Physics - Capacitor Question with Solution | TestHub

(A) At $t=0$, capacitor acts as short-circuit

At $t\to \infty$, capacitor acts as open circuit and no current flows through voltmeter.

(B) $q_x=2CV\left(1-e^{-t/2CR}\right),x=\frac{V}{R}{e}^{-t/2CR}$

$q_y=CV\left(1-e^{-t/2CR}\right),y=\frac{V}{2R}{e}^{-t/2CR}$

$\Delta V=-y\cdot 2R+\frac{q_x}{2C}$

$=V\left[1-2e^{-t/2CR}\right]=0$

(C) $\tau =1$ sec.

So, by $i=i_0{e}^{-t/\tau }$ current at $t=1$ sec. is $i_0/e$

(D) After long time no current flows since both capacitor and voltmeter does not allow.