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.
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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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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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