SSC JE Electrical Engineering 26 Sep 2019 Shift 1 Paper

Question : 161 of 200

Marks: +1, -0

A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is:

Solution:

{\dI }/{\dt}=

5A / Sec
H = 2H.
Self induced Emf =

-H{\dI }/{\dt}

= –2 × 5
= –10V

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