An uncharged capacitor is connected to a battery show that half the energy supplied by the battery is lost while charging the capacitors

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Solution

Dear Student ,
Here in this case look if V be the voltage of the battery and q charge is accumulated in the batery then the energy supplied by the battery is , E = qV .
Now as we know that the energy stored in the capacitor while charging by a battery of voltage V is ,
$E' = \frac{1}{2} C V^{2} A g a i n w e k n o w t h a t q = C V s o, E' = \frac{1}{2} C V \times V = \frac{1}{2} q V S o t h e e n e r g y s t o r e d i n t h e c a p a c i t o r i s \frac{1}{2} q V . N o w t h e e n r g y l o s t i s, E - E' = q V - \frac{1}{2} q V = \frac{1}{2} q V T h i s e n e r g y i s l o s t b y t h e b a t t e r y d u r i n g t h e c h a r g i n g o f a c a p a c i t o r d u e t o t h e h e a t i n g b y t h e i n t e r n a l r e s i s \tan c e a n d o t h e r r e s i s \tan c e s p r e s e n t i n t h e c i r c u i t .$
Regards

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