SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Question : 136 of 200

Marks: +1, -0

Two coils having self-inductance of 3 H and 2 H, respectively, have mutual inductance of 2 H. They are connected in series and carry a current of 4 A. Calculate the energy of the magnetic field when the self and mutual fluxes are in the same direction.

8 J
72 J
144 J
36 J

Validate

Solution:

\L _{1}=3 \H

\L _{2}=2 \H

\M =2 \H

\I =4 \A

Total inductance,

\L _{ \T }= \L _{1}+ \L _{2}+2 \M

=3+2+(2 ×2)=9 \H

Energy of magnetic field,

\E={1}/{2} \L_{\T} \I^{2}

={1}/{2} ×9 ×4^{2}=72 \J

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