SSC JE Electrical Engineering 2012 Paper

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Question : 187 of 200

Marks: +1, -0

Two coupled inductors

L_1 = 0.2 H

and

L_2 = 0.8 H,

have coefficient of coupling K = 0.8. The mutual inductance M is

0.16 H
0.02 H
0.32 H
0.24 H

Solution:

M=K√{L_1L_2}

=0.8√{0.2×0.8}

= 0.32 H

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