Question

In the circuit shown in the figure, the capacitor $C$ is charged to a potential $V_0$. The heat generated in the circuit when the switch S is closed, is :

• A. $\Delta q=2C{V}_0$
• B. $\Delta H=4C{V}_o^2$
• C. $\Delta q=4C{V}_0$
• D. $\Delta H=8C{V}_o^2$

Answer 1

Answer: (B)

$\Delta H=4C{V}_o^2$

Energy given by battery = $\frac{1}{2}C(3V_o{)}^2=\frac{9C{V}_o^2}{2}$

Energy stored in the capacitor = $\frac{1}{2}C{V}_o^2$

Heat $=\frac{9C{V}_o^2}{2}-\frac{1}{2}C{V}_o^2$

$\Delta H=4C{V}_o^2$