Question

In a circuit, the current falls from 14 A to 4 A in a time 0.2 ms. If the induced emf is 150 V, then the self-inductance of the circuit is:

• A. \( 6 \) H
• B. \( 6 \) mH
• C. \( 3 \) mH
• D. \( 3 \) H

Hint:

The self-inductance of a circuit determines how much voltage is induced when current changes over time.

Answer 1

The Correct Option is C

Step 1: Use the Self
-Inductance Formula The induced emf is given by: \[ V = L \frac{dI}{dt} \] where: 
- \( V = 150 \) V, 
- \( \Delta I = 14 - 4 = 10 \) A, 
- \( \Delta t = 0.2 \) ms = \( 0.2 \times 10^{-3} \) s. 
Step 2: Solve for \( L \) \[ L = \frac{V \cdot \Delta t}{\Delta I} \] \[ L = \frac{150 \times 0.2 \times 10^{-3}}{10} \] \[ L = 3 \times 10^{-3} = 3 \text{ mH} \]