Calculate the change in current in an inductor of 5mH in which the emf induced is 250vin 10s
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Explanation:
Self inductance of the coil L=5mH=0.005H
Rate of change of current
dt
di
=
0.01
2.0−2.5
⟹
dt
di
=−50 A/s
Thus induced emf E=−L
dt
di
∴ E=−0.005×(−50)
⟹ E=0.25V
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