Question

Two figures are shown as Fig. A and Fig. B. The time constant of Fig. A is \( \tau_A \), and time constant of Fig. B is \( \tau_B \). Then:

• A. \( \tau_A = \frac{1}{4} \, {s} \) and \( \tau_B = 5 \, {s} \)
• B. \( \tau_A = \frac{1}{2} \, {s} \) and \( \tau_B = \frac{1}{5} \, {s} \)
• C. \( \tau_A = 4 \, {s} \) and \( \tau_B = 5 \, {s} \)
• D. \( \tau_A = \frac{1}{3} \, {s} \) and \( \tau_B = 10 \, {s} \)

Hint:

For RL circuits, the time constant is given by \( \tau = \frac{L}{R} \), and for RC circuits, the time constant is \( \tau = RC \).

Answer 1

The Correct Option is A

We are given two circuits, Fig. A and Fig. B. We are tasked with finding the time constants \( \tau_A \) and \( \tau_B \). The time constant \( \tau \) for an RL or RC circuit is given by the formula:

\( \tau = \frac{L}{R} \quad {for RL circuits, and} \quad \tau = RC \quad {for RC circuits}. \)

Step 1: Let's analyze Fig. A:
- The resistance in Fig. A is \( R_A = 6 \, \Omega + 2 \, \Omega = 8 \, \Omega \).
- The inductance in Fig. A is \( L_A = 2 \, {H} \).
Thus, the time constant for Fig. A is:

\( \tau_A = \frac{L_A}{R_A} = \frac{2 \, {H}}{8 \, \Omega} = \frac{1}{4} \, {s}. \)

Step 2: Now, let's analyze Fig. B:
- The resistance in Fig. B is \( R_B = 10 \, \Omega + 40 \, \Omega = 50 \, \Omega \).
- The capacitance in Fig. B is \( C_B = 0.5 \, \mu{F} = 0.5 \times 10^{-6} \, {F} \).
Thus, the time constant for Fig. B is:

\( \tau_B = R_B \cdot C_B = 50 \, \Omega \cdot 0.5 \times 10^{-6} \, {F} = 5 \times 10^{-5} \, {s}. \)

Thus, the time constants are \( \tau_A = \frac{1}{4} \, {s} \) and \( \tau_B = 5 \, {s} \).