Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 3 Current Electricity are provided here with simple step-by-step explanations. These solutions for Current Electricity are extremely popular among class 12 Science students for Physics Current Electricity Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 12 Science Physics Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

Page No 16:

Question 1:

Consider a current carrying wire (current I ) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
(a) source of emf.
(b) electric field produced by charges accumulated on the surface of wire.
(c) the charges just behind a given segment of wire which push them just the right way by repulsion.
(d) the charges ahead.


Current density is basically electric current per unit area i.e. j=IA
However, electric field and current density are also inter related by a formula,
This means that current density changes due to the electric field produced by charges get collected on the surface of wire.
Hence, the correct answer is option (b).

Page No 16:

Question 2:

Two batteries of emf ε1 and ε2 (ε2 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in the given figure.

(a) The equivalent emf
εeq of the two cells is between ε1 and ε2, i.e. ε1< εeq < ε2.
(b) The equivalent emf εeq is smaller than ε1 .
(c) The εeq is given by εeq = ε1+ ε2 always.
(d) εeq is independent of internal resistances r1 and r2.


For parallel combination, equivalent emf
εeqReq=ε1r1+ε2r2εeq=ε1r1+ε2r2×Reqεeq=ε1r1+ε2r2×r1r2r1+r2      Req=r1r2r1+r2  εeq=ε1r2+ε2r1r1+r2    

It is given that,


Hence, the correct answer is option (a).

Page No 17:

Question 3:

A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100Ω. He finds the null point at l1 = 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure l1 more accurately.
(b) He should change S to 1000Ω and repeat the experiment.
(c) He should change S to 3Ω and repeat the experiment.
(d) He should give up hope of a more accurate measurement with a meter bridge.


Given, standard resistance, S = 100 Ω
l1 = 2.9 cm
When meter bridge is balanced,



For improving accuracy, the null point should be in between the meter bridge wire which means resistance for  known and unknown resistance should be nearly equal


R = S = 3Ω
Thus, the student should change S to 3Ω and repeat the experiment.
Hence, the correct answer is option (c).

Page No 17:

Question 4:

Two cells of emf’s approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
(a) The battery that runs the potentiometer should have voltage of 8 V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.
(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.


In case of potentiometer experiment, emf of a cell can be measured, if along the potentiometer wire potential drop is more than the emf of the cell. Here, emf of two cells are 5V and 10 V, so potential drop across potentiometer wire should be more than 10 V.
Hence, the correct answer is option (b).

Page No 17:

Question 5:

A metal rod of length 10 cm and a rectangular cross-section of 1cm × 12 cm is connected to a battery across opposite faces. The resistance will be
(a) maximum when the battery is connected across 1 cm × 12 cm faces.
(b) maximum when the battery is connected across 10 cm × 1  cm faces.
(c) maximum when the battery is connected across 10 cm × 12 cm faces.
(d) same irrespective of the three faces.


Resistance of the wire is given as,
R ρlA
Here, ρ is resistivity
         l is length
        A is area of cross-section
For greater value of R, l must be maximum and A should be minimum and this is only possible when battery is connected across
 1cm×12 cm faces. 
Hence, the correct answer is option (a).

Page No 17:

Question 6:

Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone.
(b) Thermal velocity alone.
(c) Both drift velocity and thermal velocity.
(d) Neither drift nor thermal velocity.


Current in a conductor is given as
I = n Ae vd
Here, I is current
         n is number of electrons per unit volume
         A is area of cross-section
         vd is drift velocity
         e is charge of electron.
Hence, the correct answer is option (a).

Page No 18:

Question 7:

Kirchhoff ’s junction rule is a reflection of
(a) conservation of current density vector.
(b) conservation of charge.
(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) the fact that there is no accumulation of charges at a junction.


Kirchhoff's junction law states that the total current into junction will be equal to the total current going out of the junction
IinIout = 0
Kirchhoff's junction rule is based on the law of conservation of charge and the fact that there is no charge accumulation at a junction.
Hence, the correct options are (b) and (d).

Page No 18:

Question 8:

Consider a simple circuit shown in the given figure stands for a variable resistance R . R can vary from R0 to infinity. r is internal resistance of the battery (<< << R0).
(a) Potential drop across AB is nearly constant as R is varied.
(b) Current through Ris nearly a constant as R is varied.
(c) Current I depends sensitively on R
(d) IVr+R always.


Internal resistance = r
Variable resistance = R'
Here, r << R << R'
Applying Ohm's law, V = IR


Since, r << R << R'


Potential difference across A and B, VABI×RR'R+R'

=V R+R'rR+R'+RR'×RR'R+R'


Therefore, VAB  is constant if R' is varied.
Hence, the correct  options are (a) and (d).

Page No 18:

Question 9:

Temperature dependence of resistivity ρ(T) of semiconductors, insulators and metals is significantly based on the following factors:
(a) number of charge carriers can change with temperature T.
(b) time interval between two successive collisions can depend on T.
(c) length of material can be a function of T.
(d) mass of carriers is a function of T.


Resistivity ρT=mne2τ
When number of charge carriers changes with temperature T, then the resistivity will also change. Also, here τ is the relaxation time which is the time interval between two successive collisions, which decrease with increase in temperature.
Hence, the correct options are (a) and (b).

Page No 18:

Question 10:

The measurement of an unknown resistance R is to be carried out using Wheatstones bridge (see Fig. 3.25 of NCERT Book). Two students perform an experiment in two ways. The first students takes R2 = 10 Ω and R1 = 5 Ω. The other student takes R2 = 1000 Ω and R1 = 500 Ω. In the standard arm, both take R3 = 5 ΩBoth find R=R2R1R3=10  within errors.
(a) The errors of measurement of the two students are the same.
(b) Errors of measurement do depend on the accuracy with which R2 and R1 can be measured.
(c) If the student uses large values of R2 and R1, the currents through the arms will be feeble. This will make determination of null point accurately more difficult.
(d) Wheatstone bridge is a very accurate instrument and has no errors of measurement.


Given for first student,
R1 = 5 Ω
R2 = 10 Ω
R3 = 5 Ω
For second student,
R1 = 500 Ω
R2 = 1000 Ω
R3 = 5 Ω

Applying wheat-stone bridge rule,


For student 1,
 R=5×105=10 Ω

For student 2,
 R=5×1000500=10 Ω  
So, wheat-stone bridge is most sensitive and accurate if resistances are of same value.
When the values of R1 and R2 becomes large current through the arms become weak which can make the determination of null point accurately more difficult.
Hence, the correct options are (b) and (c).

Page No 19:

Question 11:

In a meter bridge the point D is a neutral point in the given figure.
(a) The meter bridge can have no other neutral point for this set of resistances.
(b) When the jockey contacts a point on meter wire left of D, current flows to B from the wire.
(c) When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
(d) When R is increased, the neutral point shifts to left.


When jockey is at a neutral point D, current will not flow through galvanometer which means potentials at B and D are equal. However, potentials at different point will be different. So, the point D is unique to get null point. Now, when jockey is shifted in right of D on wire, potential in wire towards the right becomes smaller. i.e., VB > VO.
So, current flows from B to D in the wire. Now, when R is increased, potential drop across R increases which results in in-case in potential at B to get null point.
​Hence, the correct options are (a) and (c).

Page No 19:

Question 12:

Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?


In an electric circuit, when an electron approaches a junction, in addition to the uniform electric field E that it normally faces (which keep the drift velocity dv fixed), there are accumulation of charges on the surface of wires at the junction. Due to which an electric field is produced Thus these electric fields alter direction of momentum.
Hence, when charge crosses a junction in an electric circuit the momentum is not conserved at the junction.

Page No 19:

Question 13:

The relaxation time τ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of ρ with temperature. Elaborate why?


Relaxation time,

τ α vdE
This means relaxation time depends upon velocity of electrons and ions and electric field. The applied electric field affects the velocities of electron at order of 1 mm/s which can be neglected and this also supports Ohm's law.
However, change in temperature affects velocities of electrons and ions at order 102 m/s, which is significant. 
Resistivity, ρ=mne2τ

Thus, τ can be affected to an extent and hence the resistivity.

Page No 19:

Question 14:

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate Runknown by any other method?


The necessary condition for wheatstone bridge is given as PQ=RS
Here, P and Q are ratio arms, R is known resistance and S is unknown resistance-advantage of the null-point method:-

1. Resistance of galvanometer does not affect the balance point. So, no need for determination of current in resistances and internal resistance of a galvanometer.
2. By calculating least count and readings of ammeter and voltmeter, R unknown resistance can be calculated by applying kirchhoff's law and Ohm's law.

Page No 19:

Question 15:

What is the advantage of using thick metallic strips to join wires in a potentiometer?


In potentiometer, thick metallic strips are used, as they will introduce least resistance while joining the wires. Being metallic strips of least resistance, they can be ignored while measuring the length of the wire which is used for determination of null point.

Page No 19:

Question 16:

For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?


For wiring in the home, Cu or Al wires are used the main considerations involved are:
(i) Good conductivity of metal.
(ii) Cost of metal used

Page No 19:

Question 17:

Why are alloys used for making standard resistance coils?


Alloys are used for making standard resistance coils, this is because they have:
(i) Low value of temperature coefficient of resistivity.
(ii) High resistivity.

Page No 19:

Question 18:

Power P is to be delivered to a device via transmission cables having resistance RC. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.


Power delivered,
P = Voltage × Current = VI


If Rc is the resistance of transmission lines then power wasted:
Pc = I2Rc

pc=p2V2 Rcpc α 1V2

Thus, in order to reduce Pc , V should be increased.
Hence, the power wastage can be reduced by transmitting power of high voltage.

Page No 19:

Question 19:

AB is a potentiometer wire in the given figure. If the value of R is increased, in which direction will the balance point J shift?


When R is increased, the current through the wire will decrease and hence the potential gradient will also decrease. Hence, the balance length. will increase. Thus point J will shift towards B to obtain the balance point.

Page No 20:

Question 20:

While doing an experiment with potentiometer in the given figure it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased. while the jockey was moved towards the end B.
(i) Which terminal +or –ve of the cell E1, is connected at X in case (i) and how is E1 related to E ?
(ii) Which terminal of the cell E1 is connected at X in case (ii)?


(i) Current in auxiliary circuit decreases whereas potential difference across A and jockey increases. This is possible when positive terminal of E1 is connected at X and E1 > E.
(ii) Current in auxiliary circuit increases whereas potential difference across A and jockey increases. This is possible when E1 negative terminal is connected at X.

Page No 20:

Question 21:

A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, verses R.


Consider a circuit in which battery of emf E is connected to resistance R and having internal resistances r.
Let V be the potential drop across resistance and I be the current flowing through the circuit.
Applying, Kirchhoff's voltage law,
EIRIr = 0
E = I(R + r
Potential drop across the battery, V = IR
As R becomes large R + r becomes approximately equal to R and VE.


Page No 20:

Question 22:

First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’?


When n resistances each of resistance R are connected in series and then in parallel.
For series combination:
Rs = R1 + R2 + . . . . + Rn = R + R + . . . . .n times
Rs = R
For parallel combination

1Rp=1R1+1R2+. . . . +1Rn=1R+1R+ . . . . +n timeRp=Rn
In series combination, current, I ER+nR

In parallel combination, current, 10 IER+Rn

Here n = –1 (not possible)
n = 10
Hence, there are 10 resistance in combination.

Page No 20:

Question 23:

Let there be n resistors R1 ............Rn with Rmax= max (R1......... Rn) and Rmin = min {R1 ..... Rn}. Show that when they are connected in parallel, the resultant resistance RP < Rmin and when they are connected in series, the resultant resistance RS > Rmax. Interpret the result physically.


Let Rmin and Rmax be the minimum and maximium resistance among R1, R2, . . . . ., Rn. When resistance are connected in parallel,

1Rp=1R1+1R2+. . . . +1Rn

On multiplying both sides by Rmin , we get

RminRp=RminR1+RminR2+. . . .  +RminRn

Here, R1, R2 . . . , Rn there must be a resistance which is minimum.
This means there must be a term Rmin in R.H.S. which is equal to 1 and the Rmin other terms are positive.

RminRp=RminR1+RminR2+. . . . +RminRn>1Rmin>Rp

So, in parallel combination, equivalent resistance Rp is always less than any smallest resistance in the combination.

When resistance are connected in series, then equivalent resistance
Rs = R1 + R2 + . . . + Rn
In R.H.S. there must be a term Rmax which has maximum value among R1. R2 . . , Rn.
Rs = R1 + R2 + . . + Rmax + . . + Rn = R++
Rs > Rmax
So, in series combination equivalent resistance is always greater than the maximum resistance available in combination of resistance.​

Page No 20:

Question 24:

The circuit in the given figure shows two cells connected in opposition to each other. Cell E1 is of emf 6 V and internal resistance 2 Ω; the cell E2 is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B.


emf of cell E1 = 6 V
Internal resistance, r1 = 2 Ω
emf of cell E2 = 4 V
Internal resistance, r2 = 8 Ω
Effective resistance, R = 2 + 8 = 10 Ω
Effective emf, E = 6 – 4 = 2 V
Using Ohm's law,  V = IR
I=6-42+8=15 A 

The direction of flow of current is from higher potential to lower potential.
Since, VB>VA
VAB=VB-VA             =4+15×8=4+1.6             =5.6 V
Hence, the potential difference between A and B is 5.6 V.

Page No 20:

Question 25:

Two cells of same emf E but internal resistance r1 and r2 are connected in series to an external resistor R in the given figure. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.


Emf of two cells = E
​Internal resistance = r1 and r2
External resitance = R
Terminal voltage across first cell = 0
Now effective emf,

Electric current, I = E-ER+r1+r2=0

Disclaimer: According to the circuit shown in the question, current through the circuit will be zero. So, the question/image seems incorrect.

Page No 21:

Question 26:

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB.


Diameter of solid wire A, d1 = 1 mm.
Diameter of outer solid hollow tube D1 = 2 mm
Diameter of inner solid hollow tube, D2 = 1 mm
Resistance,  R=ρlA
Resistance of first conductor, RAρlπd122

Resistance of second conductor, RB=ρlπD122-D222

Ratio of two resistors, RARB=10-32-0.5×10-320.5×10-32


Hence, the ratio of resistance RA to RB is 3 : 1.

Page No 21:

Question 27:

Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.


Let n be the number of batteries in series of the circuit and Reff be the effective internal resistance of n resistors ∊eff be the effective emf of n batteries in series and R be the external resistance in the circuit.

Current through R, I εeffReff+R

On increasing voltage and resistances n-fold

εeffnew=n εeffReffnew=n Reff

Rnew=nRInew=εeffnewReffnew+Rnew           =nεeffnReff+nR=nεeffnReff+R=εeffReff+R=IInew=I
Thus, the current remains unaltered when all voltage and all resistances are increased to n-folds.

Page No 21:

Question 28:

Two cells of voltage 10 V and 2 V and internal resistances 10 Ω and 5 Ω respectively, are connected in parallel with the positive end of 10 V battery connected to negative pole of 2 V battery in the given figure. Find the effective voltage and effective resistance of the combination.


Voltage of cells, V1 = 10 V and V2 = 2 V
   But, V = IR
Applying Kirchhoff's junction rule:
I2 = I1I
In bigger outer loop having 10 V cell and resistance R:
applying Ohm's law,
10 = IR + 10I1                                    .....(1)
Now, for outer loop having 2 V cell and resistance R:
2 = 5I2RI
⇒ 2 = 5(I1I) – RI                     .....(2)
Solving equation (1) and (2) we get:
6 = 3RI + 10I
⇒ 2 = IR+103                        .....(3)
Effective potential difference due to both batteries is,
Veff = I (R + Reff)                       .....(4)
On comparing equation (3) and (4), we get:
Veff = 2 V and Reff103 Ω.

Page No 21:

Question 29:

A room has AC run for 5 hours a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of Aluminium of the same dimensions?
[ρCu = 1.7 × 10-8 m ρAl = 2.7 × 10–8 Ωm]


Energy consumed per day, E = 10 kWh
Time for using energy per day to 5 h

Power consumed, P10 kWh5 h=2 kW=2000 JS

Current supplied,
I = PV=20002209 A

Power loss in copper wire,

PI2R=I2×ρlπr2                              as, R=ρlA=ρlπr2  

Pcu=92×1.7×10-8×10227×10-324  J/s.

Fractional loss of power in Cu wire, 

Similarly power loss in Al wire,

PAl=92×2.7×10-8×10227×10-32=6.9 J/s

Fractional loss of power in Al wire, 

Page No 21:

Question 30:

In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω in the given figure. A student wanting to measure voltage E1 of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.


Variable resistance, R = 50 Ω
Let R' be the potentiometer wire resistance.
Current supplying in primary circuit be I which have EB = 10 V
Here, IVBR+R'            Using Ohm's law


Potential difference across wire of potentiometer, V = IR'

V'=10 R'R+50

Null deflection can not be obtained by 8 V.
it is possible only when:

10 R'<8 R+4002 R'<400R'<200 Ω

Similarly, for R = 10 Ω, null point is possible when,

10 R'10+R'>8

2 R'>80R'>40 Ω

Null point is obtained on 4th segment or at 34 of the total length.
10×34R'10+R'<8         At balance point7.5 R'<80+8 R'R'<160R'>160

⇒ 160 < R' < 200
Any R' between 160 Ω and 200 Ω can achieve a null point null pint is on the 4th segment of potentiometer wire.
So, 400 cm of wire > 8 V is the potential drop.
Potential gradient (k),
k × 400 cm > 8 V
 ⇒ k × 4 m > 8 V
k > 2 V/m         .....(1)
Similarly, as balance point is at 4th segment of the wire, so, no balance point at 3 m.
k × 300 cm < 8 V = k × 3 m < 8 V
k83  V/m      .....(2)
By using equation (1) and (2), we get:

​83  V/m > k > 2  V/m

Page No 21:

Question 31:

(a) Consider circuit in in the given figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

(b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 1029m31029/m3, length of circuit = 10 cm, cross-section = A = (1 mm)2


Area of cross-section, A = 1 mm= 10– 6 m2
Resistance, R = 6 Ω
Voltage, V = 6 V


Applying Ohm's law, V = IR
                              ⇒ 6 = I (6)
                              ⇒ I = 1 A
a. Current, I = Ane Vd
           ⇒ VdIAne
           ⇒ Vd = 110-6×1029×1.6×10-19
          ⇒ Vd = 10-41.6  m/s
Number off electrons in wire = n (volume of the wire)
                                               = n × Al
Here, energy is absorbed in the form of kinetic energy.

E=12 mvd2 Anl



b. Power loss, P = I2R = (1)2 × 6 = 6 j/s
Energy, E = P × t
         t=EP=2×10-1763=0.33×10-17 sec
Hence, the time scale is approximately 3 × 10–18 sec.

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