Question

For the circuit shown in figure, switch $S_1$ was closed for a long time and switch $S_2$ was open. Now the switch $S_1$ is opened and $S_2$ is closed at time $t=0$. Till the system attains the steady state.

• A. Total heat dissipated in resistor $R_2$ is $$\frac{C{V}_0^2}{6}$$
• B. Total charge flown through switch $S_2\text{is}C{V}_0$
• C. Total charge flown through switch $S_2$ is $$\frac{C{V}_0}{2}$$
• D. Total heat dissipated in resistor$R_2$ is $$\frac{C{V}_0^2}{4}$$

Answer 1

Answer: (C), (D)

Total heat dissipated in resistor$R_2$ is $$\frac{C{V}_0^2}{4}$$


After closing $S_2$:
$$\frac{q+C{V}_0}{C}+\frac{q}{C}=2V_0$$
After solving, we get:
$$q=\frac{C{V}_0}{2}$$
So, final charges on capacitors will be $\frac{C{V}_0}{2}$ and $\frac{3C{V}_0}{2}$
$$W_b=C{V}_0^2$$
$$U_i=\frac{1}{2}C{V}_0^2$$
$$U_f=\frac{1}{2}C(\frac{9V_0^2}{4}+\frac{V_0^2}{4})$$
Heat dissipated = $$\frac{C{V}_0^2}{4}$$