Question

Initially the circuit is in steady state. When the switch S is closed, the heat generated in the circuit will be:

• A. $\frac{\epsilon^2C}{2}$
• B. $2\epsilon^2C$
• C. $\frac{3\epsilon^2C}{2}$
• D. Zero

Answer 1

Answer: (D)

Zero

Answer 2

The problem asks for the heat generated in the circuit when the switch S is closed, given that the circuit is initially in a steady state with the switch S open.

1. Interpret the Circuit: The circuit consists of two identical branches connected in parallel. Each branch contains a capacitor (C) and a battery ($\epsilon$). The negative terminals of both batteries are connected to a common ground (0V). The positive terminals of the batteries are connected to the bottom plates of their respective capacitors. The switch S connects the top plates of the two capacitors.

2. Analyze the Initial State (Switch S open):

   • Since the circuit is in a steady state and the capacitors are connected directly across the batteries (positive terminal to bottom plate, negative terminal to ground), each capacitor will be fully charged.
   • The potential of the bottom plate of each capacitor is $\epsilon$ (relative to ground).
   • The potential difference across each capacitor is $\epsilon$.
   • Therefore, the potential of the top plate of each capacitor is $V_{top} = V_{bottom} + \epsilon = \epsilon + \epsilon = 2\epsilon$.
   • So, in the initial state, the top plate of the left capacitor is at $2\epsilon$, and the top plate of the right capacitor is also at $2\epsilon$.

3. Analyze the Final State (Switch S closed):

   • When the switch S is closed, it connects the top plates of the two capacitors.
   • Since both top plates were already at the same potential ($2\epsilon$), closing the switch does not create any potential difference between them.
   • Therefore, no charge will flow through the switch or redistribute within the circuit.
   • The charges on the capacitors and the energy stored in them remain unchanged.

4. Calculate Heat Generated:

   • Heat generated in a circuit is due to energy dissipation, typically when charge flows through resistance.
   • Since no charge flows when the switch is closed (because the potentials are already equal), no energy is dissipated as heat.
   • Alternatively, using the energy conservation formula: $H = W_{battery} - \Delta U$.
     • Change in stored energy, $\Delta U = U_{final} - U_{initial}$. Since no change occurs, $\Delta U = 0$.
     • Work done by batteries, $W_{battery}$. Since no charge flows through the batteries after closing the switch, $W_{battery} = 0$.
     • Therefore, $H = 0 - 0 = 0$.

The heat generated in the circuit is Zero.