engineering physics MCQ #1242

A coil has 500 turns and flux linkage 0.05 Wb. If current is 5 A, find self-inductance.

engineering physics MCQ #1242

  1. Question 1

    Q1. A coil has 500 turns and flux linkage 0.05 Wb. If current is 5 A, find self-inductance.

    • A) 5 H
    • B) 0.05 H
    • C) 50 H
    • D) 0.5 H

    Answer: 5 H

    Explanation: L = Nϕ/I = 500 * 0.05 / 5 = 5 H.