SSC JE Electrical Engineering 26 Sep 2019 Shift 2 Paper

Question : 181 of 200

Marks: +1, -0

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

0.125 H
0.15 H
0.015 H
1.5 H

Validate

Solution:

L_T = L_1 + L_2 + 2M

∴

0.6 = L_1 + L_2 + 2M

→ (1)
and 0

.1 = L_1 + L_2 – 2M

→ (2) (When direction of current is reversed)
∴ Let

L_1 = 0.2

∴

0.6 = 0.2 + L_2 + 2M

∴

0.4 = L_2 + 2M

→ (3)
and

0.1 = 0.2 + L_2 -2M

∴

-0.1= L_2 -2M

→ (4)
Solving eq (3) and (4) we get

[M =0.125H]

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