A $$16 \Omega$$ wire is bend to form a square loop. A $$9$$ V battery with internal resistance $$1 \Omega$$ is connected across one of its sides. If a $$4 \mu F$$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $$\frac{x}{2} \mu J$$, where $$x =$$ _______.

Let the square be $$ABCD$$, taken in the clockwise sense.
Since the total length of the wire has resistance $$16\ \Omega$$, each side of the square has resistance

$$R = \frac{16\ \Omega}{4} = 4\ \Omega$$

The battery (emf $$E = 9\ \text{V}$$, internal resistance $$r = 1\ \Omega$$) is connected across side $$AB$$. Thus the terminals of the battery are joined to the same two nodes $$A$$ and $$B$$ that are already connected by the 4 $$\Omega$$ side.

For easier calculation, replace the battery by its Norton equivalent across the nodes $$A$$ and $$B$$.

Step 1: Norton conversion An emf $$E$$ in series with resistance $$r$$ is equivalent to a current source $$I_N$$ in parallel with the same resistance $$r$$, where

$$I_N = \frac{E}{r} = \frac{9}{1} = 9\ \text{A}$$

Therefore, across the nodes $$A$$ (positive) and $$B$$ (negative) we now have

• a current source of $$9\ \text{A}$$ from $$A \rightarrow B$$, and
• a resistor $$1\ \Omega$$ in parallel with that source.

Step 2: Find the equivalent resistance between $$A$$ and $$B$$ of the rest of the square.

Besides the internal $$1\ \Omega$$ resistor, two purely resistive paths join $$A$$ and $$B$$:

• Direct side $$AB$$ : $$4\ \Omega$$
• The remaining three sides $$A \rightarrow D \rightarrow C \rightarrow B$$ : $$4 + 4 + 4 = 12\ \Omega$$

Hence the three resistances $$1\ \Omega, 4\ \Omega, 12\ \Omega$$ are all in parallel between the same nodes. Their combined resistance is

$$\frac{1}{R_{\text{eq}}} \;=\; \frac{1}{1} + \frac{1}{4} + \frac{1}{12} \;=\; 1 + 0.25 + 0.0833 \;=\; 1.3333$$

$$R_{\text{eq}} = \frac{1}{1.3333} = 0.75\ \Omega$$

Step 3: Node voltage between $$A$$ and $$B$$

By KCL at the parallel combination, the current source of $$9\ \text{A}$$ drives the currents through the parallel resistors, producing a node voltage

$$V_{AB} = I_N \, R_{\text{eq}} = 9\ \text{A} \times 0.75\ \Omega = 6.75\ \text{V}$$

Take $$V_B = 0$$; hence

$$V_A = +6.75\ \text{V}$$

Step 4: Currents and potentials around the other three sides

The only path from $$A$$ to $$B$$ that does not go through the direct side is the series string $$A \rightarrow D \rightarrow C \rightarrow B$$ of total resistance $$12\ \Omega$$. The current through this path is

$$I = \frac{V_{AB}}{12\ \Omega} = \frac{6.75}{12} = 0.5625\ \text{A}$$

Voltage drop on each of the three equal resistors (4 $$\Omega$$) is

$$\Delta V = I \times 4 = 0.5625 \times 4 = 2.25\ \text{V}$$

Therefore, moving clockwise:

$$V_D = V_A - 2.25 = 6.75 - 2.25 = 4.50\ \text{V}$$
$$V_C = V_D - 2.25 = 4.50 - 2.25 = 2.25\ \text{V}$$

Step 5: Potential difference across a diagonal

Place the $$4\ \mu\text{F}$$ capacitor across any diagonal; choose diagonal $$AC$$. The voltage across this diagonal is

$$V_{AC} = V_A - V_C = 6.75 - 2.25 = 4.50\ \text{V}$$

(If the capacitor had been connected across the other diagonal $$BD$$, the magnitude is the same, $$4.50\ \text{V}$$.)

Step 6: Energy stored in the capacitor

The energy in a capacitor is $$U = \tfrac{1}{2} C V^2$$.

Here $$C = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}$$, $$V = 4.50\ \text{V}$$.

Hence

$$U = \tfrac{1}{2} \times 4 \times 10^{-6} \times (4.5)^2$$ $$\qquad= 2 \times 10^{-6} \times 20.25$$ $$\qquad= 40.5 \times 10^{-6}\ \text{J}$$

Expressing in microjoules: $$U = 40.5\ \mu \text{J}$$.

The question states this energy equals $$\dfrac{x}{2}\ \mu \text{J}$$, so

$$\frac{x}{2} = 40.5 \quad\Longrightarrow\quad x = 81$$

Therefore, $$x = 81$$.