Question:
A 10 \(\mu\)F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is:
A 10 \(\mu\)F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is:
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When capacitors are connected in parallel, the total charge is conserved, and the final voltage across both capacitors is the same.
Updated On: Apr 22, 2026
- 10 \(\mu\)F
- 20 \(\mu\)F
- 30 \(\mu\)F
- 15 \(\mu\)F
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The Correct Option is D
Solution and Explanation
Step 1: Apply the principle of conservation of charge.
When capacitors are connected in parallel, the total charge before and after connection is conserved. Initially, the charge on the 10 \(\mu\)F capacitor is \( Q_1 = C_1 V_1 = 10 \times 50 = 500 \, \mu C \).
Step 2: Set up the equation.
After connection, the total charge is shared by both capacitors. The final potential difference across both capacitors is 20 V. Let \( C_2 \) be the capacitance of the second capacitor, then: \[ Q_1 = (C_1 + C_2) \times 20 \] Substitute the known values: \[ 500 = (10 + C_2) \times 20 \] \[ 500 = 200 + 20 C_2 \] \[ 20 C_2 = 300 \] \[ C_2 = 15 \, \mu F \]
Step 3: Conclusion.
The capacitance of the second capacitor is 15 µF, which is option (4).
When capacitors are connected in parallel, the total charge before and after connection is conserved. Initially, the charge on the 10 \(\mu\)F capacitor is \( Q_1 = C_1 V_1 = 10 \times 50 = 500 \, \mu C \).
Step 2: Set up the equation.
After connection, the total charge is shared by both capacitors. The final potential difference across both capacitors is 20 V. Let \( C_2 \) be the capacitance of the second capacitor, then: \[ Q_1 = (C_1 + C_2) \times 20 \] Substitute the known values: \[ 500 = (10 + C_2) \times 20 \] \[ 500 = 200 + 20 C_2 \] \[ 20 C_2 = 300 \] \[ C_2 = 15 \, \mu F \]
Step 3: Conclusion.
The capacitance of the second capacitor is 15 µF, which is option (4).
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