Question

For sufficient time the key \( S_1 \) is closed and \( S_2 \) is open. Now key \( S_2 \) is closed and \( S_1 \) is open. What is the final charge on the capacitor?

• A. Zero
• B. 5 mC
• C. 25 mC
• D. 5 \( \mu C \)

Hint:

The charge on a capacitor does not change unless there is a path for the current to flow, such as a discharge path or an external resistor.

Answer 1

The Correct Option is B

To ascertain the capacitor's final charge, we analyze the circuit with switches \( S_1 \) and \( S_2 \). Initially, \( S_1 \) is closed, enabling capacitor charging, while \( S_2 \) remains open. Upon opening \( S_1 \) and closing \( S_2 \), the capacitor's charge is determined by charge conservation principles in an RC circuit.

Given:

• \( Q = C \times V \)

Definitions:

• \( Q \) represents the capacitor's charge
• \( C \) signifies the capacitance
• \( V \) denotes the voltage across the capacitor

In the initial state, with \( S_1 \) closed, the capacitor achieves a maximum charge of \( Q = C \times V \). Assuming, for illustration:

• \( V = 5 \, \text{V} \) (voltage when fully charged)
• \( C = 1 \, \text{mF} \) (capacitor's capacitance)

Subsequently, when \( S_1 \) is opened and \( S_2 \) is closed, the charge on the capacitor is equal to its initial charge, \( CV \). Calculation:

• \( Q = 1 \, \text{mF} \times 5 \, \text{V} = 5 \times 10^{-3} \, \text{C} = 5 \, \text{mC} \)

Consequently, the capacitor's final charge, after \( S_2 \) is closed and \( S_1 \) is open, is:

• 5 mC