NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

# NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

Edited By Safeer PP | Updated on Sep 14, 2022 12:28 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 3 is from the point of view of exams. It also gives students an insight on how current flows in working of different types of devices in real life such as conductors. The previous chapters dealt with charges that were considered as at rest, whereas this chapter introduces various concepts relating to steady currents. NCERT Exemplar Class 12 Physics chapter 3 solutions includes the idea of electric energy and power, electric resistivity, and conductivity, along with a detailed explanation of carbon resistors and their colour codes. The concepts discussed in the NCERT Exemplar Class 12 Physics solutions chapter 3 are important from the point of view of board exams as the JEE Main exam. The students can also make use of NCERT Exemplar Class 12 Physics solutions chapter 3 PDF download, for further learning.

Also see - NCERT Solutions for Class 12 Physics

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQI

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.

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b)
Current density (J) is the current per unit area and is given by $\overrightarrow{J}=\sigma\overrightarrow{ E}$, where conductivity $\left (\sigma =\frac{l}{AR} \right )$ and electric field $\left ( \overrightarrow{E} \right )$.

Question:2

Two batteries of emf $\varepsilon _{1}$ and $\varepsilon _{2}\left ( \varepsilon_ {2}>\varepsilon _{1} \right )$and internal resistances r1 and r2 respectively are connected in parallel as shown in Figure.

A. The equivalent emf $\varepsilon _{eq}$ of the two cells is between $\varepsilon _{1}$ and $\varepsilon _{2}$, i.e. $\varepsilon _{1}<\varepsilon _{eq}<\varepsilon _{2}$
B. The equivalent emf $\varepsilon _{eq}$ is smaller than 1.
C. The $\varepsilon _{eq}$ is given by $\varepsilon _{eq}=1+2$always.
D. $\varepsilon _{eq}$ is independent of internal resistances r1 and r2.

The correct answer is the option (a)
Explanation: Equivalent EMF in parallel is given as:
$\varepsilon _{eq}=\frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}}$
This value will always lie between $\varepsilon _{1}$ and $\varepsilon _{2}$

Question:3

A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100. He finds the null point at $l_1 = 2.9 cm$. 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 1000and repeat the experiment.
C. He should change S to 3 and repeat the experiment.
D. He should give up hope of a more accurate measurement with a meter bridge.

The correct answer is the option (c)
Explanation:
$R=S\left (\frac{l_1}{100-l_1} \right )=100\left (\frac{2.9}{97.1} \right )=2.98\Omega$
He should aim to get the balance point near the centre (l1=50 cm)
$2.98\Omega =S\left (\frac{50}{50} \right )$
$S\cong 3\Omega$

Question:4

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

The correct answer is the option (b)
Explanation: The potential drop along the wires of the potentiometer should be greater than the emfs of cells. The battery of potentiometer must be in excess of 10V.

Question:5

A metal rod of length 10 cm and a rectangular cross-section of $1cm \times \frac{1}{2} cm$ is connected to a battery across opposite faces. The resistance will be
A. maximum when the battery is connected across $1cm \times \frac{1}{2} cm$ faces.
B. maximum when the battery is connected across $10cm \times 1 cm$ faces.
C. maximum when the battery is connected across $10cm \times \frac{1}{2} cm$faces.
D. same irrespective of the three faces.

The correct answer is the option (a)
Explanation:$R=\rho\frac{l}{A}$
Resistance is directly proportional to length and inversely proportional to the cross-sectional area. Resistance will be maximum when l is maximum and A minimum. Both these conditions are met simultaneously when the battery is connected across the $1cm \times \frac{1}{2} cm$ faces.

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

The correct answer is the option (a)
$I=Anev_d$

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 MCQII

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.

The correct answers are the options (b,d)
Explanation: According to the law of conservation of charges, the charge of a system will stay constant. From this, it can be derived that net charge flowing into a junction is equal to the net charge going out of the system.

Question:8

Consider a simple circuit shown in Figure variable resistance R’. R’ can vary from $R_0$ to infinity. r is internal resistance of the battery $(r<

A. Potential drop across AB is nearly constant as R’ is varied.
B. Current through R is nearly a constant as R’ is varied.
C. Current I depends sensitively on R’.
D.$I\geqslant \frac{V}{r+R}$ always.

The correct answers are the options (a,d)
Explanation:
$I=\frac{V}{r+R_{eq}}$
As $r+R
$I=\frac{V}{r+R_{eq}}\leq \frac{V}{r+R}$

Question:9

Temperature dependence of resistivity $\rho (T)$ 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.

The correct answers are the options (a,b)
Explanation: Resistivity is a function of relaxation time$(\tau )$ and mass of charge carrier (m). Mass of charge carrier is independent of temperature, whereas length also does not fluctuate significantly over a temperature range.

Question:10

The measurement of an unknown resistance R is to be carried out using Wheat stones bridge (see Figure of NCERT Book). Two students perform an experiment in two ways. The first students take $R_2 = 10\Omega$ and $R_1 = 5\Omega$. The other student takes $R_2 = 1000\Omega$and $R_1 = 500\Omega$. In the standard arm, both take $R_3= 5\Omega$. Both find $R=\frac{R_{2}}{R_{1}}R_{3}=10\Omega$ 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 $R_{2}$ and $R_{1}$ can be measured.
C. If the student uses large values of $R_{2}$ and $R_{1}$, 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.

The correct answers are the options (b,c)
Explanation: Using large values of $R_{1}$ and $R_{2}$ , will reduce the current flowing through the system, making it difficult to get accurate readings from the galvanometer.

Question:11

In a meter bridge the point D is a neutral point (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.

The correct answers are the options (a,c)
Explanation: No current will flow through the Galvanometer when the jockey is placed at D. Potential Difference between B and D is zero. If the jockey is placed to the right of D, Potential at B becomes greater than the potential at D, the current flows from B to D.

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 Very Short Answer

Question:12

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

As drift velocity is proportional to the electric field, when charge crosses a junction the momentum is not conserved.

Question:14

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

As no electricity will pass through the galvanometer of the Wheatstone bridge at the null point. It can be used to calculate the resistance of an unknown resistor.

Question:15

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

Thick metallic strips have large cross section and thus offer less resistance

Question:16

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

The following considerations are made to decide on the wiring:
• Conductivity of metal
• Cost
• Availability

Question:17

Why are alloys used for making standard resistance coils?

Because of their low temperature sensitivity and low temperature coefficient of resistance, alloys are used to make standard resistance coils. Also, they have a lower conductivity or higher resistivity, this reduces the length required to make coils.

Question:18

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

P (Power transmitted)=VI
H (Heat Loss)=$I^{2}R$
High current flowing through the devices will lead to higher Energy Loss (as heat), so power should be transmitted at a high voltage and low current.

Question:19

On increasing the applied resistance, the potential difference across AB will decrease. To balance the potential across the other battery, the length AJ' should increase. Balance point shift towards B.

Question:20

While doing an experiment with potentiometer (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$E_{1}$, is connected at X in case (i) and how is $E_{1}$ related to $E$?
(ii) Which terminal of the cell $E_{1}$ is connected at X in case (ii)?

i) As jockey moves closer to B, the galvanometer potential difference decreases. This requires the positive terminal of $E_{1}$ to be at X.
ii) Exactly opposite to (i), at X $E_{1}^{'}$s negative terminal is connected.

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.

$I=\frac{E}{R+r} and V=IR$
$V=\frac{ER}{R+r}=\frac{E}{1+\frac{r}{}R}$
An increase in R leads to an increase in V, till the point it approaches E
Following is the graphical representation:

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 Short Answer

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’?

$R_{Series}=nR$
$I_{Series}=\frac{E}{R+nR}=\frac{E}{(n+1)R}$
$R_{Parallel}=\frac{R}{}n$
$I_{Parallel}=\frac{E}{R+\frac{R}{}n}=\frac{E}{}R\times \frac{n}{n+1}$
$I_{Parallel}=10\times I_{Series}$
$\frac{E}{}R\times \frac{n}{n+1}=10\times \frac{E}{}R\times \frac{1}{n+1}$
$n=10$

Question:23

Let there be n resistors $R_1 ............R_n$ with $R_{max} = max (R_1......... R_n)$ and $R_{min} = min (R_1 ..... R_n)$. Show that when they are connected in parallel, the resultant resistance RP< Rmin and when they are connected in series, the resultant resistance $R_S > R_{max}$. Interpret the result physically.

$\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_{2}}+.....+\frac{1}{R_n}$
$\frac{R_{min}}{R_{parallel}}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}....+\frac{R_{min}}{R_n}$
$\frac{R_{min}}{R_{parallel}}> 1$
$R_{series}= R_1+R_2...+R_n$
$R_{series}>R_{max}$
This can be represented physically as shown below:

Question:24

Equivalent EMF of the Cells $=6V-4V=2V$
$R_{eq}=2\Omega +8\Omega=10\Omega$
$I=\frac{2V}{10 \Omega}=0.2A$
$V_B-V_A=4+8\times 0.2=5.6V$

Question:25

$EMF_{ef}=E+E=2E$
$R_{ef}=r_1+r_2$
$I=\frac{2E}{R+r_1+r_2}$
As net potential across the cell is 0,
$E-Ir_1=0$
$E-\frac{2Er_1}{R+r_1+r_2}=0$
$R=r_1-r_2$

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 2mm and inner diameter 1mm. Find the ratio of resistance to

$Resistance(R)=\rho\frac{L}{A}$
$R_A=\rho _A\left (\frac{L_A}{A_A} \right ) and R_B=\rho _B\left (\frac{L_B}{A_B} \right )$
$\frac{R_A}{R_B}=\frac{A_B}{A_A}=\frac{\pi r_A^2}{\pi (r_{out}^2-r_{in}^2)}=\frac{1^2-\left (\frac{1}{2} \right )^{2}}{\left ( \frac{1}{2} \right )^{2}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3$

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.

Original:
$EMF = V_1$
$Resistance = R + R_{eff}$
$Current = \frac{V_1}{R+R_{eff}}$
New:
$EMF = nV_1$
$Resistance =n( R + R_{eff})$
$Current =\frac{n(V_1)}{n(R+R_{eff})}= \frac{V_1}{R+R_{eff}}$
Increasing the EMF and Resistance in the same ratio (say 2), will not change the current flowing through the system.

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 Long Answer

Question:28

Kirchoff's Junction Law at $C: I_1=I_2+I_3$
Kirchoff's Loop Law for efbae:$10-I_1\times 10-I -3\times R=0$
$10-10I_2-(R+10)I_3=0$
Kirchoff's Loop Law for cbadc:- $2+5I_2-I_3R=0$
$(10-10I_2-(R+10)I_3)+(-2+5I_2-I_3R)\times 2=0$
$6-I_3R+10-2I_3R=0$
$6=3I_3R+10I_3$
$I_3=\frac{6-3I_3R}{10}$
Veff & Reff will fulfil the following:
$V_{eff}-I_3(R+R_{eff})=0$
$R_{eff}=\frac{1}{\frac{1}{10}+\frac{1}{15}}=\frac{10}{}3$
$V_{eff}-I_3\left (R+\frac{10}{}3 \right )=0$
$V_{eff}=I_3R+I_3\times \frac{10}{}3$
$V_{eff}=I_3R+\frac{6-3I_3R}{10}\times \frac{10}{}3=I_3R+6-\frac{3I_3R}{}3=2V$

Question:29

A room has AC run for 5 hours a day at a voltage of 220V. 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?

$[\rho_ {Cu} = 1.7 \times -810\Omega m, \rho _{Al} = 2.7 \times 10^{-8} \Omega m]$

Energy consumed day=10 kWh
Energy consumed hour=2 kWh
Power=2kW
P=VI
V=220V
$I=\frac{P}{}V=\frac{2000}{220}=\frac{100}{11}$
Heat $=I^2\times R_{Cu}$
$R_{Cu}=\rho\left ( \frac{L}{}A \right )=1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )$
$Heat=\left (\frac{100}{110} \right )^{2}\times 1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )=4.4 W$
fractional loss in heating
$=\frac{4.4}{2000}\times 100 =0.22\%$
$\frac{Heat_{Al}}{Heat_{Cu}}=\frac{I^2RAl}{I^2R_{Cu}}=\frac{\rho _{Al}}{\rho _{Cu }}$
HeatAl=7W
fractional loss in heating
$=\frac{7}{2000}\times1000=0.35\%$

Question:30

In an experiment with a potentiometer, $V_{B}=10V$ R is adjusted to be $50 \Omega$ (Figure). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to $10 \Omega$ 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.

Req=50+R
Veq=10V
$I=\frac{10}{50+R'}$
The potential difference across the potentiometer
$V'=IR'=\frac{10R'}{50+R'}$
V'<8
$10R'<8(50+R' )$
R'<200
$\frac{10R'}{10+R'}>8$
R'>40
At balance point:
$\frac{10\times \frac{3}{}4R'}{10+R'}<8$
R'>-160
As R’ can only be positive, R'>160
$k\times 400cm>8V$
k>2 V/m
As balance, point isn’t at 3m
$K\times 3<8$
$k<2\frac{2}{}3\frac{V}{}m$

Question:31

(a) Consider the circuit in 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 $RI_2$ per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = $10^{29}/m^3$, length of circuit = 10 cm, cross-section = A $= (1mm)^2$

a)
$A=10^{-6}m^2$
$R=6\Omega$
$n=10^{29} m^{-3}$
$e=1.6\times 10^{-19}C$
$I=\frac{V}{}R=1A$
$m_e=9.1\times10^{-31}Kg$
$L=10^{-1}m$
a) $I=Anev_d$
$v_d=\frac{I}{Ane}=6.25\times 10^{-5}m/s$
Kinetic energy=
$\frac{1}{}2\times m_e\times v_d^2=2\times 10^{-17}J$
b) Power loss=
$I^2R=6W$
$P=\frac{E}{}t$
$t=\frac{E}{}P=10^{-17}s$

## NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity Main Subtopics Covered

• Introduction
• Electric Current
• Electric Currents in Conductors
• OHM’s Law
• Drift of Electrons and the Origin of Resistivity
• Limitations of OHM’s Law
• Resistivity of Various Materials
• Temperature Dependence of Resistivity
• Electrical Energy, Power
• Combination of Resistors- Series and Parallel
• Cells, EMF, Internal Resistance
• Cells in Series and in Parallel
• Kirchhoff’s Rules
• Wheatstone Bridge
• Meter Bridge
• Potentiometer
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## NCERT Exemplar Class 12 Physics Chapter Wise Links

 Chapter 1 Electric Charges and Fields Chapter 2 Electrostatic Potential and Capacitance Chapter 4 Moving Charges and Magnetism Chapter 5 Magnetism and Matter Chapter 6 Electromagnetic Induction Chapter 7 Alternating Current Chapter 8 Electromagnetic Waves Chapter 9 Ray Optics and Optical Instruments Chapter 10 Wave Optics Chapter 11 Dual Nature of Radiation and Matter Chapter 12 Atoms Chapter 13 Nuclei Chapter 14 Semiconductor Electronics Chapter 15 Communication Systems

## Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

· NCERT Exemplar solutions for Class 12 Physics chapter 3 introduces new principles/laws as OHM’s law, Kirchhoff’s rule, etc., which display and govern the flow of current in various devices, and how it can be determined with the help of other factors.

· Students will study different circuits in which a number of resistors and cells are interconnected for which different calculations need to be done to obtain different values of current, resistance, etc., with the help of Class 12 Physics NCERT Exemplar solutions chapter 3

· NCERT Exemplar Class 12 Physics solutions chapter 3 also introduces different devices such as potentiometer, meter bridge and exhibits the methods, and formulas for understanding current, resistance, etc., through these devices.

### NCERT Exemplar Class 12 Solutions

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1. Is this chapter important from exam POV?

Yes, current electricity is one of the most crucial chapters that one needs to focus on for getting good marks in boards and entrance exams.

2. How many questions are solved in this chapter?

Our teachers have solved all questions from NCERT exemplar Class 12 Physics chapter 3 current electricity with complete steps and details.

3. How to use these solutions to the best?

One can make use of these solutions to understand the questions to the best and also understand how to solve questions in exams.

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Good Luck

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9