Question

The heat generated in the circuit after closing the switch is (Given emf of battery = 2V)

Answer 1

The heat generated in the circuit is $2C \text{ Joules}$.

Answer 2

Solution:

1. When the capacitor is charged to a final potential $V = 2 \text{ V}$, the energy stored in the capacitor is

   $$U = \frac{1}{2} C V^2 = \frac{1}{2} C (2)^2 = 2C \text{ Joules}.$$

2. The work done by the battery in charging the capacitor is

   $$W = QV = (CV)(2) = 2CV = 4C \text{ Joules}.$$

3. The difference between the work done by the battery and the energy stored in the capacitor is dissipated as heat:

   $$\text{Heat generated} = W - U = 4C - 2C = 2C \text{ Joules}.$$