A series R-C combination is connected to an AC voltage of angular frequency `omega=500 radian//s`. If the impendance of the R-C circuit is `Rsqrt(1.25)`, the time constant (in millisecond) of the circuit is

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To find the time constant of the series R-C circuit connected to an AC voltage, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Angular frequency, \( \omega = 500 \, \text{rad/s} \)
- Impedance, \( Z = R \sqrt{1.25} \)

2. **Express Impedance in Terms of R and XC**:
The impedance \( Z \) of a series R-C circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C = \frac{1}{\omega C} \).

3. **Substituting for XC**:
Substitute \( X_C \) into the impedance formula:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

4. **Square Both Sides**:
To eliminate the square root, square both sides:
\[ Z^2 = R^2 + \left(\frac{1}{\omega C}\right)^2 \]

5. **Substituting the Given Impedance**:
Substitute \( Z = R \sqrt{1.25} \):
\[ (R \sqrt{1.25})^2 = R^2 + \left(\frac{1}{500 C}\right)^2 \]
This simplifies to:
\[ 1.25 R^2 = R^2 + \frac{1}{(500 C)^2} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 1.25 R^2 - R^2 = \frac{1}{(500 C)^2} \]
\[ 0.25 R^2 = \frac{1}{(500 C)^2} \]

7. **Cross-Multiplying**:
Cross-multiply to find \( R^2 \):
\[ 0.25 R^2 (500 C)^2 = 1 \]

8. **Solving for RC**:
Rearranging gives:
\[ RC = \frac{1}{500 \sqrt{0.25}} = \frac{1}{500 \times 0.5} = \frac{1}{250} \]

9. **Calculating the Time Constant**:
The time constant \( \tau \) is given by \( \tau = RC \):
\[ \tau = \frac{1}{250} \, \text{seconds} \]

10. **Convert to Milliseconds**:
To convert seconds to milliseconds:
\[ \tau = \frac{1}{250} \times 1000 = 4 \, \text{milliseconds} \]

### Final Answer:
The time constant of the circuit is \( 4 \, \text{ms} \).

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