Animated Solution for Physics - Current Electricity: In the circuit shown, initially there is no charge on capacitors and keys $S_1$ and $S_2$ are open. The values of the capacitors are $C_1=10\mu \text{F}$, $C_2=30\mu \text{F}$ and $C_3=C_4=80\mu \text{F}$. Which of the statement(s) is/are correct ?

This problem is a beautiful masterclass in understanding the transient and steady-state behaviors of RC circuits. It tests your ability to apply the fundamental rules of capacitors at three distinct moments in time: exactly when a switch is closed ($t=0$), after a long time ($t\to \infty$), and the exact instant a new switch is introduced to a steady-state system.

Let's break down the circuit and evaluate each option systematically.

The Initial State ($t=0$)

Let's test Option (C) first. At time $t=0$, the switch $S_1$ is closed while $S_2$ remains open.

The Golden Rule: The moment a circuit is completed, an uncharged capacitor acts as a perfect short circuit (a simple wire) because there is no accumulated charge to oppose the flow of current.

Since $S_2$ is open, the middle branch is completely disconnected. The circuit simplifies to a single, large outer loop containing the $5\text{V}$ battery and three resistors in series. The total equivalent resistance is simply the sum of these resistors:

Using Ohm's law, the instantaneous current flowing through this closed circuit is:

This perfectly matches Option (C), making it a correct statement.

The Steady State ($t\to \infty$)

Now, let's evaluate Options (B) and (D). If switch $S_1$ is kept closed for a long time, the circuit reaches a steady state.

The Golden Rule: In a DC circuit, a fully charged capacitor acts as an open circuit, completely blocking the flow of direct current.

With the current dropping to zero, there is no voltage drop across any of the resistors. However, the capacitors have stored charge. Because they are all part of the same single loop, the capacitors $C_1$, $C_4$, and $C_3$ are effectively in series. We can find their equivalent capacitance:

The total charge pumped by the $5\text{V}$ battery is:

Since they are in series, this exact same charge of $40\mu \text{C}$ resides on each capacitor. We can now find the voltage across $C_1$:

This confirms that Option (D) is correct.

What about Option (B)? It asks for the voltage difference between points P and Q. Let's assume point Q (on the bottom wire) is at $0\text{V}$. Since there is no current, there is no voltage drop across the $70\Omega$ resistor. The potential at point P is entirely determined by the voltage across $C_1$, which we just calculated as $4\text{V}$. Therefore, $V_P-V_Q=4\text{V}$, not $10\text{V}$. Option (B) is incorrect.

The Transient State (Closing $S_2$)

Finally, let's tackle Option (A). After reaching the steady state, switch $S_2$ is suddenly closed.

The Golden Rule: The voltage across a capacitor cannot change instantaneously. At the exact moment a switch is closed ($t=0^{+}$), capacitors maintain their stored voltage and act mathematically like ideal batteries.

Let's draw the equivalent circuit for this specific instant: - Left Branch: $C_1$ acts as a $4\text{V}$ battery in series with the $70\Omega$ resistor. - Middle Branch: The uncharged $C_2$ acts as a short circuit, leaving the $10\text{V}$ battery in series with the $30\Omega$ resistor. - Right Branch: $C_4$ and $C_3$ both act as $0.5\text{V}$ batteries. Combining these with the $5\text{V}$ battery yields an equivalent $4\text{V}$ battery in series with a total resistance of $130\Omega$ ($30\Omega +100\Omega$).

Notice something beautiful here! Both the left and right branches have an equivalent voltage of $4\text{V}$ relative to ground. We can combine them into a single parallel branch with a $4\text{V}$ source and an equivalent resistance:

Now, we have a simple loop where the $10\text{V}$ battery in the middle branch opposes the combined $4\text{V}$ battery. The instantaneous current through the middle branch is:

This is far from the $0.2\text{A}$ claimed in Option (A), making it incorrect.

By mastering these three phases, you can confidently dismantle any complex RC circuit problem!