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 $
Show Answer
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 $
