At t = 0 , two identical capacitors A and B , each initially...

In circuit (a), the positive plate of capacitor A is connected to the p-side (anode) of the diode, and the negative plate is connected to the n-side (cathode) through the resistor. This means the diode is forward-biased and acts as a closed switch, allowing current to flow.

Capacitor A will discharge through the resistor R. The charge on the capacitor at any time $t$ during discharging is given by the equation:

$q_A(t) = q_0 e^{-\dfrac{t}{RC}}$

Given the initial charge $q_0 = CV$ and time $t = CR$, the charge on capacitor A is:

$q_A = CV e^{-\dfrac{CR}{CR}} = \dfrac{CV}{e}$

In circuit (b), the positive plate of capacitor B is connected to the n-side (cathode) of the diode, and the negative plate is connected to the p-side (anode) through the resistor. This means the diode is reverse-biased and acts as an open switch.

Since the circuit is open, no current can flow and capacitor B will not discharge. The charge on capacitor B remains constant at its initial value:

$q_B = CV$

Therefore, the respective charges on the capacitors at time $t = CR$ are $VC/e$ and $VC$.