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 $
Show Answer
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 $
BETA
