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

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: $x = 81$