Question

Two capacitors of capacitance \( 1 \mu F \) and \( 2 \mu F \) are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacitance \( C_1 \) is

• A. 1800 \( \mu J \)
• B. 1600 \( \mu J \)
• C. 1400 \( \mu J \)
• D. 7200 \( \mu J \)

Hint:

For series capacitors, the charge on each capacitor is the same, and the energy stored can be calculated using \( E = \frac{1}{2} C V^2 \).

Answer 1

The Correct Option is B

For capacitors connected in series, the same charge \( Q \) is stored on both capacitors. The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the potential difference across the capacitor. We calculate the energy stored in the first capacitor.

Step 2: Conclusion.
The energy stored in the capacitor of capacitance \( C_1 \) is 1600 \( \mu J \), corresponding to option (b).