Question

In the circuit, switch $S$ is closed at time $t=0$, Let $I_1$ and $I_2$ be the currents at any finite time $t$, then the ratio of $\frac{I_1}{I_2}$

Answer 1

For the given circuit at any instant time $t$,

$I_1=\frac{V}{R}{e}^{\left(-\frac{t}{RC}\right)}...\left(1\right)$

$I_2=\frac{V}{R}{e}^{\left(-\frac{t}{2RC}\right)}...\left(2\right)$

Fro equations (1) and (2),

$\frac{I_1}{I_2}=e^{\left(-\frac{t}{RC}+\frac{t}{2RC}\right)}$

$\Rightarrow \frac{I_1}{I_2}=e^{\left(-\frac{t}{2RC}\right)}$

$\Rightarrow \frac{I_1}{I_2}=\frac{1}{e^{\left(\frac{t}{2RC}\right)}}$

From this expression, it is clear that as $t$ increases, ratio of $\frac{I_1}{I_2}$ decreases.

Hence, option (c) is the correct answer.