Animated Solution for Physics - Current Electricity: An ideal cell of emf 10 V is connected in circuit shown in figure. Each resistance is $2\Omega$. The potential difference (in V) across the capacitor when it is fully charged is ......... .

Welcome to this fascinating journey through a classic DC circuit problem! Our mission is to uncover the potential difference across the capacitor when it is fully charged. The golden key to unlocking this puzzle lies in the phrase fully charged.

Imagine the capacitor as a bucket filling with water; once it reaches its brim, no more water can flow in. Similarly, in a DC circuit, a fully charged capacitor acts as an absolute roadblock—an open circuit. This means that absolutely no current will flow through the top branch containing the capacitor $C$ and the resistor $R_1$. For the purpose of calculating the currents flowing through the rest of the circuit, we can effectively erase this top branch from our minds.

With the top branch out of the picture, let's analyze the remaining active circuit to find its equivalent resistance. The current from the $10\text{V}$ battery flows out and passes entirely through $R_3$. After $R_3$, it reaches a junction and splits, much like a river dividing into two streams. One stream flows through $R_4$, and the other navigates through $R_2$ and $R_5$.

Notice that resistors $R_2$ and $R_5$ are connected end-to-end with no alternative paths between them, meaning they are in series. Their combined resistance is simply $2\Omega +2\Omega =4\Omega$. This $4\Omega$ combination is in parallel with $R_4$, which is $2\Omega$. Using the parallel resistance formula, their equivalent resistance is:

Finally, we add the series resistor $R_3$ to find the total equivalent resistance of the entire circuit:

With the total resistance known, Ohm's law effortlessly gives us the total current drawn from the battery:

Now comes the elegant part: tracking the electric potential at different nodes. Let's establish a baseline by assuming the negative terminal of the battery is at $0\text{V}$, which instantly makes the positive terminal $10\text{V}$.

The entire $3\text{A}$ current flows through $R_3$. The voltage drop across $R_3$ is $3\text{A}\times 2\Omega =6\text{V}$. Therefore, the potential at the node immediately after $R_3$ (let's call it Node A) drops to:

At Node A, the current splits. We need to find the potential at the node between $R_2$ and $R_5$ (let's call it Node E) because the capacitor is connected there. First, we find the current flowing through the $R_2$ branch using the current divider rule:

The voltage drop across $R_2$ is $1\text{A}\times 2\Omega =2\text{V}$. Subtracting this drop from the potential at Node A gives us the potential at Node E:

Finally, let's bring our focus back to the capacitor. Its left plate is connected to resistor $R_1$. Since we established earlier that no current flows through $R_1$, there is absolutely zero voltage drop across it. Consequently, the left plate of the capacitor sits at the exact same potential as the positive terminal, which is $10\text{V}$.

The right plate of the capacitor is connected to Node E, which we just calculated to be at $2\text{V}$. The potential difference across the capacitor is simply the difference between its two plates:

And there we have it! By systematically breaking down the circuit and tracking the potentials, we've arrived at the final answer of $8\text{V}$.