Electrostatics Question 544

Question: 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} $ as . 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}}= $

$ \frac{C _{1}+C _{2}}{C _{1}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक