Question

In the circuit given in the figure, switch S is closed at position 1 for long time. Find the total heat generated in resistor of resistance $2r_0$, when the switch S is open at position 1 and closed at position 2.

• A. $\frac{C_0{E}_0^2}{2}$
• B. $C_0{E}_0^2$
• C. $\frac{C_0{E}_0^2}{3}$
• D. $\frac{C_0{E}_0^2}{4}$

Answer 1

The correct option is C

$\frac{C_0{E}_0^2}{3}$

Step 1: Switch at position 1: Since circuit is in the steady state, the current through circuit is zero. According to the loop rule,

$E_0-\frac{q_0}{C_0}=0$

$q_0=C_0{E}_0$

Step 2: Switch at position 2: In this case, total energy stored in capacitor appears as heat energy in the resistor.

$\Delta H=I^{2}RT,\Delta H\propto R$

$\frac{\Delta H_1}{\Delta H_2}=\frac{R_1}{R_2}=\frac{r_0}{2r_0}=\frac{1}{2}or\Delta H_2=2\Delta H_1$

But $\Delta H=\Delta H_1+\Delta H_2=\frac{\Delta H_2}{2}+\Delta H_2=\frac{3}{2}\Delta H_2$

Therefore $\Delta H_2=\frac{2}{3}\Delta H=\frac{2}{3}\times \frac{1}{2}{C}_0{E}_0^2=\frac{1}{3}{C}_0{E}_0^2$