Question

In the circuit shown in figure $C_1=C_2=C$ and capacitor $C_1$ is having initial charge Q. the switch is closed at t=0. Then heat loss in the circuit till the current becomes zero is ______.

Answer 1

Refer image (1)

Initially when the switch s is then the charge in capacitor $C_1=Q$

Energy stored $=\frac{1}{2}\frac{a^2}{c}$

Now switch is closed and current becomes zero

Refer image (2)

Let us say, there is $q_1$ charge on capacitor G

and $q_2$ charge an capacitor $C_2$

Since, charge is conserved

$q_1+q_2=Q$

applying lirchoff's loop law

$\frac{q_1}{c}-\frac{q_2}{c}=0$

$\Rightarrow q_1=q_2=q$

form equation (1)

$q+q=Q$

$q=\frac{Q}{2}$

$\Rightarrow q_1=q_2=\frac{Q}{2}$

energy stored in $C_1=\frac{1}{2c}(\frac{Q}{2}{)}^2=\frac{Q^{2}w}{2\times 4c}=\frac{Q^2}{8c}$

energy stored in $C_1=\frac{1}{2c}(\frac{Q}{2}{)}^2=\frac{Q^{2}w}{2\times 4c}=\frac{Q^2}{8c}$

Total energy after s is close $=\frac{Q^2}{8c}+\frac{Q^2}{8c}$

$=\frac{Q^2}{4c}$

Energy loss in heat - Initial energy -final energy

$=\frac{Q^2}{2c}-\frac{Q^2}{4c}$

Heat loss $=\frac{Q^2}{4c}$