Question

Analyze the given circuit in the steady state condition. Charge on the capacitor is $q_o=16\mu C$.
If in the beginning the battery is removed and nodes $A$ and $C$ are shortened, then find the duration in which charge on the capacitor becomes $5.92\mu C$

Answer 1

After short circuiting the battery, we will have to find net resistance across capacitor to calculate equivalent value of ${\tau }_C$ in discharging. $3\Omega$ and $6\Omega$ are in parallel. Similarly, $4\Omega$ and $4\Omega$ are in parallel. They are then in series.

$\therefore R_{net}=4\Omega ,{\tau }_C=C{R}_{net}=(4\times 4)\mu s=16\mu s$

During discharging $q=q_o{e}^{-t/{\tau }_C}$

or $8=16e^{-t/16}$

Solving this equation, we get $t=11.1\mu s=111\times {10}^{-7}s$

So, the value of $x=7$