Physics - Capacitor Question with Solution | TestHub

Short the source and find $R_{eq}$ across the capacitor; $\tau =R_{eq}C$.

(P) $R_{eq}=\frac{6\times 3}{6+3}+1=3\Omega$, $\tau =3\times 2=6\mu s$, P$\to 1$. (Q) $R_{eq}=1+1=2\Omega$, $\tau =2\times 1=2\mu s$, Q$\to 2$.

(R) $R_{eq}=1+4=5\Omega$, $\tau =5\times 1=5\mu s$, R$\to 3$. (S) $R_{eq}=\frac{(3+2)(2+3)}{5+5}+0.6\approx 3\Omega$, $\tau \approx 3\times 5=15\mu s$, S$\to 4$.

Option (C).