posted by Anonymous on .
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?
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