Electrostatics Question 584
Question: A $ 10\mu F $ capacitor is charged to a potential difference of $ 50\ V $ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $ 20\ volt $ . The capacitance of second capacitor is [MP PET 1999; DPMT 2000]
Options:
A) $ 10\mu F $
B) $ 20\mu F $
C) $ 30\mu F $
D) $ 15\mu F $
Show Answer
Answer:
Correct Answer: D
Solution:
$ V=\frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}+C _{2}} $
therefore $ 20=\frac{10\times 50+C _{2}\times 0}{10+C _{2}} $
therefore $ 200+20C _{2}=500 $
therefore $ C _{2}=15\mu F $
