Physics - Capacitor Question with Solution | TestHub

This is the case of DC RC circuit for charging of a capacitor option

(A) is correct because in steady state ${\mathrm{I}}_{\mathrm{R}}=0$ hence ${\mathrm{V}}_{\mathrm{R}}=0$ option

(B) is incorrect ${\mathrm{V}}_{\mathrm{R}}={\mathrm{V}}_0{\mathrm{e}}^{-\mathrm{t}/\tau }$ where ${\mathrm{V}}_0=10\mathrm{V}$ and $\tau =\mathrm{RC}$ option

(C) is correct $\tau =\mathrm{RC}=1\times 10^3\times 1\times 10^{-6}=1\mathrm{ms}$

Hence after one time constant, ${\mathrm{e}}^{-1}=0.63$ (approx)

$\therefore {\mathrm{V}}_{\mathrm{R}}=10\times 0.63=6.3\mathrm{V}$ (approx) option

(D) correct Initially capacitor behaves as a short circuit

Hence $I=\frac{V}{R}=\frac{10}{1000}=10\mathrm{mA}$