JEE Physics 2024 Question with Solution

Given: A square loop is made from a wire of total resistance $16 \, \Omega$, so each side has resistance $4 \, \Omega$. A battery of emf $9 \, \text{V}$ with internal resistance $1 \, \Omega$ is connected across one side, and a capacitor of capacitance $4 \, \mu \text{F}$ is connected across a diagonal.

Find: The value of $x$ if stored energy is $x^2 \, \mu \text{J}$.

In steady state, no current flows through the capacitor branch, so we first find the potential difference across the diagonal of the square.

Between the two battery terminals, one branch is the side of the square having resistance $4 \, \Omega$. The other branch is the remaining three sides in series, giving $12 \, \Omega$. Hence the external resistance across the battery is

Including internal resistance, total resistance becomes

So the current supplied by the battery is

The terminal voltage across the square loop is

Now the branch of resistance $12 \, \Omega$ carries current

Hence the potential drop across one side of that branch is

Therefore, the potential difference across the diagonal ends comes out to be

The energy stored in the capacitor is

Substituting $C = 4 \times 10^{-6} \text{ F}$ and $V_C = \frac{9}{4} \text{ V}$,

This does not match the source solution text. The provided page states the final form as $x^2 \, \mu \text{J}$ with answer $81$, and the provided correct answer field also gives $(81)$.

Therefore, from the provided source, the recorded answer is $x = 81$.

The provided the solution is internally inconsistent. It marks Option C while the question is numerical-value type, and it also shows an incorrect direct use of $V = 9 \, \text{V}$ across the capacitor before later stating $25.92 \, \mu \text{J}$ without valid circuit justification.

Because the solution's itself is inconsistent, the final answer has been taken from the answer key, which states $(81)$.