An Uncharged Capacitor Is Connected To A Battery. Show That Half The Energy Supplied By Battery Is Lost As Heat While Charging The Conductor

The charge stored on capacitor = Q

If Q charge has passed through battery work done by battery = QV

Energy stored in capacitor = \(frac{1}{2} CV^{2} = frac{1}{2} QV\)

Since Q = CV


The energy lost as heat = Work done by battery − Energy store in a capacitor

= \(QV – frac{1}{2} QV

= \(frac{1}{2} QV\)

Explore more such questions and answers at BYJU’S.

Was this answer helpful?


0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




Free Class