Electrostatics Question 566

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

Options:

A) $ 15\mu F $

B) $ 5\mu F $

C) $ 10\mu F $

D) $ 16.6\mu F $

Show Answer

Answer:

Correct Answer: A

Solution:

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

therefore $ 40=\frac{10\times 100+C _{2}\times 0}{10+C _{2}} $

therefore $ C _{2}=15\mu F $



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