Electrostatics Question 559

Question: $ 2\mu F $ 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. becomes$ 20\ volts $ . Then the capacity of another capacitance will be [CPMT 1991; DPMT 2001]

Options:

A) $ 2\mu F $

B) $ 4\mu F $

C) $ 18\mu F $

D) $ 10\mu F $

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Answer:

Correct Answer: C

Solution:

By using, common potential $ V=\frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}+C _{2}} $

therefore $ 20=\frac{2\times 200+C _{2}\times 0}{2+C _{2}} $

therefore $ C _{2}=18\mu F $



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