JEE Advanced 2023 Paper 1 Q03 Physics Electrostatics & Circuits RC Circuits Hard

JEE Advanced 2023 Paper 1 · Q03 · RC Circuits

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 $1\,\Omega$ resistor. The key is closed at time $t = t_0$. Which of the following statement(s) is(are) correct? [Given: $e^{-1} = 0.36$]

  1. A. The value of the resistance $R$ is $3\,\Omega$
  2. B. For $t < t_0$, the value of current $i_1$ is $2$ A
  3. C. At $t = t_0 + 7.2\,\mu s$, the current in the capacitor is $0.6$ A
  4. D. For $t \to \infty$, the charge on the capacitor is $12\,\mu C$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C, D

Solution

Before switch closure: current through $1\,\Omega$ is $1$ A, so voltage across it = $1$ V. By KVL around the relevant loops, with $15$ V source and $6\,\Omega$ branch, solving gives $R = 3\,\Omega$ — (A) correct. Total current from $15$ V source through $6\,\Omega + 3\,\Omega$ branches = $2$ A — (B) correct. After switching, Thevenin equivalent across capacitor: $V_{\text{eq}} = 6$ V, $R_{\text{eq}} = 18/5\,\Omega$, so steady-state charge = $C V_{\text{eq}} = 2\,\mu F \cdot 6$ V $= 12\,\mu C$ — (D) correct. Time constant $\tau = R_{\text{eq}} C = (18/5)(2) = 36/5\,\mu s$. At $t - t_0 = 7.2\,\mu s = \tau$, $i = (V_{\text{eq}}/R_{\text{eq}})e^{-1} = (5/3)(0.36) = 0.6$ A — (C) correct.

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