Electrostatics Question 578

Question: In the circuit shown here $ C _{1}=6\mu F,\ C _{2}=3\mu F $ and battery$ B=20V $ . The switch $ S _{1} $ is first closed. It is then opened and afterwards $ S _{2} $ is closed. What is the charge finally on $ C _{2} $

Options:

A) $ 120\mu C $

B) $ 80\mu C $

C) $ 40\mu C $

D) $ 20\mu C $

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Answer:

Correct Answer: C

Solution:

Common potential $ V=\frac{6\times 20+3\times 0}{(6+3)}=\frac{120}{9}Volt $

So, charge on $ 3\mu F $ capacitor

$ Q _{2}=3\times {{10}^{-6}}\times \frac{120}{9}=40\mu C $



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