Post a New Question


posted by .

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?

  • Physics -

    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question