Question

A capacitor of capacitance $6\mu \text{F}$ and initial charge $192\mu \text{C}$ is connected with a key $S$ and resistance as shown in figure. Point $M$ is earthed. If key is closed at $t=0$, then the current (in Ampere) through resistance $R\left(1\Omega \right)$, at $t=16\text{ln}2\mu \text{s}$ is

Answer 1

Equivalent resistance of the resistors connected in series in left side of the capacitor is, $2+2+4=8\Omega$.

Equivalent resistance of the resistors connected in series in the right side of the capacitor is, $1+1+2=4\Omega$.


$8\Omega$ and $4\Omega$ are connected in parallel, their equivalent resistance, $R_{eq}=\frac{8\times 4}{8+4}=\frac{8}{3}\Omega$

Discharging current of the capacitor, $i=\frac{Q}{R_{eq}C}{e}^{\frac{-t}{R_{eq}C}}$

Current is divided in inverse ratio of resisatnce, when they are connected in parallel.
Current through $4\Omega$ resistance,$i_1=i\times \frac{R_{eq}}{4}=\frac{2i}{3}$
$i_1=\frac{2}{3}\times \frac{192\times {10}^{-6}}{\frac{8}{3}\times 6\times {10}^{-6}}{e}^{\frac{-16ln2\times {10}^{-6}}{\frac{8}{3}\times 6\times {10}^{-6}}}=4\text{A}$
This $4\text{A}$ current will flow through all three resistance connected in the right side of the capacitor, since they are connected in series.