Question

A 16Ω wire is bent to form a square loop. A 9 V battery with internal resistance 1Ω is connected across one of its sides. If a 4μF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be \(\frac{x}{2}\) μJ, where \(x =\) _____.

Answer 1

Correct Answer: 81

Step 1: Calculate Equivalent Resistance:

• The square loop consists of four \(4 \, \Omega\) resistors, each forming the sides of the square.
• The equivalent resistance \( R_{eq} \) between points \( A \) and \( B \) (opposite sides of the square) is:

\[ R_{eq} = \frac{12 \times 4}{12 + 4} = 3 \, \Omega \]

- Including the internal resistance of the battery, the total resistance is \( R = 3 + 1 = 4 \, \Omega \).

Step 2: Calculate the Current \( I \):

\[ I = \frac{V}{R} = \frac{9}{4} = 2.25 \, A \]

Step 3: Determine Current Through Each Side:

• Due to symmetry, the current through each \(4 \, \Omega\) resistor in parallel with the capacitor is \( I_1 \):

\[ I_1 = \frac{9}{16} = 0.5625 \, A \]

Step 4: Calculate Voltage Across the Capacitor:

\[ V_{AB} = I_1 \times 8 = 4.5 \, V \]

Step 5: Calculate Energy Stored in the Capacitor:

• Energy stored in a capacitor is given by:

\[ U = \frac{1}{2} C V_{AB}^2 \]

- Substitute values:

\[ U = \frac{1}{2} \times 4 \times (4.5)^2 = \frac{81}{2} \, \mu J \]

Step 6: Determine \( x \):

- Since \( U = \frac{x}{2} \, \mu J \), we find \( x = 81 \).

So, the correct answer is: \( x = 81 \)