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Step-by-Step Solution
Step 1: Determine the Resistance of Each Side
The total resistance of the wire is 16\,\Omega and it is formed into a square, so the square has 4 equal sides. Hence, each side of the square has a resistance:
R_{\text{side}} = \frac{16\,\Omega}{4} = 4\,\Omega.
Step 2: Circuit Configuration
The 9 V battery with internal resistance 1\,\Omega is connected across one side of the square. That side has a resistance of 4\,\Omega . The remaining three sides (each also 4\,\Omega ) are connected in series with each other, giving 4 + 4 + 4 = 12\,\Omega , and this 12\,\Omega path is in parallel with the single 4\,\Omega side.
Step 3: Parallel Equivalent Resistance
The 4\,\Omega side is in parallel with the 12\,\Omega side. The equivalent resistance R_{\text{parallel}} of the parallel combination is:
R_{\text{parallel}} = \frac{4 \times 12}{4 + 12}
= \frac{48}{16}
= 3\,\Omega.
Step 4: Total Circuit Resistance and Current
The internal resistance of the battery ( 1\,\Omega ) is in series with R_{\text{parallel}} = 3\,\Omega , so the total circuit resistance is:
R_{\text{total}} = 3\,\Omega + 1\,\Omega = 4\,\Omega.
The current leaving the battery is then given by Ohm’s law:
I = \frac{9\,\mathrm{V}}{4\,\Omega} = 2.25\,\mathrm{A}.
Step 5: Voltage Across the Square Loop
The voltage drop across the internal resistor ( 1\,\Omega ) is:
I \times r = 2.25\,\mathrm{A} \times 1\,\Omega = 2.25\,\mathrm{V}.
Therefore, the voltage remaining across the parallel network (the square loop) is:
V_{\text{square}} = 9\,\mathrm{V} - 2.25\,\mathrm{V} = 6.75\,\mathrm{V}.
Step 6: Node Voltages
Label the corners of the square as A, B, C, D, in clockwise order, with the battery connected across side AB. Let B be at 0 V and A be at 6.75\,\mathrm{V} (since AB sees the 6.75 V). The other path from B to C to D to A has 12\,\Omega total in series and is also across 6.75\,\mathrm{V}.
The current through the 12\,\Omega path is:
I_{\text{BCD}} = \frac{6.75\,\mathrm{V}}{12\,\Omega} = 0.5625\,\mathrm{A}.
The voltage drop across each 4 \,\Omega segment in that path is:
0.5625\,\mathrm{A} \times 4\,\Omega = 2.25\,\mathrm{V}.
Hence the potentials at each point are:
C is at 2.25\,\mathrm{V} (relative to B which is 0 V).
D is at 2.25 + 2.25 = 4.5\,\mathrm{V} .
A is at 4.5 + 2.25 = 6.75\,\mathrm{V} (consistent with side AB).
The capacitor is placed across the diagonal AC, so the voltage across AC is:
V_{AC} = V(A) - V(C) = 6.75\,\mathrm{V} - 2.25\,\mathrm{V} = 4.50\,\mathrm{V}.
Step 7: Energy Stored in the Capacitor
The capacitor has capacitance 4\,\mu\mathrm{F} and charges up to 4.50\,\mathrm{V}. The energy stored in a capacitor is given by:
U = \frac{1}{2} C V^2.
Substituting C = 4 \times 10^{-6}\,\mathrm{F} and V = 4.50\,\mathrm{V} , we get:
U
= \frac{1}{2} \times 4 \times 10^{-6}\,\mathrm{F} \times (4.50\,\mathrm{V})^2.
Since (4.50)^2 = 20.25,
U = \frac{1}{2} \times 4 \times 10^{-6}\,\mathrm{F} \times 20.25\,\mathrm{V}^2
= \frac{1}{2} \times 81 \times 10^{-6}\,\mathrm{J}
= 40.5 \times 10^{-6}\,\mathrm{J}
= 40.5\,\mu\mathrm{J}.
Step 8: Final Value of x
The question states the energy is \frac{x}{2}\,\mu\mathrm{J} . From our calculation, U = 40.5\,\mu\mathrm{J} , so:
\frac{x}{2} = 40.5 \quad \Longrightarrow \quad x = 81.
Final Answer
x = 81
