Question

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time t=0. The time constant (in seconds) of the circuit for t>0 is___.

Hint:

In RL circuits, always find the equivalent resistance first, considering both series and parallel combinations, to compute the time constant accurately.

Answer 1

Correct Answer: 0.75

To find the time constant of the circuit for t > 0, we follow these steps: 

1. After the switch S is closed at t = 0, the inductor becomes active in the circuit. The inductor (3 H) and the resistors (2 Ω, 4 Ω, and 4 Ω) are connected.
2. Identify the total resistance in the loop affecting the inductor. The two 4 Ω resistors are in parallel since they are across the same nodes relative to the inductor.
3. Calculate the equivalent resistance of the parallel resistors:
   \[ R_{\text{eq}} = \frac{1}{\frac{1}{4} + \frac{1}{4}} = 2 \, \Omega \]
4. Add the series resistances:
   \[ R_{\text{total}} = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega \]
5. The time constant \( \tau \) is given by:
   \[ \tau = \frac{L}{R_{\text{total}}} = \frac{3 \, \text{H}}{4 \, \Omega} = 0.75 \, \text{s} \]
6. Validate:
   The calculated time constant \(0.75 \, \text{s}\) is within the given range (0.75, 0.75).

The time constant for the circuit is 0.75 seconds.