Animated Solution for Physics - Current Electricity: Comprehension Passage

Imagine you are looking at a city's water supply network, where pipes represent wires and water pressure represents voltage. In our circuit, we have two batteries acting as pumps, a capacitor acting as a storage tank, and a switch that can change the flow of the entire system.

The problem asks us to find the charge on the capacitor in two different steady states. The key phrase here is "for a long time".

In a DC circuit, after a long time, a capacitor becomes fully charged. Once it's full, it acts as an open circuit, meaning no steady current can flow through its branch. This simplifies our analysis immensely!

The Switch at Position P

Let's analyze the first scenario. The switch is connected to position P. This brings the $1\text{ V}$ battery into our active circuit.

To find the charge on the capacitor, we first need to determine the potential difference across it. We can use Nodal Analysis to do this quickly. Let's set the bottom continuous wire as our reference ground, meaning its potential is $0\text{ V}$. We will call the junction directly above the capacitor Node A, with an unknown potential $V_A$.

Now, we apply Kirchhoff's Current Law (KCL) at Node A. Since the capacitor branch is an open circuit, the sum of the currents flowing left and right must be zero.

Let's solve this equation. Multiplying the entire equation by 2 to clear the denominator, we get:

Expanding and simplifying this gives:

With the potential at Node A known, the potential difference across the capacitor is simply $\frac{4}{3}\text{ V}$. The charge $q_1$ is the capacitance multiplied by this voltage.

That's our first answer!

The Switch at Position Q

Now for the second part of the journey. The switch is moved to position Q.

Look closely at the circuit diagram. The $1\Omega$ resistor is now directly connected to the bottom wire, which is at $0\text{ V}$. The $1\text{ V}$ battery is completely bypassed and is no longer part of the active circuit.

We apply Kirchhoff's Current Law at Node A once again for this new steady state. The left end of the $1\Omega$ resistor is now at $0\text{ V}$.

Solving this is just as straightforward. Multiplying by 2 gives:

This simplifies to:

Finally, the new charge $q_2$ is the capacitance times this new voltage.

Notice how changing the switch position redistributed the potentials and exactly halved the stored charge on the capacitor.

This problem beautifully demonstrates the power of Nodal Analysis in steady-state DC circuits. By simply identifying the nodes and applying KCL, we bypassed complex loop equations and arrived at the solution elegantly.