Electrostatics Question 573
Question: A condenser of capacity $ C _{1} $ is charged to a potential $ V _{0} $ . The electrostatic energy stored in it is $ U _{0} $ . It is connected to another uncharged condenser of capacity $ C _{2} $ in parallel. The energy dissipated in the process is [MP PMT 1994]
Options:
A) $ \frac{C _{2}}{C _{1}+C _{2}}U _{0} $
B) $ \frac{C _{1}}{C _{1}+C _{2}}U _{0} $
C) $ ( \frac{C _{1}-C _{2}}{C _{1}+C _{2}} )U _{0} $
D) $ \frac{C _{1}C _{2}}{2(C _{1}+C _{2})}U _{0} $
Show Answer
Answer:
Correct Answer: A
Solution:
Loss of energy during sharing =$ \frac{C _{1}C _{2}{{(V _{1}-V _{2})}^{2}}}{2(C _{1}+C _{2})} $ In the equation, put $ V _{2}=0,\ V _{1}=V _{0} $ Loss of energy $ =\frac{C _{1}C _{2}V _{0}^{2}}{2(C _{1}+C _{2})} $
$ =\frac{C _{2}U _{0}}{C _{1}+C _{2}} $
$ [ \because \ U _{0}=\frac{1}{2}C _{1}V _{0}^{2} ] $
BETA
