Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is Q A and Q B respectively, then (Here e is the base of natural logarithm)

Solution

Physics sir · 3 steps

Trace which way the discharge current must flow, then check whether each diode lets it through.

1. Discharge direction. The left plate is positive, so the discharge current leaves it, goes down through R, along the bottom wire, and up the right side into the negative plate. In both circuits it must flow upward through the diode.

   Circuit	Diode arrow points	Diode state	Charge at t = CR
   A	down, against the current	reverse biased, no current	stays CV
   B	up, along the current	forward biased, conducts	discharges through R

2. Discharge in B. With an ideal diode conducting, B discharges through R alone, so Q(t) = CV e^(-t/CR).

   Q_B = CV e^(-CR/CR) = CV/e

3. So. Q_A = VC because no charge can leave A, and Q_B = VC/e. That is option (c).

The idea to keep. A diode in a capacitor circuit is just a one-way switch. First find which way the discharge current wants to go. If the diode points the other way, nothing happens. If it points the same way, use Q = Q₀ e^(-t/RC).

Answer from the NTA final answer key. Solution by PrepWiser.