why the total energy drawn from the battery be 2 times the energy stored in the capacitor?

Open in App

Solution

Dear student
$L e t a c a p a c i t o r o f c a p a c i \tan c e C i s c h a r g e d w i t h t h e c e l l o f e m f E v o l t . C h a r g e s u p p l i e d b y t h e c e l l Q = C \times E E n e r g y s u p p l i e d b y t h e c e l l = W o r k d o n e b y t h e c e l l i s = E \times Q = E \times C \times E = C E^{2} E n e r g y s t o r e d i n t h e c a p a c i t o r = \frac{1}{2} C \times E^{2} E n e r g y s u p p l i e d b y t h e c e l l = I n c r e a s e i n e n e r g y o f c a p a c i t o r + H e a t p r o d u c e d C E^{2} = \frac{1}{2} C \times E^{2} + H e a t p r o d u c e d H e a t p r o d u c e d = \frac{1}{2} C \times E^{2} R e g a r d s$

Suggest Corrections

Similar questions

Q. A 10

μ

F capacitor is charged by a battery of emf 200 V. The energy drawn from the battery, and the energy stored in the capacitor, are respectively

A capacitor is connected with a battery and stores energy U. After removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

Q. A capacitor is charged with a battery and energy stored is

U

. After disconnecting the battery, another identical uncharged capacitor is connected in parallel with it. The energy stored in each capacitor is now

Q. Two capacitor of capacitance

10

μ F

and

20

μ F

are connected in series with

6 v

battery. If E is the energy stored in

20

μ F

capacitor what will be the total energy supplied by the battery in terms of E.

Q. A capacitor of capacitance

C_{1}

is charged by connecting it to a battery. The battery is now removed and this capacitor is connected to a second uncharged capacitor of capacitance

C_{2}

. If the charge gets distributed equally on the two capacitors after connection, the ratio of the total energy stored in the capacitors after connection to the total energy stored in them before connection is

Join BYJU'S Learning Program