PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 4 marks · Case-based

CBSE Class 12 Physics 2025: question 29

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedElectrostatic Potential and Capacitance

A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key $S_{1}$ (keeping $S_{2}$ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference $V_{c}$ (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as $q = Q [1 - e^{- t / R C}]$
The charging current can be obtained by differentiating it and using $\frac{d}{d x} (e^{m x}) = m e^{m x}$ .
Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V.

(i)
The final charge on the capacitor, when key $S_{1}$ is closed and $S_{2}$ is open, is
1. (A)5 μC
2. (B)5 mC
3. (C)25 mC
4. (D)0.1 C
1 mark
(ii)
For sufficient time the key $S_{1}$ is closed and $S_{2}$ is open. Now key $S_{2}$ is closed and $S_{1}$ is open. What is the final charge on the capacitor ?
1. (A)Zero
2. (B)5 mC
3. (C)2.5 mC
4. (D)5 μC
1 mark
(iii)
The dimensional formula for RC is
1. (A) $[M L^{2} T^{- 3} A^{- 2}]$
2. (B) $[M^{0} L^{0} T^{- 1} A^{0}]$
3. (C) $[M^{- 1} L^{- 2} T^{4} A^{2}]$
4. (D) $[M^{0} L^{0} T A^{0}]$
1 mark
(iv)
The key $S_{1}$ is closed and $S_{2}$ is open. The value of current in the resistor after 5 seconds, is
1. (A) $\frac{1}{2 e}$ mA
2. (B) $e$ mA
3. (C) $\frac{1}{e}$ mA
4. (D) $\frac{1}{2 e}$ mA
1 mark
OR
(iv)
The key $S_{1}$ is closed and $S_{2}$ is open. The initial value of charging current in the resistor, is
1. (A)5 mA
2. (B)0.5 mA
3. (C)2 mA
4. (D)1 mA
1 mark

Figure in the paper: RC circuit: a capacitor C in the top branch, a resistor R on the left, a battery V at the bottom right, and two keys S1 and S2 meeting at a junction near the bottom left; S1 connects to the battery and S2 connects via a wire across to the right side, shorting the capacitor-resistor loop. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

When fully charged, the capacitor voltage equals the battery emf. When the battery is removed, there is a closed path through R.

Hint 2

Differentiate $q = Q (1 - e^{- t / R C})$ to get $I = (Q / R C) e^{- t / R C} = (V / R) e^{- t / R C}$ , and find RC first.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Final charge $Q = C V = 500 \times 1 0^{- 6} \times 10 = 5 \times 1 0^{- 3}$ C $= 5$ mC. Option (B).
1 mark
(ii) With $S_{1}$ open and $S_{2}$ closed, the battery is removed and the capacitor discharges completely through R. Final charge = zero. Option (A).
1 mark
(iii) RC is the time constant, so it has dimensions of time: $[M^{0} L^{0} T A^{0}]$ . Option (D).
1 mark
(iv)(a) $τ = R C = 20 \times 1 0^{3} \times 500 \times 1 0^{- 6} = 10$ s. $I = \frac{d q}{d t} = \frac{V}{R} e^{- t / R C} = \frac{10}{20 \times 1 0 ^{3}} e^{- 5/10} = 0.5 e^{- 1/2}$ mA $= \frac{1}{2 e}$ mA. Option (A).
1 mark
(iv)(b) [alternative] At $t = 0$ : $I_{0} = \frac{V}{R} = \frac{10}{20 \times 1 0 ^{3}} = 0.5$ mA. Option (B).

Answer: (i) B, 5 mC; (ii) A, zero; (iii) D, [M⁰L⁰TA⁰]; (iv)(a) A, 1/(2√e) mA; (iv)(b) B, 0.5 mA

Where the marks are

(i) Q = CV = 5 mC, option B (1 mark)
(ii) Capacitor fully discharges, option A (1 mark)
(iii) RC has dimensions of time, option D (1 mark)
(iv) $I = (V / R) e^{- t / R C} = \frac{1}{2 e}$ mA, option A; or OR part $I_{0} = V / R = 0.5$ mA, option B (1 mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 29

Hints

Step-by-step solution

Revise the chapter

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