Two copper coils A and B are wound over a plastic pipe. Coil A is connected to a sinusoidal voltage source of angular frequency 50 rad/s so that a current of 100 mA passes through it. The voltage acro

Question

Two copper coils A and B are wound over a plastic pipe. Coil A is connected to a sinusoidal voltage source of angular frequency 50 rad/s so that a current of 100 mA passes through it.

The voltage across coil B is 5 volts. Now, if coil B is short-circuited, there is a change of current of 2 mA in coil A.Then, the mutual inductance between the two coils and the percentage change in the impedance of coil A are respectively:

TestHub · Accepted Answer

Mutual inductance M = ω I V ​ = 50 × 0.1 5 ​ = 1 H . Change in impedance Z Δ Z ​ = I Δ I ​ = 0.1 0.002 ​ = 0.02 = 2% . Therefore, M = 1 H and change in impedance is 2% .

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