The combined inductance of two coils connected in a series is 0.6H and 0.1H, depending on the relative directions of the currents to the coils. If one of the coils when isolated has a self-inductance of 0.2 H, then calculate the natural inductance?

• A. 0.015 H
• B. 0.15 H
• C. 1.5 H
• D. 0.125 H

Answer: (B)

The combined inductance of two coils connected in series
L = L1 + L2 + 2M
In series adding case
L1 + L2 + 2M = 0.6 H ——–(1)
In series opposing case
L1 + L2 − 2M = 0.1 H ——–(2)
Subtracting eqn (2) from eqn (1) we get
4M = 0.5 H
M = 0.125 H
Let L1 = 0.2 H (since the coil when isolated, its self-inductance is 0.2 H)
Putting the value of  M &  L1 in equation (2)
0.2 + L2 + 2 ×0.125 = 0.06
0.2 + L2  = 0.35
L2  = 0.15 H