Animated Solution for Physics - Current Electricity: In the given circuit, with steady current, the potential difference across the capacitor must be

Welcome to this fascinating circuit problem! At first glance, the circuit might look a bit intimidating with its multiple branches, batteries, and a capacitor sitting right in the middle. But don't worry, the key to unlocking this problem lies in understanding the magic words: steady current.

In a DC circuit, a capacitor acts like a sponge for electrical charge. When you first connect the circuit, current flows as the capacitor charges up. However, once it reaches its maximum capacity—a state we call the steady state—it stops accepting any more charge. At this point, the capacitor acts exactly like an open switch. This is a crucial realization because it means absolutely zero current will flow through the middle branch containing the capacitor $C$.

With the middle branch effectively removed from our current calculations, the circuit simplifies beautifully into a single, outer rectangular loop. Let's analyze this loop to find the steady current $i$.

We have two batteries in this loop: a $2V$ battery on the bottom and a $V$ battery on the top. If you trace the loop, you'll notice that their positive terminals are facing the same direction (leftwards), which means they are connected in opposition. They are fighting each other to push the current! The stronger $2V$ battery wins, so the net electromotive force (EMF) driving the current is:

The total resistance in this outer loop is simply the sum of the two resistors, as they are now in series:

Using Ohm's Law, the steady current $i$ flowing through the outer loop is:

Now, we need to find the potential difference across the capacitor. Let's call the nodes where the capacitor connects $A$ (top) and $B$ (bottom). The voltage across the capacitor is simply $V_A-V_B$.

By carefully applying Kirchhoff's Voltage Law (KVL) and tracing the potential drops along the branches, we can determine this difference. The geometry of the circuit and the balanced potential drops dictate that the potential difference between nodes $A$ and $B$ exactly equals the voltage drop across the top resistor $R$.

Therefore, the voltage across the capacitor $V_C$ is:

Substituting the value of the current $i$ we found earlier:

And there we have it! The potential difference across the capacitor in the steady state is $\frac{V}{3}$. This elegant result shows how understanding the physical behavior of components like capacitors can drastically simplify complex circuit analysis.