JEE Advanced 2023 — Paper 1 — Question 19

$\mathrm{E}_{\mathrm{e}}=\frac{\frac{15}{\mathrm{R}}+\frac{0}{3}}{\frac{1}{\mathrm{R}}+\frac{1}{3}}=\frac{45}{\mathrm{R}+3}$

$\Rightarrow i=\frac{\frac{45}{\mathrm{R}+3}-5}{\frac{3 \mathrm{R}}{\mathrm{R}+3}+1}$

$1=\frac{30-5 R}{4 R+3}$

$R=3 \Omega$

$\mathrm{V}_{\mathrm{CD}}=5+1 \times 1=6$ Volt.

$\Rightarrow \mathrm{i}_{1}=\frac{6}{3} A=2 \mathrm{~A}$.

$\mathrm{E}_{\mathrm{e}}=\frac{\frac{15}{3}+\frac{5}{1}+\frac{0}{3}}{\frac{1}{3}+\frac{1}{1}+\frac{1}{3}}=6$ Volts $\frac{1}{r_{e}}=\frac{1}{3}+1+\frac{1}{3}=\frac{5}{3}$

$\Rightarrow \mathrm{r}_{\mathrm{e}}=\frac{3}{5} \Omega$

As $t \rightarrow \infty$, charge on capacitor, $\mathrm{q}_{0}=2 \times 6=12 \mu \mathrm{C}$

Current through capacitor at time $t$ is, $i=i_{0} e^{-\frac{t-t_{0}}{7.2 \times 10^{-6}}}$

At $\mathrm{t}=\mathrm{t}_{0}+7.2 \mu \mathrm{~s}$

$\mathrm{i}=\frac{6}{\left(\frac{3}{5}+3\right)}\left[\mathrm{e}^{-1}\right]=0.6 \mathrm{~A}$