Question

How fast can the 155 A current through a 0.320 H inductor be shut off if the induced emf cannot exceed 70.0 V?

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Answer 1

emf = L ×( current change / time required)

70 = 0.320 × (155-0)/(time)

hence, time = (155 × 0.320 )/ 70

time required= 49.6/70 = 0.708 sec. ans