ElectroMagnetic InductionHard

Question

In a coil of self inductance of 5henry, the rate of change of current is 2 ampere per second, the e.m.f. induced in the coil is :

Options

A.5 V
B.-5 V
C.-10 V
D.10 V

Solution

e.m.f. = −L  = −5 × 2 = - 10 V

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