AIPMT PRELIMS 2004 Physics RL Circuits MCQ Question

To solve the problem regarding the time constant of an RL circuit, we need to understand the relationship between inductance (L), resistance (R), and the time constant ($\tau$) of the circuit.

Step 1: Understanding Time Constant in RL Circuits

The time constant ($\tau$) for an RL circuit is defined as the time it takes for the current to reach approximately 63.2% of its maximum value after a voltage is applied. It can be calculated using the formula:

$$\tau = \frac{L}{R}$$

where:

• $L$ is the inductance in henries (H),
• $R$ is the resistance in ohms (Ω).

Step 2: Given Values

From the problem, we have:

• Inductance, $L = 40 \, \text{H}$
• Resistance, $R = 8 \, \Omega$

Step 3: Calculating the Time Constant

Using the formula for the time constant, we can substitute the given values:

$$\tau = \frac{L}{R} = \frac{40 \, \text{H}}{8 \, \Omega} = 5 \, \text{seconds}$$

Step 4: Conclusion

Therefore, the time constant of the circuit is $5$ seconds, which corresponds to option D.

Explanation of Why Other Options are Incorrect

Let's evaluate the other options briefly:

• A) $1/5$ seconds: This value is significantly lower than what we computed. It does not fit the relationship between $L$ and $R$ that we derived.

• B) $40$ seconds: This option suggests a much larger time constant, which could only be true if the resistance was much smaller or the inductance was much larger in a different configuration, neither of which applies here.

• C) $20$ seconds: Similar to option B, this value does not align with our calculation. It would imply a much different ratio of $L$ to $R$.

Summary

The correct answer is $D$ (5 seconds), which we derived using the formula for the time constant in an RL circuit. The other options do not satisfy the relationship between the given inductance and resistance in the circuit. Thus, they are incorrect.