/R) = (1/R1)+ (1/R2) + (1/R3) ……+ (1/Rn)

Therefore, equivalent resistance = 0.999 ohm or 1 ohm

Solution 2

Resistance of electric lamp, R1 = 100 ohm

Resistance of toaster, R2 = 50 ohm

Resistance of water filter, R3 = 500 ohm

Voltage of the source, V = 220 V

These are connected in parallel, as shown in the following figure.



Concept Insight:- For parallel combination, equivalent resistance R is given by

(1/R) = (1/R1)+ (1/R2) + (1/R3)

According to Ohm's law,
V = IR

I =

where,

I = Current flowing through the circuit

begin mathsize 14px style straight I equals fraction numerator 220 over denominator begin display style 500 over 16 end style end fraction equals fraction numerator 220 cross times 16 over denominator 500 end fraction equals 3520 over 500
therefore straight I equals 7.04 space straight A end style

7.04 A of current is drawn by all the three given appliances.

Therefore, current drawn by an electric iron connected to the same voltage source of 220 V = 7.04 A

Let R’ be the resistance of the electric iron. According to Ohm's law,

V = IR'

 

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

Therefore, the resistance of the electric iron is 31.25Wand the current flowing through it is 7.04 A.

Solution 3

The advantages of connecting electrical devices in parallel with the battery instead of connecting them in series are:

1. In parallel circuit, if one electrical device stops working, then all other devices keep working normally. This is not the case when devices are connected in series.

2. There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage. In series circuit, the applied voltage is shared by all the appliances.

3. The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel. In series combination, the total effective resistance of the circuit increases.

Solution 4

There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors to get a total resistance of 4 Ω.

Concept Insight:- Here, 6 Ω and 3 Ω resistors are connected in parallel.

Therefore, their equivalent resistance will be given by


This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) The following circuit diagram shows the connection of the three resistors to get a total resistance of 1 Ω.




Concept Insight:- All the resistors are connected in parallel. Therefore, their equivalent resistance will be given as



Therefore, the total resistance of the circuit is 1 Ω.

Solution 5

There are four coils of resistances 4 ohm, 8 ohm, 12 ohm, and 24 ohm respectively.

(a) Concept Insight:- For getting the highest resistance from a combination of resistances, connect them in series.

If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 ohm.

Therefore, 48 ohm is the highest total resistance that can be secured by the combination of given resistances.

(b) Concept Insight:- For getting the lowest resistance from a combination of resistances, connect them in parallel.

If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by


Therefore, 2 ohm is the lowest total resistance that can be secured by the combination of given resistances.


NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 218

Solution 1

Concept Insight:- The amount of heat produced in a conductor is proportional to its resistance.

The resistance of the heating element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord of the heater is much lower, so it does not become very hot and does not glow when current flows through it.

Solution 2

Concept Insight:- The amount of heat (H) produced is given by the Joule’s law of heating as

H = VIt

where,

Voltage, V = 50 V

Time, t = 1 h = 1 × 60 × 60 s

Amount of current,  I = A


Therefore, the heat generated is 4.8 x 106 J.

Solution 3

Concept Insight:- The amount of heat (H) produced is given by the joule’s law of heating as

H = VIt =I2Rt 

where,

Current, I = 5 A

Resistance, R = 20 ohm

Time, t = 30 s

H = 52 x 20 x 30 = 1.5 x 104 J
 
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.

Concept Insight:- The amount of heat (H) produced is given by the joule’s law of heating as

H = VIt =I2Rt

where,

Current, I = 5 A

Resistance, R = 20 Ω

Time, t = 30 s

Therefore, the amount of heat developed in the electric iron is

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 220

Solution 1

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Concept Insight:- Power of an appliance determines the rate at which electrical energy is delivered to it.

Solution 2

Concept Insight:- Power of an appliance can be determined by the rate at which electrical energy is delivered to it.

Power (P) is given by the expression,

P = VI

where,

Voltage, V = 220 V

Current, I = 5 A

P = 220 × 5 = 1100 W

Energy consumed by the motor = P t

where,

Time, t = 2 h = 2 × 60 × 60 = 7200 s

P = 1100 × 7200 = 7.92 × 106 J

Therefore, power of the motor = 1100 W

Energy consumed by the motor = 7.92 × 106 J

NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 221

Solution 1

(d) 25

Resistance of a piece of wire is proportional to its length.

The given piece of wire has a resistance R. The wire is cut into five equal parts.

Therefore, resistance of each part=

All the five parts are connected in parallel. Hence, equivalent resistance (R') is given as

Concept Insight:- For parallel combination, equivalent resistance R' is given by

 
 

Solution 2

Concept Insight:-  Power of an appliance determines the rate at which electrical energy is delivered to it.

Electrical power is given by the expression, P = VI … (i)

According to Ohm's law, V = IR … (ii)

where,

V = Potential difference

I = Current

R = Resistance

So, it can be written that

P = (IR) × I

P = I2 R


From equation (ii), it can be written



Power P cannot be expressed as IR2.

Solution 3

(d)  25 W

Concept Insight:-  Power of an appliance determines the rate at which electrical energy is delivered to it.

Power of an appliance is given by the expression,

P = VI =
R =

where,

Power rating, P = 100 W

Voltage, V = 220 V


Resistance, R = = 484 ohm

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as

 

Therefore, the power consumed will be 25 W.

Solution 4

(c) 1:4

Both the wires have the same resistance because they are made of the same material and have equal lengths and diameters. Let the resistance be R.

For series combination, equivalent resistance Rs is

Rs = R + R = 2R

For parallel combination, equivalent resistance Rp is

Let the current through the series combination be Is and heat produced in the circuit be Hs.

Let the current through the parallel combination be Ip and heat produced in the circuit be Hp.

Therefore, the ratio of heat produced in series and parallel combinations is 1:4.

Solution 5

To measure the potential difference between two points, a voltmeter should be connected in parallel across these points.

Concept Insight:-  Voltmeter is always connected in parallel to the element of any electrical circuit across which potential difference is to be measured.

Solution 6

Resistance (R) of the copper wire of length (l) and cross-section (A) is given by the expression,

begin mathsize 14px style straight R equals straight rho l over straight A end style

where,

Resistivity of copper, begin mathsize 14px style straight rho equals 1.6 cross times 10 to the power of negative 8 end exponent space straight capital omega space straight m end style

Area of cross-section of the wire, begin mathsize 14px style straight A equals straight pi open parentheses Diameter over 2 close parentheses squared end style

Diameter = 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

We know that,

begin mathsize 14px style straight R equals straight rho l over straight A
l equals fraction numerator R A over denominator straight rho end fraction
l equals fraction numerator 10 cross times 3.14 cross times open parentheses begin display style fraction numerator 0.0005 over denominator 2 end fraction end style close parentheses squared over denominator 1.6 cross times 10 to the power of negative 8 end exponent end fraction
l equals fraction numerator 10 cross times 3.14 cross times 25 over denominator 4 cross times 1.6 end fraction
l equals 122.72 space straight m end style

Therefore, the length of the wire is 122.72 m.

So, if the diameter of the wire is doubled, the new diameter = 2 × 0.0005 = 0.001 m. 

Let the new resistance be R'.

begin mathsize 14px style straight R apostrophe equals straight rho l over straight A
straight R apostrophe equals fraction numerator 1.6 cross times 10 to the power of negative 8 end exponent cross times 122.72 over denominator straight pi open parentheses begin display style 1 half cross times 10 to the power of negative 3 end exponent end style close parentheses squared end fraction
straight R apostrophe equals fraction numerator 1.6 cross times 10 to the power of negative 8 end exponent cross times 122.72 cross times 4 over denominator 3.14 cross times 10 to the power of negative 6 end exponent end fraction
straight R apostrophe equals 250.2 cross times 10 to the power of negative 2 end exponent equals 2.5 space straight capital omega end style

Therefore, the new resistance is 2.5 Ω.

Solution 7

The VI graph is shown below. The voltage is plotted on x -axis and current is plotted on y -axis.

Solution 8

Resistance (R) of a resistor is given by Ohm's law as,

V = IR
R =  
where,

Potential difference, V = 12 V

Current in the circuit, I = 2.5 mA = 2.5 x 10-3 A

Concept Insight: - Convert all the quantities in the same unit system and then proceed to calculations.


Solution 9

Concept Insight:- In a series combination, current flowing through all the components of the circuit is the same.

There is no current division occurring in a series circuit. Current flow through all the components is the same, given by Ohm's law as

V = IR

where,

V = Potential difference

I = Current through the circuit

R = Resistance of the circuit

Let, R be the equivalent resistance of resistances 0.2 ohm, 0.3 ohm, 0.4 ohm, 0.5 ohm and 12 ohm.

These are connected in series. Hence, the sum of the resistances will give the value of R.

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 ohm

Potential difference, V = 9 V

Therefore, the current that would flow through the circuit and hence 12 ohm resistor is

I = V/R = 9/13.4 = 0.67 A

Solution 10


Solution 11


Solution 12


Solution 13


Solution 14


NCERT Solution for Class 10 Physics Chapter 12 - Electricity Page/Excercise 222

Solution 15

Concept Insight:- The voltage across each component of a parallel circuit remains the same.

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.

We know, Power = Voltage x Current

Current drawn by the bulb of rating 100 W is given by,

Current =
Similarly, current drawn by the bulb of rating 100 W is given by,


Current =
Hence, current drawn from the line = 

Solution 16

Concept Insight: Energy consumed by an electrical appliance is given by the expression,

H = Pt

where,

P = Power of the appliance

T = Time

Energy consumed by a TV set of power 250 W in 1 h = 250 x 3600 = 9 x 105 J

Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 x 600 = 7.2 x 10 5 J

Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Solution 17

Concept Insight:- Rate of heat produced by a device is given by the expression for power as

 P = I2R

Resistance of the electric heater, R = 8 ohm

Current drawn, I = 15 A

P = (15)2 × 8 = 1800 J/s

Therefore, heat is produced by the heater at the rate of 1800 J/s.

Solution 18

(a) The melting point of tungsten is very high, so the tungsten filament can be kept white-hot without melting away. Hence, tungsten is mainly used almost exclusively for filament of incandescent lamps.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is generally more than that of pure metals of which it is made. It produces large amount of heat. Moreover, at high temperatures, the alloys do not melt readily.

(c) In a series arrangement, if any one of the appliances fails or is switched off, then the flow of current through the entire circuit stops and all other appliances stop working. Thus series arrangement is not used for domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e.,


(e) Copper and aluminium wires have low resistivity. They are good conductors of electricity. Hence, they are usually employed for electricity transmission.


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