Question

A series R-C combination is connected to an AC voltage of angular frequency \( \omega = 1000 \) rad/s. If the impedance of the RC circuit is \( R/\sqrt{2} \), the time constant (in milliseconds) of the circuit is:

• A. 3
• B. 1
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{3} \)

Hint:

For a series R-C combination, the impedance and time constant are key to solving problems. Use the formula for impedance \( Z = \sqrt{X_L^2 + R^2} \) to find relationships between \( R \), \( L \), and \( C \).

Answer 1

The Correct Option is B

The impedance \( Z \) of a series R-C combination is given by: \[ Z = \sqrt{X_L^2 + R^2} \] where \( X_L = \omega L \) is the inductive reactance and \( R \) is the resistance. In this case, we are given: \[ Z = \frac{R}{\sqrt{2}} \] Since \( X_L = 0 \), the equation simplifies to: \[ Z = R \] Now, the time constant \( \tau \) for an R-C circuit is given by: \[ \tau = R \cdot C \] Using the relation for the impedance of the circuit, we have: \[ Z = \frac{1}{\omega C} = \frac{R}{\sqrt{2}} \] So, we can solve for \( C \) as follows: \[ C = \frac{1}{\omega R \sqrt{2}} \] Substituting \( \omega = 1000 \) rad/s and simplifying: \[ C = \frac{1}{1000 R \sqrt{2}} = \frac{1}{1000 R} \quad \text{(in milliseconds)} \] Thus, the time constant \( \tau \) is: \[ \tau = R \cdot \frac{1}{1000 R} = 1 \, \text{ms} \] Thus, the correct answer is \( \tau = 1 \, \text{ms} \).