Exercises — Current Electricity

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12 marksNCERT Exercises, Chapter 3
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is $0.4 Ω$ , what is the maximum current that can be drawn from the battery?
Hint. Maximum current is drawn when the external resistance is zero, leaving only the internal resistance.
The current in a circuit is $I = \frac{ε}{R + r}$ , so it is largest when the external resistance $R$ is as small as possible.

Setting $R = 0$ , which corresponds to short-circuiting the terminals, leaves only the internal resistance to limit the current:

$I_{m a x} = \frac{ε}{r} = \frac{12}{0.4} = 30 A$

This is why a car battery can deliver the very large currents a starter motor needs, and also why shorting its terminals is dangerous.

✦ I_max = 30 A
23 marksNCERT Exercises, Chapter 3
A battery of emf 10 V and internal resistance $3 Ω$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Hint. Use emf = I(R + r) for the first part, then terminal voltage = emf - I r.
The emf drives current through the external resistance and the internal resistance in series, so $ε = I (R + r)$ .

Rearranging for $R$ :

$R = \frac{ε}{I} - r = \frac{10}{0.5} - 3 = 20 - 3 = 17 Ω$

The terminal voltage is less than the emf because some potential is dropped across the internal resistance:

$V = ε - I r = 10 - (0.5) (3) = 8.5 V$

Equivalently $V = I R = 0.5 \times 17 = 8.5$ V, which confirms the result.

✦ R = 17 ohm; terminal voltage = 8.5 V
33 marksNCERT Exercises, Chapter 3
At room temperature ( $27. 0^{\circ}$ C) the resistance of a heating element is $100 Ω$ . What is the temperature of the element if the resistance is found to be $117 Ω$ , given that the temperature coefficient of the material is $1.70 \times 1 0^{- 4}^{\circ}$ C $^{- 1}$ ?
Hint. Apply R2 = R1[1 + alpha(T2 - T1)] and solve for the temperature difference.
Resistance varies with temperature according to $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$ .

Substituting the given values:

$\frac{117}{100} = 1 + 1.70 \times 1 0^{- 4} (T_{2} - 27)$

$0.17 = 1.70 \times 1 0^{- 4} (T_{2} - 27)$

$T_{2} - 27 = \frac{0.17}{1.70 \times 1 0 ^{- 4}} = 1000$

Therefore $T_{2} = 102 7^{\circ}$ C.

The very small value of $α$ is why such a large temperature rise is needed to increase the resistance by only 17 per cent.

✦ T2 = 1027 degrees Celsius
42 marksNCERT Exercises, Chapter 3
A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 1 0^{- 7}$ m $^{2}$ , and its resistance is measured to be $5.0 Ω$ . What is the resistivity of the material at the temperature of the experiment?
Hint. Rearrange R = rho L / A for the resistivity.
Resistance depends on the material and the geometry through $R = \frac{ρ L}{A}$ .

Rearranging for the resistivity:

$ρ = \frac{R A}{L} = \frac{5.0 \times 6.0 \times 1 0 ^{- 7}}{15} = \frac{3.0 \times 1 0 ^{- 6}}{15}$

$ρ = 2.0 \times 1 0^{- 7} Ω m$

The current is specified as negligibly small so that no appreciable heating occurs, which keeps the temperature and hence the resistivity constant during the measurement.

✦ rho = 2.0 x 10^-7 ohm metre
53 marksNCERT Exercises, Chapter 3
A silver wire has a resistance of $2.1 Ω$ at $27. 5^{\circ}$ C, and a resistance of $2.7 Ω$ at $10 0^{\circ}$ C. Determine the temperature coefficient of resistivity of silver.
Hint. Rearrange the same temperature relation, this time solving for alpha.
Starting again from $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$ and solving for $α$ :

$α = \frac{R _{2} - R _{1}}{R _{1} ( T _{2} - T _{1} )}$

Substituting $R_{1} = 2.1 Ω$ at $27. 5^{\circ}$ C and $R_{2} = 2.7 Ω$ at $10 0^{\circ}$ C:

$α = \frac{2.7 - 2.1}{2.1 \times ( 100 - 27.5 )} = \frac{0.6}{2.1 \times 72.5} = \frac{0.6}{152.25}$

$α \approx 3.9 \times 1 0^{- 3}^{\circ} C^{- 1}$

Note that $R_{1}$ must be the resistance at the reference temperature $T_{1}$ , since $α$ is defined relative to that value.

✦ alpha = 3.9 x 10^-3 per degree Celsius
64 marksNCERT Exercises, Chapter 3
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is $27. 0^{\circ}$ C? The temperature coefficient of resistance of nichrome is $1.70 \times 1 0^{- 4}^{\circ}$ C $^{- 1}$ .
Hint. Convert both currents into resistances using Ohm's law first, then apply the temperature relation.
The current falls as the element heats up, because its resistance rises. Convert each current into a resistance using Ohm's law.

At room temperature: $R_{1} = \frac{230}{3.2} = 71.9 Ω$ .

At the steady temperature: $R_{2} = \frac{230}{2.8} = 82.1 Ω$ .

Now apply $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$ :

$\frac{82.1}{71.9} = 1.143 = 1 + 1.70 \times 1 0^{- 4} (T_{2} - 27)$

$T_{2} - 27 = \frac{0.143}{1.70 \times 1 0 ^{- 4}} \approx 840$

Therefore $T_{2} \approx 86 7^{\circ}$ C.

The initial current is the larger one because the element is still cold when the supply is switched on.

✦ T2 is approximately 867 degrees Celsius

75 marksNCERT Exercises, Chapter 3

Determine the current in each branch of the network shown in Fig. 3.20 — a bridge with A-B $= 10 Ω$ , B-C $= 5 Ω$ , A-D $= 5 Ω$ , D-C $= 10 Ω$ , a bridge arm B-D $= 5 Ω$ , and a 10 V battery in series with $10 Ω$ closing the circuit from C back to A.

Hint. The bridge is not balanced, so current does flow through the middle arm. Apply Kirchhoff's rules and solve.

First check whether the bridge is balanced. For balance we would need $\frac{R _{A B}}{R _{B C}} = \frac{R _{A D}}{R _{D C}}$ , that is $\frac{10}{5} = 2$ against $\frac{5}{10} = 0.5$ .

These are unequal, so the bridge is not balanced and current does flow through the middle arm BD. The network cannot be reduced by simple series and parallel combinations.

Applying Kirchhoff's junction and loop rules to the network and solving the resulting simultaneous equations gives the branch currents:

Branch	Current
A to B (10 $Ω$ )	$\frac{4}{17} \approx 0.235$ A
A to D (5 $Ω$ )	$\frac{6}{17} \approx 0.353$ A
D to B (5 $Ω$ , bridge arm)	$\frac{2}{17} \approx 0.118$ A
B to C (5 $Ω$ )	$\frac{6}{17} \approx 0.353$ A
D to C (10 $Ω$ )	$\frac{4}{17} \approx 0.235$ A
Total through the battery	$\frac{10}{17} \approx 0.588$ A

The junction rule checks the result. At A the total $\frac{10}{17}$ splits into $\frac{4}{17}$ and $\frac{6}{17}$ . At B the incoming $\frac{4}{17}$ plus $\frac{2}{17}$ arriving from D gives the $\frac{6}{17}$ leaving towards C.

Note the bridge current flows from D to B, not from B to D — the direction emerges from the solution rather than being assumed.

✦ Total current 10/17 A (about 0.588 A); AB = DC = 4/17 A (0.235 A); AD = BC = 6/17 A (0.353 A); bridge arm carries 2/17 A (0.118 A) flowing from D to B

84 marksNCERT Exercises, Chapter 3
A storage battery of emf 8.0 V and internal resistance $0.5 Ω$ is being 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?
Hint. During charging the current is driven backwards through the battery, so the terminal voltage exceeds the emf.
During charging the external supply drives current into the battery, against its own emf. The net driving voltage is therefore the difference between them.

$I = \frac{120 - 8.0}{15.5 + 0.5} = \frac{112}{16} = 7.0 A$

Because the current is forced backwards through the battery, the potential drop across its internal resistance now adds to the emf rather than subtracting from it:

$V = ε + I r = 8.0 + (7.0) (0.5) = 11.5 V$

This sign reversal is the key difference from the discharging case in Exercise 3.2.

Purpose of the series resistor. It limits the charging current to a safe value. Without it the current would be $\frac{120 - 8}{0.5} = 224$ A, which would overheat and destroy the battery.

✦ Terminal voltage = 11.5 V during charging; the series resistor limits the charging current to a safe value, which would otherwise reach 224 A
94 marksNCERT Exercises, Chapter 3
The number density of free electrons in a copper conductor is $8.5 \times 1 0^{28}$ 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 cross-section is $2.0 \times 1 0^{- 6}$ m $^{2}$ and it carries a current of 3.0 A.
Hint. Find the drift velocity from I = n A e v_d, then divide the length by it.
The current is carried by electrons drifting slowly through the conductor, related by $I = n A e v_{d}$ .

Rearranging for the drift velocity:

$v_{d} = \frac{I}{n A e} = \frac{3.0}{8.5 \times 1 0 ^{28} \times 2.0 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19}}$

$v_{d} = \frac{3.0}{2.72 \times 1 0 ^{4}} \approx 1.1 \times 1 0^{- 4} m/s$

The time to traverse the wire is then:

$t = \frac{L}{v _{d}} = \frac{3.0}{1.1 \times 1 0 ^{- 4}} \approx 2.7 \times 1 0^{4} s$

which is about 7.5 hours.

This strikingly long time is worth pausing on. Electrons drift extremely slowly, yet a lamp lights instantly because the electric field is established through the wire at nearly the speed of light, setting every electron in motion at once.

✦ v_drift = 1.1 x 10^-4 m/s; time = 2.7 x 10^4 s, about 7.5 hours

Solutions written by the tuition.in editorial team and checked against leph103.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.

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