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During a 74-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 4.6-mA current in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12 Ù. The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?

1 answer

The voltage in the second coil is
V2 = I2*R2 = 4.6*10^-3*12 = 5.52*10^-2 V

This equals M12*dI1/(dt) caused by mutual inductance M12

M12 is the mutual inductance and dt = 74*10^-3 s.

Solve for the current change dI1

dI1 = (5.52*10^-2 V)*74*10^-3s/(3.2*10^-3 H) = 1.27 Amp
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