JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Current Electricity

Before the key is closed ($t<t_0$), the capacitor branch is open. Current through the $1\Omega$ resistor is 1 A, so the voltage across it is 1 V. This means the node voltage is $V=5+1=6$ V.

For the branch with $R$: $15-IR=6$, giving $IR=9$.

For the $3\Omega$ branch: $6-3I_1=0$, giving $I_1=2$ A. Option (B) is correct.

By KCL: $I=1+2=3$ A, so $R=9/3=3\Omega$. Option (A) is correct.

After the key closes, the Thevenin equivalent voltage is $\epsilon =6$ V with equivalent resistance $R_{eq}=3\Omega \parallel 1\Omega \parallel 3\Omega =0.6\Omega$.

Time constant: $\tau =RC=3.6\times 2=7.2\mu s$.

Current in capacitor: $i_C=\frac{6}{3.6}{e}^{-t/\tau }=\frac{5}{3}{e}^{-t/7.2}$.

At $t=t_0+7.2\mu s$: $i_C=\frac{5}{3}{e}^{-1}=\frac{5}{3}\times 0.36=0.6$ A. Option (C) is correct.

For $t\to \infty$, the capacitor is fully charged to 6 V: $Q=CV=2\times 6=12\mu C$. Option (D) is correct.