Question:
A \(10\ \mu\mathrm{F}\) capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :
A \(10\ \mu\mathrm{F}\) capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :
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Charge is conserved when the charged capacitor is connected in parallel to the uncharged one.
Updated On: Nov 5, 2026
- \(20\ \mu\mathrm{F}\)
- \(15\ \mu\mathrm{F}\)
- \(10\ \mu\mathrm{F}\)
- \(30\ \mu\mathrm{F}\)
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The Correct Option is B
Solution and Explanation
Initial charge \(Q=C_1V_1=10\ \mu\mathrm{F}\times50\ \mathrm{V}=500\ \mu\mathrm{C}\).
After connecting in parallel, charge is conserved: \((C_1+C_2)V=Q\Rightarrow(10+C_2)\times20=500\).
\(10+C_2=25\Rightarrow C_2=15\ \mu\mathrm{F}\).
Answer: (2) \(15\ \mu\mathrm{F}\)
After connecting in parallel, charge is conserved: \((C_1+C_2)V=Q\Rightarrow(10+C_2)\times20=500\).
\(10+C_2=25\Rightarrow C_2=15\ \mu\mathrm{F}\).
Answer: (2) \(15\ \mu\mathrm{F}\)
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