Question:
In the circuit shown, the energy stored in the capacitor is \( n \, \mu \text{J} \). The value of \( n \) is:
In the circuit shown, the energy stored in the capacitor is \( n \, \mu \text{J} \). The value of \( n \) is:
Updated On: Mar 26, 2026
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Correct Answer: 75
Solution and Explanation
To find the energy stored in the capacitor, we first analyze the circuit to determine the potential difference across the capacitor.
The total current in the circuit can be calculated as:
\[
I_s = \frac{12}{3 + 9} = \frac{12}{12} = 1 \, \text{A}.
\]
The current through the upper branch (\( AB \)) is:
\[
I_1 = \frac{12}{4 + 2} = \frac{12}{6} = 2 \, \text{A}.
\]
The voltage drop across the \( 3 \, \Omega \) resistor is:
\[
V_A - V_C = 3I_1 = 3 \cdot 1 = 3 \, \text{V}.
\]
The voltage at point \( A \) with respect to point \( D \) is:
\[
V_A - V_D = 3 \cdot 1 + 2 \cdot 2 = 3 + 4 = 8 \, \text{V}.
\]
The voltage across the capacitor is:
\[
V_A - V_B = 5 \, \text{V}.
\]
The energy stored in the capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2,
\]
where \( C = 6 \, \mu \text{F} \) and \( V = 5 \, \text{V} \).
Substitute the values:
\[
U = \frac{1}{2} \cdot 6 \cdot 5^2 = \frac{1}{2} \cdot 6 \cdot 25 = 75 \, \mu \text{J}.
\]
Thus, the energy stored in the capacitor is \( \boxed{75 \, \mu \text{J}} \).
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