A coil has 200 turns and flux linkage 0.1 Wb. If current is 2 A, find self-inductance.
Question 1
Q1. A coil has 200 turns and flux linkage 0.1 Wb. If current is 2 A, find self-inductance.
Answer: 10 H
Explanation: L = Nϕ/I = 200 * 0.1 / 2 = 10 H.