NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for class 12 Chapter 3 Current Electricity is essential resource material that will make you score good marks in class 12 Physics examinations and entrance examinations. Current Electricity class 12 NCERT solutions PDF are provided here for free download and students can get answers to the questions provided in the textbook along with exemplary problems, numerical on current electricity class 12 PDF, worksheets and exercises. Current Electricity NCERT solutions class 10 aids you in preparation of Physics class 12 Chapter 3 notes which will come in handy while doing revisions.

The NCERT solutions class 12 Physics Chapter 3 Current Electricity is one of the most common topics for class 12 board examination. Students must prepare for exams using the NCERT solutions provided here to score well in the examination. The NCERT class 12 Physics solutions are one of the best tools to prepare Physics for board examination.

Class 12 Physics NCERT solutions for Current Electricity

The important questions in Chapter 3 of class 12 Physics that are frequently asked are definitions of resistance, conductance, drift velocity, cell EMF, resistivity, conductance and internal resistance. The graphing of resistivity for metals, semiconductors and alloys must be practiced thoroughly. The conversion of complex circuits into simple series and parallel circuits must be well versed. Students should practice a wide range of problems to get better at this. Questions from potentiometer or meter bridge are regularly asked. The resistivity chart for different colour codes can be asked in the exams.

Topics covered in NCERT class 12 chapter 3 of Physics Current Electricity

Section Number Topic
3.1 Introduction
3.2 Electric Current
3.3 Electric Currents In Conductors
3.4 Ohm’s Law
3.5 Drift Of Electrons And The Origin Of Resistivity
3.5.1 Mobility
3.6 Limitations Of Ohm’s Law
3.7 Resistivity Of Various Materials
3.8 Temperature Dependence Of Resistivity
3.9 Electrical Energy, Power
3.10 Combination Of Resistors – Series And Parallel
3.11 Cells, Emf, Internal Resistance
3.12 Cells In Series And In Parallel
3.13 Kirchhoff’s Rules
3.14 Wheatstone Bridge
3.15 Meter Bridge
3.16 Potentiometer

Class 12 Physics NCERT Solutions Current Electricity Important Questions


Question 3.1:

The car has a storage battery with an emf of 12 V and internal resistance 0.4Ω. Calculate the maximum current that can be extracted from the battery?

Answer 3.1:

In the given question,

The EMF of the battery is given as E = 12 V

The internal resistance of the battery is given as R = 0.4 Ω

The amount of maximum current drawn from the battery is given by  = I

According to Ohm’s law,

E = IR

Rearranging, we get

I=ERI = \frac{ E }{ R }

Substituting values in the above equation, we get

I=120.4I = \frac{ 12 }{ 0.4 } = 30 A

Therefore, the maximum current drawn from the given battery is 30 A.

 

Question 3.2:

A battery with an emf of 10 V and an internal resistance 3Ω is connected to a resistor.  Calculate the resistance of the resistor if the current flowing in the circuit is 0.5 A. Also, calculate the voltage of the battery when the circuit is closed.

Answer 3.2:

Given:

The EMF of the battery (E = 10 V)

The internal resistance of the battery (R = 3 Ω)

The current in the circuit (I = 0.5 A)

Consider the resistance of the resistor to be R.

The current in the circuit can be found out using Ohm’s Law as,

I=ER+rI = \frac{ E }{ R + r }

Consider the Terminal voltage of the resistor to be V.

Then, according to Ohm’s law,

V = IR

Substituting values in the equation, we get

V = 0.5 × 17

V = 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.

 

Question 3.3 :

a) A series combination of three resistors with the following resistance r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω is made. Calculate the total resistance of the combination.

b) Calculate the potential drop across each resistor given above, when this combination is connected to a battery of emf 12 V with negligible internal resistance.

Answer 3.3 :

  1. a) We know that resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are combined in series.

The total resistance of the above series combination can be calculated by the algebraic sum of individual resistances as follows:

Req = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus calculated Req = 6 Ω

b )Let us consider I to be the current flowing the given circuit

Also,

The emf of the battery is E = 12 V

Total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law, relation for current can be obtained as

I=ERI = \frac{ E }{ R }

Substituting values in the above equation, we get

I=126I = \frac{ 12 }{ 6 } = 2 A

Therefore, the current calculated is 2 A

Let the Potential drop across 1 Ω resistor = V 1

The value of V 1 can be obtained from Ohm’s law as :

V 1 = 2 x 1 = 2 V

Let the Potential drop across 2 Ω resistor = V 2

The value of V 2 can be obtained from Ohm’s law as :

V 2 = 2 x 2 = 4 V

Let the Potential drop across 3 Ω resistor = V 3

The value of V 3 can be obtained from Ohm’s law as :

V 3 = 2 x 3 = 6 V

Therefore, the potential drops across the given resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are calculated to be

V 1 = 2 x 1 = 2 V

V 1 = 2 x 1 = 4 V

V 1 = 2 x 1 = 6 V

 

Question 3.4 :

a) A parallel combination of three resistors with the following resistance r 1 = 2 Ω , r 2 = 4 Ω and r 3 = 5 Ω is made. Calculate the total resistance of the combination.

b) Calculate the current drawn from the battery when the above combination of resistors is given is connected to a battery of emf 20 V with negligible internal resistance.

Answer 3.4 :

A ) Resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are combined in parallel

Hence the total resistance of the above circuit can be calculated  by the following formula :

1R=1R1+1R2+1R3\frac{ 1 }{ R } = \frac{ 1 }{ R1 } + \frac{ 1 }{ R2 } + \frac{ 1 }{ R3 } 1R=12+14+15\frac{ 1 }{ R } = \frac{ 1 }{ 2 } + \frac{ 1 }{ 4 } + \frac{ 1 }{ 5 } 1R=10+5+420\frac{ 1 }{ R } = \frac{ 10 + 5 + 4 }{ 20 } 1R=1920\frac{ 1 }{ R } = \frac{ 19 }{ 20 }

Therefore, total resistance of the parallel combination given above is given by :

R = 1920\frac{ 19 }{ 20 }

 

B ) Given that emf of the battery , E = 20 V

Let the current flowing through resistor R 1 be I 1

I 1 is given by :

I1=VR1I_{ 1 } = \frac{ V }{ R_{ 1 }} I1=202I_{ 1 } = \frac{ 20 }{ 2 } I1=10AI_{ 1 } = 10 A

Let the current flowing through resistor R 2 be I 2

I 2 is given by :

I2=VR2I_{ 2 } = \frac{ V }{ R_{ 2 }} I2=204I_{ 2 } = \frac{ 20 }{ 4 } I2=5AI_{ 2 } = 5 A

 

Let the current flowing through resistor R 3 be I 3

I 3 is given by :

I3=VR1I_{ 3 } = \frac{ V }{ R_{ 1 }} I3=205I_{ 3 } = \frac{ 20 }{ 5 } I3=4AI_{ 3 } = 4 A

Therefore , the total current can be found by the following formula :

I = I 1 + I 2 + I 3 = 10 + 5 + 4 = 19 A

therefore the current flowing through each resistors is calculated to be :

I1=10AI_{ 1 } = 10 A I2=5AI_{ 2 } = 5 A I3=4AI_{ 3 } = 4 A

and the total current is calculated to be, I = 19 A

 

Question 3.5 :

Resistance of a heating element is observed to be 100 Ω at a room temperature of ( 27 ° C ) . Calculate the resistance of this element if the resistance is 117 Ω, given that the temperature coefficient  of the material used for the element is 1.70 x 10 – 4 C – 1

Answer 3.5 :

Given that the room temperature , T = 27 ° C

The heating element has a resistance of , R = 100 Ω

Let the increased temperature of the filament be T 1

At T 1 , the resistance of the heating element is  R 1 = 117 Ω

Temperature coefficient of the material used for the element is 1.70 x 10 – 4 C – 1

α = 1.70 x 10 – 4 C – 1

α is given by the relation ,

α=R1RR(T1T)\alpha = \frac{ R_{ 1 } – R }{ R \left ( T _{ 1 } – T \right )} T1T=R1RRαT _{ 1 } – T = \frac{ R _{ 1 } – R }{ R \alpha } T127=117100100(1.7x104)T _{ 1 } – 27 = \frac{ 117 – 100 }{ 100 ( 1.7 x 10_{-4}) } T127=1000T _{ 1 } – 27 = 1000

T 1 = 1027 ° C

Therefore, the resistance of the element is 117 Ω at T 1 = 1027 ° C

 

Question 3.6 :

The length of a wire is 15 m and uniform cross-section is 6.0 x 10 – 7 m 2.  A negligibly small current is passed through it with a resistance of 5.0 Ω. Calculate the resistivity of the material at the temperature at which the experiment is conducted.

Answer 3.6 :

Given that the length of the wire , L = 15 m

Area of cross – section is given as , a = 6.0 x 10 – 7 m 2

Let the resistance of the material of the wire be , R , ie. , R = 5.0 Ω

Resistivity of the material is given as ρ

R=ρLAR = \rho \frac{ L }{ A } ρ=RxAL=5×6×10715=2×107\rho = \frac{ R x A }{ L } = \frac{ 5 \times 6 \times 10 ^{ – 7 }}{ 15 } = 2 \times 10 ^{ – 7 }

Therefore, the resistivity of the material is calculated to be 2×1072 \times 10 ^{ – 7 }

 

Question 3.7 :

A silver wire is observed to have a resistance of 2.1 Ω at a temperature of 27.5 ° C and a resistance of 2.7 Ω at a temperature of 100 ° C. Calculate the temperature coefficient of resistivity of silver.

Answer 3.7 :

Given :

Given that temperature T 1 = 27.5 ° C

Resistance R 1 at temperature T 1 is given as :

R 1 =  2.1 Ω ( at T 1 )

Given that temperature T 2 = 100 ° C

Resistance R 2 at temperature T 2 is given as :

R 2 =  2.7 Ω ( at T 2 )

Temperature coefficient of resistivity of silver = α

α=R2R1R1(T2T1)\alpha = \frac{ R _{ 2 } – R _{ 1 }}{ R _{ 1 } \left ( T _{ 2 } – T _{ 1 }\right )} α=2.72.12.1(10027.5)=0.0039C1\alpha = \frac{ 2.7 – 2.1 }{ 2.1 \left ( 100 – 27.5 \right )} = 0.0039 ^{\circ} C ^{ – 1 }

Therefore, the temperature coefficient of resistivity of silver is 0.0039C10.0039 ^{\circ} C ^{ – 1 }

 

Question 3.8 :

A heating element made up of nichrome is connected to a 230 V supply draws an initial current of 3.2 A  which after a few seconds, settles to a steady value of 2.8 A. calculate the steady temperature of the heating element at a room temperature of 27.0 ° C. The temperature coefficient of nichrome ( material used to make the above  heating device ) averaged over the temperature range involved is 1.70 x 10 – 4 ° C  – 1   

Answer 3.8 :

In the given problem,

The supply voltage is V = 230 V

The initial current drawn is I 1 = 3.2 A

Consider the initial resistance to be R 1, which can be found by the following relation :

R1=VIR _{ 1 } = \frac{ V }{ I }

Substituting values, we get

R1=2303.2R _{ 1 } = \frac{ 230 }{ 3.2 } = 71.87 Ω

Value of current at steady state , I 2 = 2.8 A

Value of resistance at steady state = R 2

R 2 can be calculated by the following equation :

R2=2302.8R _{ 2 } = \frac{ 230 }{ 2.8 } R2=82.14ΩR _{ 2 } = 82.14 Ω

The temperature coefficient of nichrome averaged over the temperature range involved is 1.70 x 10 – 4 ° C – 1

Value of initial temperature of nichrome , T 1 = 27.0 ° C

Value of steady state temperature reached by nichrome = T 2

This temperature T 2 can be obtained by the following formula :

α=R2R1R1(T2T1)\alpha = \frac{ R _{ 2 } – R _{ 1 }}{ R _{ 1 } \left ( T _{ 2 } – T _{ 1 }\right )} T227=82.1471.8771.87x(1.7x104)T _{ 2 } – 27 = \frac{ 82.14 – 71.87 }{ 71.87 x ( 1.7 x 10_{-4}) } T227=840.5T_{ 2 } – 27 = 840.5

T 2 = 840.5 + 27 = 867.5 ° C

Hence, the steady temperature of the heating element is 867.5 ° C

 

Question 3.9 :

Calculate the current in each branch of the network shown in the figure shown below :

Current Electricity

Answer 3.9 :

The current flowing through various branches of the network is shown in the figure given below :

network

Let I 1 be the current flowing through the outer circuit

Let I 2 be the current flowing through AB branch

Let I 3 be the current flowing through AD branch

Let I 2 – I 4 be the current flowing through branch BC

Let I 3 + I 4 be the current flowing through branch BD

Let us take closed-circuit ABDA into consideration, we know that potential is zero.

i.e ,  10 I 2 + 5 I 4 – 5 I 3 = 0

2 I 2 +  I 4 –  I 3 = 0

I 3 = 2 I 2 +  I 4                                                                                               . . . . . . . . . . . . . . . . . . . . . . eq ( 1 )

 

Let us take closed circuit BCDB into consideration , we know that potential is zero.

i.e ,  5 ( I 2 – I 4 ) – 10 ( I 3 + I 4 ) – 5 I 4 = 0

5 I 2 + 5 I 4 – 10 I 3 – 10 I 4 –  5 I 4 = 0

5 I 2 – 10 I 3 –  20 I 4 = 0

I 2 = 2 I 3 – 4 I 4                                                                                          . . . . . . . . . . . . . . . . . . . . . . eq ( 2 )

 

Let us take closed circuit ABCFEA into consideration , we know that potential is zero.

i.e ,  – 10 + 10 ( I 1 ) + 10 ( I 2 ) + 5 ( I 2 –  I 4 ) = 0

10 = 15 I 2 + 10 I 1 – 5 I 4

3 I 2 + 2 I 2 – I 4 = 2                                                                                    . . . . . . . . . . . . . . . . . . . . . . eq ( 3 )

From equation ( 1 )  and ( 2 ) , we have :

I 3 = 2 ( 2 I 3 + 4 I 4 ) + I 4

I 3 = 4 I 3 + 8 I 4 + I 4

– 3 I 3 = 9 I 4

– 3 I 4 = + I 3                                                                                            . . . . . . . . . . . . . . . . . . . . . . eq ( 4 )

Putting equation  ( 4 ) in equation ( 1 ) , we have  :

I 3 = 2 I 2 + I 4

– 4 I 4 = 2 I 2

I 2 = – 2 I 4                                                                                                . . . . . . . . . . . . . . . . . . . . . . eq ( 5 )

From the above equation , we infer that :

I 1 = I 3 + I 2                                                                                                   . . . . . . . . . . . . . . . . . . . . . . eq ( 6 )

Putting equation ( 4 ) in equation ( 1 ) , we obtain

3 I 2 + 2 ( I 3 + I 2 ) – I 4 = 2

5 I 2 + 2 I 3 – I 4 = 2                                                                                  . . . . . . . . . . . . . . . . . . . . . . eq ( 7 )

Putting equations ( 4 ) and ( 5 ) in equation ( 7 ) , we obtain

5 ( – 2 I 4 ) + 2 ( – 3 I 4 ) – I 4 = 2

– 10 I 4 –  6 I 4 – I 4 = 2

17 I 4 = – 2

I4=217AI _{ 4 } = \frac{ – 2 }{ 17 } A

Equation ( 4 ) reduces to

I 3 = – 3 ( I 4 )

I3=3(217)=617AI_{ 3 } = – 3 \left ( \frac{ – 2 }{ 17 } \right ) = \frac{ 6 }{ 17 } A

I 2 = – 2 ( I 4 )

I2=2(217)=417AI_{ 2 } = – 2 \left ( \frac{ – 2 }{ 17 } \right ) = \frac{ 4 }{ 17 } A I2I4=417217=617AI _{ 2 } – I _{ 4 } = \frac{ 4 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 6 }{ 17 } A I3+I4=617217=417AI _{ 3 } + I _{ 4 } = \frac{ 6 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 4 }{ 17 } A I1=I3+I2I _{ 1 } = I _{ 3 } + I _{ 2 } I1=617+417=1017AI _{ 1 } = \frac{ 6 }{ 17 } + \frac{ 4 }{ 17 } = \frac{ 10 }{ 17 } A

Therefore, current in each branch is given as :

In branch AB=417AAB = \frac{ 4 }{ 17 } A

In branch BC=617ABC = \frac{ 6 }{ 17 } A

In branch CD=417ACD = \frac{ – 4 }{ 17 } A

In branch AD=617AAD = \frac{ 6 }{ 17 } A

In branch BD=217ABD = \frac{ – 2 }{ 17 } A

Total current  =417+617+417+617+217=1017A= \frac{ 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 2 }{ 17 } = \frac{ 10 }{ 17 } A

 

Question 3. 10 :

A ) In the metre bridge shown below in the figure, 39.5 cm is found to be the balance point from the end A with N resistor having a value of 12.5 Ω. Calculate the value of the resistor M. Also, state the reason behind making the connections between the resistors in a Wheatstone bridge or meter bridge of thick copper strips.

B ) Calculate the new balance point of the bridge above if M and N are interchanged.

C ) Would the galvanometer show any deflection if the cell and galvanometer are interchanged at the balance point of the Wheatstone bridge?    

Answer 3. 10 :

A meter bridge with resistors M and N are shown in the figure.

Meter Bridge

( a ) Let L 1 be the balance point from end A ,

Given that , L 1 = 39.5 cm

Given that resistance of the resistor N = 12.5 Ω

We know that , condition for the balance is given by the equation :

MN=100L1L1\frac{ M }{ N } = \frac{ 100 – L _{ 1 } }{ L _{ 1 } } M=10039.539.5×12.5=8.2ΩM = \frac{ 100 – 39.5 }{ 39.5 } \times 12.5 = 8.2 \Omega

Thus calculated the resistance of the resistor M , M =  8.2Ω8.2 \Omega

 

Question 3.11 :

A storage battery has an emf of 8.0 V and internal resistance of 0.5 Ω is allowed by charged by a 120 V, DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

 

Answer 3.11 :

Given :

The EMF of the given storage battery is E = 8.0 V

The Internal resistance of the battery is given by r = 0.5 Ω

The given DC supply voltage is V = 120 V

The resistance of the resistor is R = 15.5 Ω

Effective voltage in the circuit = V 1

R is connected to the storage battery in series.

Hence, it can be written as

V 1 = V – E

V 1 = 120 – 8 = 112 V

Current flowing in the circuit = I , which is given by the relation ,

I=V1R+rI = \frac{ V ^{ 1 }}{ R + r } I=11215.5+5I = \frac{ 112 }{ 15.5 + 5 } I=11216I = \frac{ 112 }{ 16 } I=7AI = 7 A

We know that Voltage across a resistor R given by the product,

I x R = 7 × 15.5 = 108.5 V

We know that ,

DC supply voltage = Terminal voltage + voltage drop across R

Terminal voltage of battery = 120 – 108.5 = 11.5 V

A series resistor when connected in a charging circuit limits the current drawn from the external source.

The current will become extremely high in its absence. This is extremely dangerous.

 

Question 3.12 :

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm

length of the wire. If the cell is replaced by another cell and the balance point shifts to

63.0 cm, what is the emf of the second cell?

 

Answer 3.12 :

Emf of the cell, E 1 = 1.25 V

The balance point of the potentiometer, l 1 = 35 cm

The cell is replaced by another cell of emf E 2.

New balance point of the potentiometer, l 2 = 63 cm

The balance condition is given by the relation ,

E1E2=I1I2\frac{ E _{ 1 } }{ E _{ 2 } } = \frac{ I _{ 1 } }{ I _{ 2 } } E2=E1×I2I1E _{ 2 } = E _{ 1 } \times \frac{ I_{ 2 } }{ I _{ 1 } } E2=1.25×6335=2.25VE _{ 2 } = 1.25 \times \frac{ 63 }{ 35 } = 2.25 V

 

Question 3.13 :

A copper conductor has a number density of free electrons estimated in Example 3.1 of

8.5 × 10 28 m – 3. How long does an electron take to drift from one end of a wire 3.0 m

long to its other end? The area of the cross-section of the wire is 2.0 × 10 – 6 m 2 and it is

carrying a current of 3.0 A.

 Answer 3.13 :

Given that Number density of free electrons in a copper conductor , n =  8.5 x 10 28 m – 3

Let the Length of the copper wire be l

Given , l = 3.0 m

Let the area of cross – section of the wire be A = 2.0 x 10 – 6 m 2

Value of the current carried by the wire , I = 3.0 A  , which is given  by the equation ,

I = n A e V d

Where,

e = electric charge = 1.6 x 10 – 19 C

Vd=Driftvelocity=Lengthofthewire(l)timetakentocoverl(t)V _{ d } = Drift velocity = \frac{ Length of the wire \left ( l \right ) }{ time taken to cover l \left ( t \right )} I=nAeltI = n A e \frac{ l }{ t } t=n×A×e×lIt = \frac{ n \times A \times e \times l }{ I } t=3×8.5×1028×2×106x1.6×10193.0t = \frac{ 3 \times 8.5 \times 10 ^{ 28 } \times 2 \times 10 ^{ – 6 } x 1.6 \times 10 ^{ – 19 }}{ 3.0 } t=2.7×104sect = 2.7 \times 10 ^{ 4 } sec

 

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