A current of I A passing through the primary coil is cut off in 10^-3 s and this induces an emf of 160 V in th
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A current of I A passing through the primary coil is cut off in 10^-3 s and this induces an emf of 160 V in th

[From: ] [author: ] [Date: 12-03-10] [Hit: ]
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A current of I A passing through the primary coil is cut off in 10^-3 s and this induces an emf of 160 V in the secondary coil. If the mutual inductance is 40 mH, calculate the initial current I in the primary.

-
e = - M (di /dt)

dI /dt = - 160 / 40 e – 3 = - 4000
dI = -4000 * 1.e-3 = -4 A

It is given that dI = 0 – I = - I

- I = - 4 A
Hence
I = 4 A
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