Current Electricity Class 12 - NCERT Solutions with Important Formulas, Questions and Free PDF

3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM’s law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm’s law

I = $\frac{E}{(R+r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $\times 17=8.5V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

3.3 (a) The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

(b) From Ohm’s law, I = $\frac{V}{R}$ we get I = $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I $\times R$ = 2 $\times 1$ = 2 V

Potential drop across 2 Ω resistor = I $\times R$ = 2 $\times 2$ = 4 V

Potential drop across 3 Ω resistor = I $\times R$ = 2 $\times 3$ = 6 V

3.4 (a) Let $R_1$ = 2 Ω, $R_2$ = 4 Ω, $R_3$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ + $\frac{1}{R_3}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10+5+4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

(b) The EMF of the battery = 20 V

Current through $R_{1,}$ $I_1$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_2$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_3$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_1$ + $I_2$ + $I_3$ = 10 + 5 + 4 = 19 A

3.5 Let T be the room temperature and R be the resistance at room temperature and $T_1$ be the required temperature and $R_1$ be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27 $?,$ R = 100 Ω, $T_1$ = ?, $R_1$ = 117 Ω, α = 1.70 $\times {10}^{-4}$ ° ${\mathit{C}}^{-1}$

We know the relation of α can be given as

α = $\frac{R_1-R}{R(T_1-T)}$ = $\frac{117-100}{100(T_1-27)}$ = 1.70 $\times {10}^{-4}$

or ( $T_1$ - 27) = $\frac{117-100}{100\times 1.70\mathrm{}\times {10}^{-4}}$ = 1000

$T_1$ = 1000 +27 = 1027 $?$$$

3.6 Given, length of the wire, l = 15 m

Uniform cross section, A = 6.0 $\times {10}^{-7}$ $m^2$

Resistance measured, R = 5.0 Ω

From the relation of

R = $\rho \frac{l}{A}$ , where $\rho$ is the resistivity of the material, we get

$\rho =\frac{AR}{l}$ = $\frac{6.0\mathrm{}\times {10}^{-7}\times 5}{15}$ = 2 $\times {10}^{-7}$ Ωm.

3.7 Let us assume R = 2.1 Ω, T = 27.5 °C, $R_1$ = 2.7 Ω, $T_1$ = 100 $?$ , α = resistivity of silver

We know the relation of α can be given as

α = $\frac{R_1-R}{R(T_1-T)}$ = $\frac{2.7-2.1}{2.1(100-27.5)}$ = 3.94 $\times {10}^{-3}$ ${?}^{-1}$

3.8 Given

Supply voltage, V = 230 V

Initial current drawn, I= 3.2 A

Final current drawn, $I_1$ = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature, $T_1$ = ?

From Ohm’s law, we get initial resistance, $R$ = $\frac{V}{I}$ = $\frac{230}{3.2}$ Ω = 71.875 Ω

Final resistance, $R_1$ = $\frac{V}{I_1}$ = $\frac{230}{2.8}$ Ω = 82.143 Ω

From the relation of α = $\frac{R_1-R}{R(T_1-T)}$ , where α is the temperature coefficient of resistance, we get

1.7 $\times {10}^{-4}$ = $\frac{82.143-71.875}{71.875(T_1-27)}$

$T_1-27=$ 840.34

$T_1=867.34?$

Therefore the steady temperature of heating element required is $867.34?$$$

3.9

Let us assume

$I_1$ = Current flowing through the outer circuit

$I_2$ = Current flowing through the branch AB

$I_3$ = Current flowing through the branch AD

$I_4$ = Current flowing through the branch BD

( $I_2$ - $I_4)$ = Current flowing through the branch BC

( $I_3$ + $I_4)$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10 $I_2$ + 5 $I_4$ - 5 $I_3$ = 0

$I_3=I_4+2I_2$ …………(1)

For the close circuit BCDB, 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

$I_2-2I_3$ - 4 $I_4$ = 0

$I_2=2I_3$ +4 $I_4$ …………….(2)

For the close circuit ABCFEA, potential is zero, i.e.

-10 + 10 $I_1+$ 10 $I_2$ + 5( $I_2$ - $I_4)$ = 0

10 $I_1$ + 15 $I_2$ - 5 $I_4$ = 10

2 $I_1$ + 3 $I_2$ - $I_4$ = 2 …………(3)

Solving equations (1) and (2) we get

$I_2=-2I_4$

$I_3=-3I_4$

Substituting these values in equation (3), we get

$I_4=-\frac{2}{17}$ A

$I_2=\frac{4}{17}$ A

$I_3=\frac{6}{17}$ A

$I_1$ = $I_2+I_3$ = $\frac{10}{17}$ A

3.10 (a) Balance point from end A, $l_1$ = 39.5 cm

Resistance of the resistor, S = 12.5 Ω

Condition for the balance is given as,

$\frac{R}{S}$ = $\frac{l_1}{{100-l}_1}$

$R=S\times \frac{l_1}{100-l_1}$ = 12.5 $\times \frac{39.5}{100-39.5}$ = 8.16 Ω

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{a}\mathrm{}\mathrm{W}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{e}$ bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

(b) If R & S are interchanged, then $l_1$ and (100 - $l_1$ ) will also get interchanged. The balance point will be (100- $l_1)$ from A. So the new balance point is 100 – 39.5 = 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection, no current will flow through galvanometer.

3.11 EMF of the battery, E = 8 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Let the effective voltage in the circuit be $V_1$

Then $V_1$ = V – E = 120 – 8 = 112 V

If the current flown in the circuit be I then from the relation

I = $\frac{V_1}{R+r}$ we get I = $\frac{112}{15.5+0.5}$ = 7 A

Voltage across the resistor R = IR = 7 $\times 15.5=$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherwise, the current will be extremely high.

3.12 EMF of the cell,  $E_1$ = 1.25 V

Let the EMF of the replaced cell be $E_2$

Existing balance point,  $l_1$ = 35 cm

New balance point,  $l_2$ = 63 cm

From the relation of balance condition, we get

$\frac{E_1}{E_2}$ = $\frac{l_1}{l_2}$ , we get $E_2$ = $\frac{E_1*l_2}{l_1}$ = $\frac{1.25*63}{35}$ = 2.25 V

Therefore the emf of the another cell is 2.25V

3.13 Given, number of free electron in a copper conductor, n = 8.5 $\times {10}^{28}$ $m^{-3}$

Length of the copper wire, l = 3.0 m

The area of cross section, A = 2 $\times {10}^{-6}$ $m^2$

Current carried by the wire, I = 3.0 A

From the relation I = nAe $V_d$ where

e = Electric charge = 1.6 $\times {10}^{-19}$ C

$V_d=\mathrm{D}\mathrm{r}\mathrm{i}\mathrm{f}\mathrm{t}\mathrm{}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}$

We get $V_d$ = $\frac{I}{nAe}$

Again, Drift velocity ( $V_d)$ = $\frac{Lengthofthewire\left(l\right)}{Timetakentocoverl\left(t\right)}$

t = $\frac{l}{V_d}$ = $\frac{lnAe}{I}$ = $\frac{3\times 8.5\mathrm{}\times {10}^{28}\times 2\mathrm{}\times {10}^{-6}\times 1.6\mathrm{}\times {10}^{-19}}{3}$ = 27.2 $\times {10}^3$ seconds

3.14 Surface charge density of the earth, $\sigma$ = ${10}^{-9}$ C $m^{-2}$

Current over entire Globe, I = 1800 A

Radius of the earth, r = 6.37 $\times {10}^6$ m

Hence, surface area of the earth, A = 4 $\pi r^2$ = 4 $\times \pi \times {(6.37\mathrm{}\times {10}^6)}^2$ = 5.1 $\times {10}^{14}$ $m^2$

Charge on the earth surface, q = $\sigma$ $\times A$ = 0.51 $\times {10}^6$ C

If t is time taken to neutralize the earth surface, then q = I $\times t$

t = $\frac{q}{I}$ = $\frac{0.51\mathrm{}\times {10}^6}{1800}$ = 283.28 seconds

3.15 (a) Number of secondary cells, n = 6

emf of each secondary cell = 2 V

Internal resistance of each secondary cell, r = 0.015 Ω

Resistance of the resistor, R = 8.5 Ω

Current drawn from the supply, I is given as I = $\frac{Totalvoltage}{R+totalinternalresistance}$

= $\frac{6\times 2}{8.5+6\times 0.015}$ = 1.396 A

(b) Terminal voltage = I $\times R$ = 1.396 $\times 8.5$ = 11.87 V

After long use emf = 1.9 V

Internal resistance of the cell, r = 380 Ω

Maximum current can be drawn = $\frac{1.9}{380}$ = 5 $\times {10}^{-3}$ A

No, the cell cannot drive the starter motor of the car as it requires high current.

3.16 Given, resistivity of aluminium, ${\rho }_1$ = 2.63 $\times {10}^{-8}$ Ωm

Resistivity of copper, ${\rho }_2$ = 1.72 $\times {10}^{-8}$ Ωm

Relative density of aluminium = 2.7

Relative density of copper = 8.9

For Aluminium wire: Let

$l_1$ = length, $m_1$ = mass, $R_1$ = resistance, $A_1$ = cross-section area, ${\rho }_1$ = density

For Copper wire: Let

$l_2$ = length, $m_2$ = mass, $R_2$ = resistance, $A_2$ = cross-section area, ${\rho }_2$ = density

From the relation R = $\frac{\rho l}{A}$ , we get

$R_1=\frac{{\rho }_1{l}_1}{A_1}$ ………(1), and

$R_2=\frac{{\rho }_2{l}_2}{A_2}$ ………(2)

It is given $R_1$ = $R_2$ and $l_1$ = $l_2$ . Hence

$\frac{A_1}{A_2}$ = $\frac{{\rho }_1}{{\rho }_2}$ = $\frac{2.63}{1.72}$ ( ${10}^{-8}$ cancel each other)

From the relation mass = volume $\times \mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}$ = length $\times$ cross sectional area $\times$ relative density, we get

$m_1=l_1{A}_1{d}_1$ and $m_2=l_2{A}_2{d}_2$ , where $d_1$ = relative density of aluminium and $d_2$ = relative density of copper

$\frac{m_1}{m_2}=\frac{l_1{A}_1{d}_1}{l_2{A}_2{d}_2}$ = $\frac{A_1{d}_1}{A_2{d}_2}$ = $\frac{2.63}{1.72}\times \frac{2.7}{8.9}$ = 0.464

It can be inferred from this ratio that $m_1$ is less than $m_2$ . Hence, aluminium is preferred as overhead power cables over copper.

3.17 From Ohm’s law, R = $\frac{V}{A}$

From the given table we get $\frac{3.94}{0.2}$ = $\frac{7.87}{0.4}$ = $\frac{11.8}{0.6}$ …………… $\frac{158}{8}$ $\approx 19.7\Omega$ = constant

Hence manganin is an ohmic conductor.

Q.3.18  (a) When a steady current flows in a metallic conductor of non-uniform cross section, only the current flowing is constant. Current density, Electric field and Drift speed are inversely proportional to the cross section area, hence not constant.

(b) Ohm's law is not applicable to all conductors, vacuum diode semi-conductor is a non-ohmic conductor.

(c) According to ohm's law V = IR, voltage is directly proportional to current, hence to draw high current from a low voltage source, internal resistance ®, needs to be low.

(d) To prevent the drawing of extra current, which can cause short circuit, the internal resistance for a high voltage system needs to be high.

3.19 (a) Alloys of metal 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 increase 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 10²².

3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{eff}$ is maximum. $R_{eff}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{eff}$ is minimum. $R_{eff}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{nR}{\frac{R}{n}}$ = $n^2$

Let us assume, $R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$Ω

From the circuit the equivalent resistance is given by

R = $\frac{2\times 1}{2+1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2\times 3}{2+3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1\times 2\times 3}{1\times 2+2\times 3+3\times 1}$ = $\frac{6}{11}$ Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2 $\Omega$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2\times 4}{2+4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $\times 4=\frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $\times R=5R$

3.21 Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by

R’ = 2 + $\frac{R\text{'}}{(R^{\text{'}}+1)}$ or R’ = $\frac{2R^{\text{'}}+2+R\text{'}}{(R^{\text{'}}+1)}$ or ${R\text{'}}^2$ + R’ = 2R’ + 2 + R’

${R\text{'}}^2-2R^{\text{'}}-2=0$

R’ = $\frac{2\pm \sqrt{4+8}}{2}$ = 1 $\pm \surd 3$

Since R’ cannot be negative, hence R’ = 1+ $\surd 3$ = 2.73 Ω

Internal resistance, r = 0.5Ω

Total resistance = 2.73 + 0.5 = 3.23 Ω

Current drawn from the source = $\frac{12}{3.23}$ A= 3.72 A

3.22 Constant emf of the standard cell, $E_1$ = 1.02 V

Balance point on the wire, $l_1$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_1}{l_1}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_1}\times E_1$ = $\frac{82.3}{67.3}\times 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The method would not work if the emf of the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

3.23 Internal resistance of the cell = r

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

An external resistance R is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, $l_2$ = 64.8 cm

Current flowing through circuit = I

The relation connecting resistance and emf is,

r = ( $\frac{l_1-l_2}{l_2}$ )R = $\frac{76.3-64.8}{64.8}$ $\times 9.5=$ 1.69 Ω

Therefore, the internal resistance is 1.69 Ω.

No, in fact, it is one of the easiest chapter of class 12 Physics. Other chapters which are considered comparatively easy are Ray Optics and Electric Charges and Fields.

In simple words, current electricity can be defined as the electric charge continuously moving from one place to another along a pathway. It is measured in amperes (A). Electric current is needed for electrical devices to work.

The current's SI unit is ampere. The symbol for ampere is A. The term ampere is named after the French physicist André-Marie Ampère.

There are two types of electricity - Static and Current electricity. The electric charges buildup on a material's surface is called the static electricity. The continuous flow of electric charge is termed as the current electricity. Current electricity is of two types - Alternating Current (AC) and Direct Current (DC). In AC, the charge direction reverses periodically and in DC, charge flows in one direction.

According to this chapter, a galvanometer is used to find and measure the small electric currents in a circuit. The principle that works in a galvanometer is the electromagnetic induction.