Animated Solution for Physics - Current Electricity: At $t=0$, switch $S$ is closed. The charge on the capacitor is varying with time as $Q=Q_0(1-e^{-\alpha t})$. Obtain the value of $Q_0$ and $\alpha$ in the given circuit parameters.

Welcome to a classic and highly conceptual problem from JEE Advanced! This question beautifully intertwines the concepts of DC steady-state analysis and transient RC circuits. It challenges us to look beyond the standard single-loop RC circuit and analyze a slightly more complex network.

Imagine you are an electrical engineer tasked with understanding how this specific capacitor charges over time. The charging equation is given as $Q=Q_0(1-e^{-\alpha t})$. Our mission is to decode this equation and find the exact physical meaning and mathematical values of $Q_0$ and $\alpha$. Let's dive in!

When we close the switch $S$ at $t=0$, the battery starts pushing charge through the circuit. The capacitor $C$ begins to charge, and the current dynamically splits between the capacitor branch and the resistor $R_2$.

The equation $Q=Q_0(1-e^{-\alpha t})$ is the standard charging profile. Here, $Q_0$ represents the absolute maximum charge the capacitor will eventually hold, and $\alpha$ dictates how fast it gets there.

Let's first hunt down $Q_0$. We know that $Q_0$ is achieved at the steady state, which occurs when $t\to \infty$.

Why? Because once it's full, it strongly opposes any further flow of charge. With the capacitor branch effectively "open", the current from the battery has only one path to follow: through the outer loop containing $R_1$ and $R_2$.

Let's calculate this steady-state current, $I_{\text{steady}}$. Using Ohm's law for the outer loop:

Now, look at the circuit geometry. The capacitor $C$ is connected perfectly in parallel with the resistor $R_2$. This means the potential difference across the capacitor, $V_C$, must be identical to the potential difference across $R_2$.

Since the charge on a capacitor is given by $Q=CV$, we can easily find the maximum charge $Q_0$:

Now, let's move forward and find $\alpha$. In the exponent of our charging equation, $\alpha$ is the reciprocal of the circuit's time constant, ${\tau }_c$.

Here is the catch: $R_{\text{net}}$ is not just any random resistance. It is the Thevenin equivalent resistance as seen from the terminals of the capacitor.

To find this equivalent resistance, we must turn off all independent voltage sources. We do this by short-circuiting the ideal battery $V$.

Imagine standing at the capacitor's location and looking back into the circuit with the battery replaced by a plain wire. The left end of $R_1$ is now directly connected to the bottom wire, which is also connected to the bottom end of $R_2$. Since their top ends are already joined at the node above the capacitor, $R_1$ and $R_2$ are perfectly in parallel!

The equivalent resistance of two parallel resistors is:

Substitute this back into our expression for $\alpha$:

And there we have it! We have successfully decoded the circuit and found both parameters. This problem is a brilliant reminder that complex RC circuits can always be tamed by breaking them down into steady-state analysis and Thevenin equivalent resistance. Keep practicing, and never let these circuits intimidate you!