Question

An uncharged capacitor is connected to a battery.Show that half the energy supplied by battery is lost as heat while charging the conductor

Answer 1

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 .

$dU = v dq$

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

Then, energy supplied by the battery

$\int dE = \int Vdq = V*Q \\ \\ =V*CV=CV^2$

3) Energy stored in the capacitor ,

$\int dU = \int vdq \\ \\ = \int \frac{q}{C}dq \\ \\=\frac{Q^2}{2C}$

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

$\frac{C^2V^2}{2C}=\frac{1}{2}CV^2$

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.