An ideal cell of emf $10\ \text{V}$ is connected in the circuit shown in figure. Each resistance is $2\ \Omega$. The potential difference (in V) across the capacitor when it is fully charged is ______.
Numerical value type. Enter your answer.
Answer: 8
When the capacitor is fully charged no current flows through $C$ or $R_5$, so the right end of $C$ is at the potential of the right junction (the negative side of the cell, take it as $0$ V); the left wire is at $10$ V.
$R_1 + R_2 = 4\ \Omega$ is in parallel with $R_3 = 2\ \Omega$ (both between the left wire and the middle junction), giving $\tfrac43\ \Omega$, in series with $R_4$: total $\tfrac{10}{3}\ \Omega$, current $3$ A.
Middle junction: $3\times2 = 6$ V. Current through $R_1R_2$ branch $= \dfrac{10-6}{4} = 1$ A, so the junction of $R_1$ and $R_2$ is at $10 - 2 = 8$ V.
Voltage across the capacitor $= 8 - 0 = 8$ V.
Solution by Sreeraj P, M.Sc Physics