Animated Solution for Physics - Current Electricity: A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed, the total amount of charge that flows from $Y$ to $X$ is

Unraveling a Classic RC Circuit Puzzle

Imagine a circuit where two distinct worlds exist side-by-side: a branch of capacitors storing static energy, and a branch of resistors guiding a steady flow of current. What happens when you suddenly bridge these two worlds? This classic JEE problem explores exactly that scenario, testing your grasp of steady-state behavior, potential dividers, and the fundamental law of charge conservation.

Before the switch $S$ is closed, the circuit consists of two parallel branches connected across a $9\text{V}$ battery. The top branch contains two capacitors ($3\mu \text{F}$ and $6\mu \text{F}$) in series. In a DC circuit, once the capacitors are fully charged (steady state), they act as open circuits, meaning no current flows through the top branch.

Let's find the initial charge on these capacitors. Since they are in series, their equivalent capacitance is:

The total charge drawn from the battery is:

Now, let's focus on the isolated node $X$ between the two capacitors. The left side of the battery is the positive terminal. Therefore, the left plate of the $3\mu \text{F}$ capacitor is positive ($+18\mu \text{C}$), and its right plate (connected to $X$) is negative ($-18\mu \text{C}$). Similarly, the left plate of the $6\mu \text{F}$ capacitor (also connected to $X$) is positive ($+18\mu \text{C}$).

A Shift in Potential

When we close switch $S$, we create a direct conductive path between node $X$ and node $Y$. This forces the two nodes to share the exact same electrical potential ($V_X=V_Y$). To find this new potential, we must look at the resistor branch.

Unlike the capacitors, the resistors allow a steady current to flow. The total resistance of the series combination is $R=3\Omega +6\Omega =9\Omega$. Using Ohm's law, the steady-state current is:

Let's assign $0\text{V}$ to the negative terminal (right side) and $9\text{V}$ to the positive terminal (left side). As the $1\text{A}$ current flows through the $3\Omega$ resistor, there is a potential drop of $V=I\times R=1\times 3=3\text{V}$.

Therefore, the potential at node $Y$ becomes:

Because the switch is closed, the potential at node $X$ is now firmly anchored at $6\text{V}$.

The Charge Redistribution

With the potential at node $X$ fixed at $6\text{V}$, the capacitors must adjust their charges to match the new potential differences across them.

For the $3\mu \text{F}$ capacitor, the potential difference is now $9\text{V}-6\text{V}=3\text{V}$. Its new charge is:

For the $6\mu \text{F}$ capacitor, the potential difference is $6\text{V}-0\text{V}=6\text{V}$. Its new charge is:

Let's re-examine node $X$. The right plate of the first capacitor now holds $-9\mu \text{C}$, and the left plate of the second capacitor holds $+36\mu \text{C}$. The total final charge at node $X$ is:

Since node $X$ started with a net charge of zero and ended with $+27\mu \text{C}$, this extra charge must have traveled through the only available path: the switch. Therefore, exactly $27\mu \text{C}$ of charge flowed from node $Y$ to node $X$.