Animated Solution for Physics - Semiconductor Electronics: Two identical capacitors $A$ and $B$, charged to the same potential $5\text{V}$ 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)

Welcome to a fascinating exploration of a classic JEE Main problem that beautifully marries the concepts of electrostatics and semiconductor physics. At first glance, this problem might seem like a straightforward application of RC circuits, but it hides a subtle trap that tests your physical intuition over mere diagram reading.

Let's embark on this journey by breaking down the physical setup and the underlying mathematics.

We are presented with two identical capacitors, $A$ and $B$, each with capacitance $C$. Both capacitors have been pre-charged to a potential difference of $V=5\text{V}$.

From the fundamental definition of capacitance, the initial charge stored on both capacitors is given by:

At time $t=0$, these fully charged capacitors are placed into two distinct circuits. Our goal is to determine the remaining charge on each capacitor after a specific time interval, $t=CR$.

The Trap of the Diagram

Let's focus our attention on Circuit A. The diagram displays a resistor $R$, a capacitor $A$, and a diode $D_1$.

First, we must determine the biasing of the diode. By observing the polarity markings on the capacitor, we see that the top plate is positively charged ($+$) and the bottom plate is negatively charged ($-$).

Tracing the wires, the positive top plate is connected to the flat bar of the diode symbol, which represents the n-side (cathode). Conversely, the negative bottom plate is connected to the triangle, representing the p-side (anode).

When the n-side is at a higher potential than the p-side, the diode is reverse-biased. An ideal reverse-biased diode acts as an open circuit, offering infinite resistance.

The JEE Trap: Visually, the diagram resembles three parallel branches. If the resistor were truly in parallel with the capacitor, the capacitor would simply discharge through the resistor, completely ignoring the open diode branch. However, the physical reality intended by the problem's solution dictates that the resistor and diode are in series with the capacitor.

Because the diode is reverse-biased and acts as an open switch, the entire series loop is broken. No current can flow. Therefore, the charge on capacitor $A$ remains perfectly trapped and constant over time.

The RC Discharge

Now, let's shift our focus to Circuit B. The setup is identical, except the diode $D_2$ has been flipped.

The positive top plate of the capacitor is now connected to the p-side (anode) of the diode, and the negative bottom plate is connected to the n-side (cathode).

This configuration means the diode is forward-biased. An ideal forward-biased diode acts as a perfect conductor, or a short circuit, offering zero resistance.

With the diode allowing current to pass freely, the circuit is closed. Capacitor $B$ will now begin to discharge its stored energy through the resistor $R$.

The discharging of a capacitor in an RC circuit follows a classic exponential decay model. The charge $Q(t)$ at any given time $t$ is described by the master equation:

We are asked to find the charge on capacitor $B$ at the specific time $t=CR$. This time interval is known as the time constant ($au$) of the circuit.

Let's substitute our known values into the decay equation. We know the initial charge is $Q_0=CV$, and the time is $t=CR$:

The $RC$ terms in the exponent beautifully cancel each other out, leaving us with:

Which can be rewritten as:

By carefully analyzing the biasing of the diodes and understanding the physical intent of the circuit diagrams, we have successfully determined the final states of both capacitors.

Capacitor $A$ remains fully charged at $Q_A=VC$, while capacitor $B$ discharges to $Q_B=\frac{VC}{e}$.

Comparing our derived values with the given options, we find a perfect match. This problem serves as a brilliant reminder to always let physical principles guide your interpretation of schematic diagrams!