JEE Advanced 2023 Paper 1, Physics Section 1 Q3: Charge & discharge of a condenser

JEE Advanced2023Paper 1Physics Section 1

In a circuit shown in the figure, the capacitor $C$ is initially uncharged and the key $K$ is open. In this condition, a current of $1$ A flows through the $1 Ω$ resistor. The key is closed at time $t = t_{0}$ . Which of the following statement(s) is(are) correct?

[Given: $e^{- 1} = 0.36$ ]

A
The value of the resistance $R$ is $3 Ω$ .
B
For $t < t_{0}$ , the value of current $I_{1}$ is $2$ A.
C
At $t = t_{0} + 7.2 μ$ s, the current in the capacitor is $0.6$ A.
D
For $t \to \infty$ , the charge on the capacitor is $12 μ$ C.

Solution

Before the key is closed ( $t < t_{0}$ ). The capacitor branch is disconnected, so only the top two loops carry current. Let $V$ be the potential of the common node (relative to the other rail). Through the $1 Ω$ branch the given current is $1$ A flowing in a fixed direction, so $V = 5 + 1 \cdot 1 = 6$ V.

For the $3 Ω$ branch in series with the lower rail, the current $I_{1}$ satisfies $V - 3 I_{1} = 0 \Rightarrow I_{1} = \frac{6}{3} = 2$ A. Option (B) holds.

Resistance $R$ . KCL at the node: total current into the node from the $15$ V branch equals the sum of currents leaving through the other two branches: $I = 1 + 2 = 3$ A. Then $15 - I R = V = 6$ , giving $I R = 9$ and $R = 3 Ω$ . Option (A) holds.

After the key is closed. Replace the circuit external to the capacitor with its Thevenin equivalent at the capacitor terminals. The three sources combine to give $ε_{e q} = \frac{15/3 + 5/1 + 0/3}{1/3 + 1/1 + 1/3} = \frac{10}{5/3} = 6$ V. The Thevenin resistance is $(\frac{1}{3} + \frac{1}{1} + \frac{1}{3})^{- 1} = \frac{3}{5} = 0.6 Ω$ . The capacitor sits in series with the additional $3 Ω$ in its branch, so the total series resistance is $3 + 0.6 = 3.6 Ω$ .

Steady-state charge. $Q_{\infty} = C ε_{e q} = 2 μ F \times 6 V = 12 μ$ C. Option (D) holds.

Time constant and current. $τ = R C = 3.6 \times 2 = 7.2 μ$ s. The charging current is $i (t) = \frac{ε _{e q}}{R _{e f f}} e^{- t / τ} = \frac{6}{3.6} e^{- t / τ}$ . At $t = t_{0} + 7.2 μ$ s, $t / τ = 1$ , so $i = \frac{6}{3.6} \times 0.36 \approx 0.6$ A. Option (C) holds.

Correct options: (A), (B), (C), (D).

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