In the question 42, if the switch is opened after the capacitor has been charged, it will discharge with a time constant

Options

A. RC
B. 2 RC
C. `(1)/(2)RC`
D. RC log2

Explanation

To solve the problem of finding the time constant for the discharge of a capacitor after the switch is opened, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- Initially, the capacitor is charged when the switch is closed. When the switch is opened, the circuit changes, and we need to analyze the discharging process of the capacitor.

2. **Identify the Components**:
- In the circuit, we have a capacitor (C) and two resistors (R1 and R2) that are connected in series when the switch is opened.

3. **Determine the Equivalent Resistance**:
- Since R1 and R2 are in series, the equivalent resistance (R_eq) can be calculated as:
\[ R_{eq} = R_1 + R_2 = R + R = 2R \]

4. **Use the Formula for Time Constant**:
- The time constant (τ) for an RC circuit is given by the formula:
\[ \tau = R_{eq} \times C \]
- Substituting the equivalent resistance we found:
\[ \tau = (2R) \times C \]

5. **Final Result**:
- Therefore, the time constant for the discharging capacitor is:
\[ \tau = 2RC \]

### Conclusion:
The time constant for the discharge of the capacitor after the switch is opened is \( 2RC \).

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