# NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for class 12 physics chapter 3 Current Electricity- Current is the flow of electric charge through a given area. If we want to utilize the charge flow we must have a closed path. Solutions of NCERT class 12 physics chapter 3 current electricity explains questions about the factors affecting the current flow, how current is flowing in a conductor, basic laws related to current electricity and circuit analysis and a few measuring devices. The solutions of NCERT  will help you boost your knowledge. CBSE NCERT solutions for class 12 physics chapter 3 current electricity explains questions related to Ohm’s law and Kirchoff's law which will help in solving circuit related problems in exams.  The chapter also gives an idea about ammeter, voltmeter, potentiometer, whetstones bridge and meter bridge. The NCERT solutions for class 12 physics chapter 3 current electricity also explains problems related to heat developed in a resistor.

## NCERT solutions for class 12 physics chapter 3 Current Electricity Exercise:

Given,

Emf of battery, E = 12 V

Internal resistance of battery, r = 0.4 $\Omega$

Let I be the maximum current drawn from the battery.

We know, according to Ohm's law:

E = Ir

I = E/r = 12/0.4

$\implies$I = 30 A

Hence the maximum current drawn from the battery is 30 A.

Given,

Emf of the battery, E = 10 V

The internal resistance of the battery, r = 3 $\Omega$

Current in the circuit, I = 0.5 A

Let R be the resistance of the resistor.

Therefore, according to Ohm's law:

E = IR' = I(R + r)

10 = 0.5(R + 3)

R = 20 - 3 = 17 $\Omega$

Also,

V = IR (Across the resistor)

= 0.5 x 17 = 8.5 V

Hence, terminal voltage across the resistor = 8.5 V

We know that when resistors are combined in series, the effective resistance is the sum of that resistance.

Hence, total resistance of the three resistance combination = 1 + 2 + 3 = 6 $\Omega$

Since the resistances are in series, the current through each one of them will be equal to the current through the circuit but voltage/ potential drop will be different.

Total resistance, R = 6 $\Omega$

Emf, V = 12 V

According to Ohm's law:

V = IR

$\dpi{80} \implies$12 = I x  6

$\dpi{80} \implies$ I = 12/6 = 2 A

Now, using the same relation, voltage through resistors:

1$\Omega$ : V(1) = 2 x 1 = 2V

2$\Omega$ : V(2) = 2 x 2 = 4V

3$\Omega$ : V(3) = 2 x 3 = 6V

(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )

We know that when resistances are in parallel combination, the total resistance R is given by:

$\frac{1}{R} = \sum \frac{1}{R_{i}}$

Therefore, total resistance of the given three resistances in parallel combination is

$\frac{1}{R} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$\implies \frac{1}{R} = \frac{1}{2}+\frac{1}{4}+\frac{1}{5} = \frac{10+5+4}{20}=\frac{19}{20}$

$\implies R= \frac{20}{19} = 1.05 \Omega$

Hence, the total resistance is 1.05 $\dpi{80} \Omega$

Since the resistances are in parallel, the voltage across each one of them will be equal.

Emf, V = 20 V

According to Ohm's law:

V = IR $\dpi{80} \implies$ I = V/R

Therefore, current across each one of them is:

2$\dpi{80} \Omega$:  I = 20/2 = 10 A

4$\dpi{80} \Omega$:  I = 20/4 = 5 A

5$\dpi{80} \Omega$:  I = 20/5 = 4 A

Given,

temperature coefficient of filament, $\alpha$ = $1.70\times 10^{-4}\degree C^{-1}$

$\dpi{100} T_1= 27\degree C$ ;  $\dpi{100} R_1 = 100 \Omega$

Let $\dpi{100} T_2$ be the temperature of element, $\dpi{100} R_2= 117\Omega$

(Positive $\alpha$ means that the resistance increases with temperature. Hence we can deduce that $\dpi{100} T_2$ will be greater than  $\dpi{100} T_1$)

We know,

$\dpi{100} R_2 = R_1[1 + \alpha \Delta T]$

$\dpi{80} \implies$ $117 = 100[1 +(1.70\times 10^{-4})(T_2 - 27) ]$

$\\ \Rightarrow T_2-27=\frac{117-100}{1.7\times10^{-4}}\\\Rightarrow T_2-27=1000\\\Rightarrow T_2=1027^\circ C$

Hence, the temperature of the element is 1027 °C.

Given,

Length of the wire, l = 15 m

The cross-sectional area of the wire, A = $6.0\times 10^{-7}\;m^2$

The resistance of the wire, R = 5 Ω

We know,

$R = \rho l/ A$ , where $\rho$ is the resistivity of the material

$\dpi{100} \implies \rho = RA/l = 5 \times6\times10^{-7}/15$

$\dpi{100} \implies \rho = 2\times10^{-7}$

Hence, the resistivity of the material of wire is $\dpi{100} \rho = 2\times10^{-7}\ m$

Given,

$\dpi{100} T_1= 27.5 \degree C$ ;  $\dpi{100} R_1 = 2.1\Omega$

$\dpi{100} T_2= 100\degree C$ ;  $\dpi{100} R_2= 2.7\Omega$

We know,

$\dpi{100} R_2 = R_1[1 + \alpha \Delta T]$

2.7 = 2.1[1 + $\alpha$ (100 -  27.5) ]

$\alpha$  = (2.7 - 2.1) / 2.1(100 -27.5)

$\alpha$ = 0.0039 $\dpi{80} \degree C^{-1}$

Hence, the temperature coefficient of silver wire is 0.0039 $\dpi{80} \degree C^{-1}$

For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.

Given,

Voltage, V = 230 V

$I_1= 3.2 A$  and $I_2 = 2.8 A$

Using Ohm's law:

$R_1 = 230/3.2 = 71.87\Omega$

and

$R_2= 230/2.8 = 82.14 \Omega$

Now, the temperature coefficient of filament,

$\alpha$ = $1.70\times 10^{-4}\degree C^{-1}$

$\dpi{100} T_1= 27\degree C$

Let $\dpi{100} T_2$ be the steady temperature of the heating element.

We know,

$\dpi{100} R_2 = R_1[1 + \alpha \Delta T]$

$\dpi{80} \implies$ 230/2.8 = 230/3.2[1 + ($\dpi{80} 1.70\times 10^{-4}$) ( T2 -  27) ]

$\dpi{80} \implies$ 3.2/2.8 = 1 + ($\dpi{80} 1.70\times 10^{-4}$) ( T2 -  27)

$\dpi{80} \implies$ 1.14 - 1 =  ($\dpi{80} 1.70\times 10^{-4}$) ( T2 -  27)

$\dpi{80} \implies$ T2 -  27 = 0.14 / ($\dpi{80} 1.70\times 10^{-4}$)

$\dpi{80} \implies$ T2  = (840.5 + 27) °C

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

## 3.9 Determine the current in each branch of the network shown in Fig. 3.30:

Let current in the circuit is distributed like

where I1, I2, and I3 are the different current through shown branches.

Now, applying KVL in Loop

$10-I10-I_25-(I_2+I_3)10=0$

Also, we have $I=I_1+I_2$

so putting it in kvl equation

$10-(I_1+I_2)10-I_25-(I_2+I_3)10=0$

$10-10I_1-10I_2-5I_2-10I_2-10I_3=0$

$10-10I_1-25I_2-10I_3=0$.................................(1)

Now let's apply kvl in the loop involving I1 I2 AND I3

$5I_2-10I_1-5I_3=0$.................................(2)

now, the third equation of KVL

$5I_3-5(I_1-I_3)+10(I_2+I_3)=0$

$-5I_1+10I_2+20I_3=0$..............................(3)

Now we have 3 equation and 3 variable, on solving we get

$I_1= \frac{4}{17}A$

$I_2= \frac{6}{17}A$

$I_3= \frac{-2}{17}A$

Now the total current

$I=I_1+I_2=\frac{4}{17}+\frac{6}{17}=\frac{10}{17}$

Balance point from the end A, l1 = 39.5 cm

Resistance of Y = 12.5 Ω

We know, for a meter bridge, balance condition is:

$\frac{X}{Y} = \frac{l_1}{l_2} = \frac{l_1}{100 - l_1}$

$\implies X = \frac{39.5}{100 - 39.5}\times12.5 = 8.2\Omega$

The connections between resistors in a Wheatstone or meter bridge made of thick copper strips to minimise the resistance of the connection which is not accounted for in the bridge formula.

(b) Determine the balance point of the bridge above if X and Y are interchanged.

If X and Y are interchanged.

Then, X = 12.5 Ω , Y = 8.2 Ω

We know, for a meter bridge, balance condition is:

$\frac{X}{Y} = \frac{l_1}{l_2} = \frac{l_1}{100 - l_1}$

$\\ \implies \frac{12.5}{8.16} = \frac{l_1}{100 - l_1} \\ \\ \implies 1.53(100 - l_1) = l_1 \\ \implies 2.53l_1 = 153$

$\therefore l_1 =60.5 cm (from \ point\ A)$

if the galvanometer and cell are interchanged the galvanometer will show no current and hence no deflection.

Given,

Emf of battery, E = 8 V

Internal resistance of battery, r = 0.5 Ω

Supply Voltage, V = 120 V

The resistance of the resistor, R = 15.5 Ω

Let V' be the effective voltage in the circuit.

Now, V' = V - E

V' = 120 - 8  = 112 V

Now, current flowing in the circuit is:

I = V' / (R + r)

$\implies I = \frac{112}{15.5 + 0.5} = 7 A$

Now, using Ohm ’s Law:

Voltage across resistor R is v = IR

v = 7 x 15.5 = 108.5 V

Now, the voltage supplied, V = Terminal voltage of battery + v

$\therefore$ Terminal voltage of battery = 120 -108.5  = 11.5 V

The purpose of having a series resistor is to limit the current drawn from the supply.

Given,

$\dpi{100} E_1$ = 1.25 V ,  $\dpi{100} l_1$ = 35 cm

And, $\dpi{100} l_2$ = 63 cm

Let  $\dpi{100} E_2$ be the voltage in the second case.

Now, the balance condition is given by :

$\dpi{100} \frac{E_1}{E_2} = \frac{l_1}{l_2}$

$\dpi{100} \implies E_2 = \frac{l_2}{l_1}\times E_1 = \frac{63}{35}\times1.25$

$\dpi{100} \implies E_2 = 2.25 V$

Therefore, the emf of the second cell = 2.25 V

We know,

$I = neAv_d$

$\dpi{90} V_d$:drift Velocity = length of wire(l) / time taken to cover

$I = neA \frac{l}{t}$

by substituting the given values

$\dpi{90} \implies$ t = 2.7 x $\dpi{90} 10^4$ s

Therefore, the time required by an electron to drift from one end of a wire to its other end is  $2.7\times 10^4$ s.

## NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity Additional Exercises:

Given,

The surface charge density of earth $\rho$ = $\dpi{100} 10^{-9}C m^{-2}$

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x $\dpi{100} 10^6$ m

$\dpi{100} \therefore$ The surface area of earth A = $\dpi{100} 4\pi r^2$

=  $\dpi{80} 4\pi (6.37\times10^6 )^2$   = $\dpi{100} 5.09 \times 10^{14} m^2$

Now, charge on the earth surface,   $q=\rho\times A$

Therefore,

$q=\dpi{100} 5.09 \times 10^{5} C$

Let the time taken to neutralize earth surface be  t

$\dpi{100} \therefore$ Current I = q / t

$\dpi{100} \implies$  t = 282.78 s.

Therefore, time take to neutralize the Earth's surface is 282.78 s

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

$\dpi{100} I = \frac{nE}{R + nr}$

$\dpi{100} \implies I = \frac{6(2)}{8.5 + 6(0.015)} = \frac{12}{8.59}$

$\dpi{100} \implies$ I = 1.4 A

Hence current drawn from the supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

Given,

Emf , E = 1.9 V

Internal resistance, r =380 Ω

The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A

The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.

We know,

R = $\dpi{100} \rho$l / A

The wires have the same resistance and also are of the same length.

Hence,

$\dpi{100} \frac{\rho_{Al}}{A_{Al}} = \frac{\rho_{Cu}}{A_{Cu}}$

$\dpi{100} \implies \frac{A_{Al}}{A_{Cu}} = \frac{\rho_{Al}}{\rho_{Cu}} = \frac{2.63}{1.73}$

Now, mass = Density x Volume = Density x Area x length

Taking the ratio of their masses for the same length

$\dpi{100} \implies \frac{m_{Al}}{m_{Cu}} = \frac{d_{AL}A_{Al}}{d_{Cu}A_{Cu}} = \frac{2.7\times2.63}{8.9\times1.73} < 1$

Hence, $\dpi{100} m_{Al} < m_{Cu}$

Therefore, for the same resistance and length, the aluminium wire is lighter.

Since aluminium wire is lighter, it is used as power cables.

Current

A

Voltage

V

Current

A

Voltage

V

0.2 3.94 3.0 59.2
0.4 7.87 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.2
2.0 39.4 8.0 158.0

The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.

Hence the resistor made of alloy manganin follows Ohm's law.

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

No. Ohm’s law is not universally applicable for all conducting elements.

A semiconductor diode is such an example.

Ohm's law states that: V = I x R

Hence for a low voltage V, resistance R must be very low for a high value of current.

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of $\dpi{80} \boldsymbol{10^{22}}$ .

3.20 (a)

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

3.20 (a)

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

3.20 (a)

The ratio is nR/(R/n) = $n^2$

3.20 (b)

We have, equivalent resistance = 11/3

Let's break this algebraically so that we can represent it in terms of 1, 2 and 3

$\frac{11}{3}=\frac{9+2}{3}=3+\frac{2}{3}=3+\frac{1*2}{1+2}$

this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

3.20 (b)

Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it

Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1

R = 11/5 Ω

3.20 (b)

1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.

3.20 (b)

Connect all three resistors in parallel.

Equivalent resistance is  R =  1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)

R = 6/11 Ω

3.20 (c)

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,

so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is

$Equivalent\ R_{loop}=\frac{2*4}{2+4}=\frac{8}{6}=\frac{4}{3}$

now we have 4 such loops in series so,

$Total\ Equivalent\ R_{loop}=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=\frac{16}{3}$

Hence equivalent resistance of the circuit is 16/3 ohm.

3.20 (c)

It can be seen that all 5 resistors are in series, so

Equivalent Resistance = R + R + R + R + R = 5R

Hence equivalent resistance is 5R.

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

Here from the figure, We can consider  the box as a resistance of R'

Now, we can write,

equivalent resistance = R'

' =[( R')Parallel with (1)] + 1 + 1

$\frac{R'*1}{R'+1}+2=R'$

$R'+2R'+2=R'^2+R'$

$R'^2-2R'-2=0$

$R'=1+\sqrt{3},or1-\sqrt{3}$

Since resistance can never be negative we accept

$R'=1+\sqrt{3}$

, We have calculated the equivalent resistance of infinite network,

Now

Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network

= 0.5+1+1.73

=3.23 ohm

$V=IR$

$I=\frac{V}{R}=\frac{12}{3.23}=3.72A$

Hence current drawn from the 12V battery is 3.72 Ampere.

(a) What is the value of $\epsilon$

Given

maintained  constant emf of standard cell  = 1.02V, balanced point of this cell = 67.3cm

Now when the standard cell is replaced by another cell with emf = $\varepsilon$. balanced point for this cell = 82.3cm

Now as we know the relation

$\frac{\varepsilon}{l} =\frac{E}{L}$

$\varepsilon =\frac{E}{L}*l=\frac{1.02}{67.3}*82.3=1.247V$

hence emf of another cell is 1.247V.

(b) What purpose does the high resistance of 600 k$\Omega$ have?

If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k$\Omega$.

(c) Is the balance point affected by this high resistance?

No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, their won't be any balance point in the wire

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.

Given,

the balance point of cell in open circuit = $l_1=76.3cm$

value of external resistance added = $R=9.5\Omega$

new balance point = $l_2=64.8cm$

let the internal resistance of the cell be $r$.

Now as we know,  in a potentiometer,

$r=\frac{l_1-l_2}{l_2}*R$

$r=\frac{76.3-64.8}{64.8}*9.5=1.68\Omega$

hence the internal resistance of the cell will be 1.68 $\Omega$

## NCERT Solutions for Class 12 Physics - Chapter Wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Subject wise solutions-

## Importance of solutions of NCERT for class 12 chapter 3 current electricity in board exams:

As far as the CBSE board exam and competitive exams are considered the solutions of class 12 NCERT chapter 3 current electricity is important. In NEET and JEE Main previous year papers 7 to 10% questions are asked from this chapter.  In 2019 CBSE board exam 6 marks questions are asked from current electricity. Learn CBSE NCERT solutions for class 12 physics chapter 3 current electricity for a better score in board and competitive exams.

• Q.

### How many questions are asked from Current Electricity in NEET exam?

You can get 3 to 4 questions from current electricity in NEET Exam. This chapter is very important for NEET exam. If we analyse previous year question papers, on an average 7 to 9 % questions are asked from this chapter. To study this chapter NCERT along with previous year questions will help you. The chapter is simple and easily you can solve the questions if you had enough practice.

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