Question

If an uncharged capacitor is connected to a battery, show that half of the energy supplied by the battery is used to charge the capacitor. ​

Answer 1

Explanation:

We have, battery of potential 'V' connected in circuit with capacitor of capacitance 'C'.

Initially, capacitor is uncharged.

So, when 'dq' charge passes the capacitor .

Energy supplied by the battery , dE = Vdq. (V is constant of battery )

Energy stored in capacitor, v is potential across the capacitor .

dU = v dqdU=vdq

2) Now, when capacitor is charged up-to 'Q'.

Then, energy supplied by the battery

\begin{lgathered}\int dE = \int Vdq = V*Q \\ \\ =V*CV=CV^2\end{lgathered}∫dE=∫Vdq=V∗Q=V∗CV=CV2

3) Energy stored in the capacitor ,

\begin{lgathered}\int dU = \int vdq \\ \\ = \int \frac{q}{C}dq \\ \\=\frac{Q^2}{2C}\end{lgathered}∫dU=∫vdq=∫Cqdq=2CQ2

This is also, written as (Q= CV) :

\frac{C^2V^2}{2C}=\frac{1}{2}CV^22CC2V2=21CV2

4) By using conservation of energy,

E = U + lost

=> lost = E - U = 1/2CV^2 .

Hence,

We can say half the energy supplied by the battery is lost as heat or in other forms.

Answer 2

Answer: Let us consider that the voltage of battery be V.

Charge on capacitor will be Q = CV

Now the work done by the battery should be V*Q = CV^2

And, Potential energy stored in capacitor is [ CV^2 ]/2

So you can clearly see that only half of energy is stored in the capacitor rest half is is released as heat or electromagnetic radiations.