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Question: Answered & Verified by Expert
A 10 μF 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. What is the capacitance value (in μF) of the second capacitor?
PhysicsElectrostaticsJEE Main
Solution:
1281 Upvotes Verified Answer
The correct answer is: 15
q1=10×50=500 μC, C1=10 μF, C2=?, q2=0

As, V=q1+q2C1+C2

C1+C2=q1+q2V=500+020=25 μF

C2=25-C1=25-10=15 μF

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