Question

An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :

• A. 0.2
• B. 0.4
• C. 0.8
• D. 0.125

Hint:

A "coil" in AC or DC problems is almost always an RL series circuit. Finding L and R individually is the standard path to finding the time constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A real inductor has both inductance (\(L\)) and resistance (\(R\)). The time constant (\(\tau\)) of an RL circuit is the ratio of \(L\) to \(R\).
Step 2: Key Formula or Approach:
1. Energy stored \(U = \frac{1}{2} L I^2\)
2. Power dissipated \(P = I^2 R\)
3. Time constant \(\tau = \frac{L}{R}\)
Step 3: Detailed Explanation:
1. From energy: \(64 = \frac{1}{2} L (8)^2 \implies 64 = 32 L \implies L = 2 \text{ H}\).
2. From power: \(640 = (8)^2 R \implies 640 = 64 R \implies R = 10 \text{ } \Omega\).
3. Time constant: \(\tau = \frac{L}{R} = \frac{2}{10} = 0.2 \text{ s}\).
Step 4: Final Answer:
The time constant of the circuit is 0.2 s.