Electrostatics Question 544
A capacitor of capacity $ C _{1} $ is charged to the potential of $ V _{o} $ . On disconnecting with the battery, it is connected with a capacitor of capacity $ C _{2} $ . The ratio of energies before and after the connection of switch $ S $ will be
Options:
A) $ (C _{1}+C _{2})/C _{1} $
B) $ C _{1}/(C _{1}+C _{2}) $
C) $ C _{1}C _{2} $
D) $ C _{1}/C _{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Energy (U) $ =\frac{q^{2}}{2C} $ .
q remains same so $ U\propto \frac{1}{C} $
therefore $ \frac{U _{Before}}{U _{After}}= $ 1
$ \frac{C _{1}+C _{2}}{C _{1}} $
BETA
