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 $
BETA
