A $16\ \Omega$ wire is bent to form a square loop. A $9\ \text{V}$ battery with internal resistance $1\ \Omega$ is connected across one of its sides. If a $4\ \mu\text{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\dfrac{x}{2}\ \mu\text{J}$, where $x =$ ______.
Numerical value type. Enter your answer.
Answer: 81
Each side has $4\ \Omega$. Across the side PQ where the battery is connected, $4\ \Omega$ is in parallel with the other three sides ($12\ \Omega$):
$$R_{ext} = \frac{4\times12}{16} = 3\ \Omega,\qquad I = \frac{9}{3+1} = 2.25\ \text{A}$$
$V_{PQ} = 2.25\times3 = 6.75\ \text{V}$. The path P–S–R–Q ($12\ \Omega$) carries $\dfrac{6.75}{12} = 0.5625\ \text{A}$, so each side of it drops $2.25\ \text{V}$.
Take $V_P = 6.75\ \text{V}$, $V_Q = 0$. Then $V_S = 4.5\ \text{V}$ and $V_R = 2.25\ \text{V}$. Diagonal PR: $6.75 - 2.25 = 4.5\ \text{V}$ (diagonal QS also gives $4.5\ \text{V}$).
$$U = \frac12CV^2 = \frac12\times4\times(4.5)^2 = 40.5\ \mu\text{J} = \frac{81}{2}\ \mu\text{J}$$
So $x = 81$.
Solution by Sreeraj P, M.Sc Physics