Animated Solution for Physics - Current Electricity: In an $L-C-R$ circuit as shown below, both switches are open initially. Now, switch $S_1$ is closed and $S_2$ kept open ($q$ is charge on the capacitor and $\tau =RC$ is capacitance time constant). Which of the following statement is correct?

Imagine you are standing in front of a control panel with two switches, $S_1$ and $S_2$. Before you, lies an intricate $L-C-R$ circuit. Initially, both switches are open, meaning the circuit is completely dead. No current flows, and the capacitor holds zero charge.

Now, the problem states that we close switch $S_1$ while keeping $S_2$ open. Let's visualize what happens the moment $S_1$ makes contact.

Because $S_2$ remains open, the entire bottom branch containing the inductor $L$ is completely cut off from the power source. It is as if the inductor doesn't even exist in this scenario!

The active part of our circuit is strictly the top loop. If we trace the path from the battery, the current flows through the closed switch $S_1$, into the capacitor $C$, through the resistor $R$, and back to the battery. What we have beautifully isolated is a classic $R-C$ charging circuit.

In an $R-C$ charging circuit, the capacitor doesn't charge instantly. It builds up charge exponentially over time. The master equation governing this growth is:

Here, $q_0$ represents the maximum steady-state charge the capacitor can hold, which is simply the capacitance $C$ multiplied by the battery voltage $V$ ($q_0=CV$). The term $\tau$ is the time constant of the circuit, defined as $\tau =RC$. It dictates how fast the capacitor charges.

Substituting $q_0$, our equation becomes:

Now, let's evaluate the given options. Option (c) asks us to find the charge at a specific time, $t=2\tau$.

Let's carefully substitute this time into our master equation:

Notice how elegantly the $\tau$ in the numerator and denominator cancel each other out!

This result perfectly matches option (c).

As a quick bonus, let's look at option (a) which talks about energy. The total work done by the battery is $C{V}^2$. The energy stored in the capacitor is $\frac{1}{2}C{V}^2$. By conservation of energy, the remaining $\frac{1}{2}C{V}^2$ is dissipated as heat in the resistor. Therefore, the work done by the battery is actually twice the energy dissipated, making option (a) incorrect.

Physics is all about isolating the active components and applying the fundamental laws. Great job navigating this circuit!