A square loop of resistance 16 Ω is connected with battery of 9 V and internal resistance of 1 Ω. In steady state, find energy stored in capacitor of capacity C = 4 μF as shown (at steady state current divides symmetrically)

Correct Option: 25.92 μJ

Step-by-Step Explanation:

1. Determine the Resistance of the Sides of the Square Loop:
The total resistance of the square loop is given as $R=16\Omega$.
Since the loop is a square consisting of four equal sides, the resistance of each side is:
$R_{\text{side}}=\frac{16}{4}=4\Omega$

2. Analyze the Loop at Steady State:
In the steady state, the capacitor behaves as an open circuit (infinite resistance), meaning no current flows through the diagonal branch containing the capacitor $C$.
Let the connection points of the external circuit be at the midpoints of the left and right sides of the square:
- Let $P_1$ be the midpoint of the left side (connecting point).
- Let $P_2$ be the midpoint of the right side (connecting point).
- Let the corners of the square loop be labeled as: Bottom-Left (BL), Top-Left (TL), Top-Right (TR), and Bottom-Right (BR).

The current enters at $P_1$ and divides into two parallel paths to reach $P_2$:
- Top Path: $P_1\to \text{TL}\to \text{TR}\to P_2$
The resistance of this path is:
$R_{\text{top}}=2\Omega +4\Omega +2\Omega =8\Omega$
- Bottom Path: $P_1\to \text{BL}\to \text{BR}\to P_2$
The resistance of this path is:
$R_{\text{bottom}}=2\Omega +4\Omega +2\Omega =8\Omega$

3. Calculate the Equivalent Resistance and Circuit Current:
The equivalent resistance $R_{\text{eq}}$ of the parallel combination is:
$R_{\text{eq}}=\frac{8\times 8}{8+8}=4\Omega$
With a battery of EMF $E=9\text{V}$ and internal resistance $r=1\Omega$, the total current $I$ in the circuit is:
$I=\frac{E}{R_{\text{eq}}+r}=\frac{9}{4+1}=1.8\text{A}$

By symmetry, the current divides equally between the top and bottom paths:
$I_1=I_2=\frac{1.8}{2}=0.9\text{A}$

4. Calculate the Potential Difference across the Capacitor:
The capacitor is connected across the diagonal between the Bottom-Left corner (BL) and the Top-Right corner (TR).
Let us calculate the electric potentials at BL and TR relative to the entry node $P_1$ (assuming $V_{P_1}=0\text{V}$):
- Potential at BL:
$V_{\text{BL}}=V_{P_1}-I_2{R}_{P_1\to \text{BL}}=0-0.9\times 2=-1.8\text{V}$
- Potential at TR (following the top path):
$V_{\text{TR}}=V_{P_1}-I_1{R}_{P_1\to \text{TL}}-I_1{R}_{\text{TL}\to \text{TR}}=0-0.9\times 2-0.9\times 4=-1.8-3.6=-5.4\text{V}$

The potential difference $V_C$ across the capacitor is:
$V_C=V_{\text{BL}}-V_{\text{TR}}=-1.8-(-5.4)=3.6\text{V}$

5. Find the Energy Stored in the Capacitor:
The energy $U$ stored in a capacitor of capacity $C=4\mu \text{F}$ is:
$U=\frac{1}{2}C{V}_C^2$
$U=\frac{1}{2}\times 4\mu \text{F}\times {\left(3.6\text{V}\right)}^2$
$U=2\times 12.96\mu \text{J}=25.92\mu \text{J}$