Electrostatics Question 559

Question: 2μF capacitance has potential difference across its two terminals 200 volts . It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes20 volts . Then the capacity of another capacitance will be [CPMT 1991; DPMT 2001]

Options:

A) 2μF

B) 4μF

C) 18μF

D) 10μF

Show Answer

Answer:

Correct Answer: C

Solution:

By using, common potential V=C1V1+C2V2C1+C2

therefore 20=2×200+C2×02+C2

therefore C2=18μF



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