Electrostatics Question 566

Question: A condenser of capacitance 10μF has been charged to 100volts . It is now connected to another uncharged condenser in parallel. The common potential becomes 40volts . The capacitance of another condenser is [MP PET 1992]

Options:

A) 15μF

B) 5μF

C) 10μF

D) 16.6μF

Show Answer

Answer:

Correct Answer: A

Solution:

By using V=C1V1+C2V2C1+C2

therefore 40=10×100+C2×010+C2

therefore C2=15μF

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