Electrostatics Question 584

Question: A 10μF 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 volt . The capacitance of second capacitor is [MP PET 1999; DPMT 2000]

Options:

A) 10μF

B) 20μF

C) 30μF

D) 15μF

Show Answer

Answer:

Correct Answer: D

Solution:

V=C1V1+C2V2C1+C2

therefore 20=10×50+C2×010+C2

therefore 200+20C2=500

therefore C2=15μF

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