Question

Two coils A and B have mutual inductance 2 × 10–2 henry. If the current in the primary is i = 5 sin (10 πt) then the maximum value of e.m.f. induced in coil B is

(a) π volt (b) π/2 volt (c) π/3 volt (d) π/4 volt

Answer 1

Answer:

Dear student,

emf induced = −MdIdte = −2×10−2xddt[5 sin (10πt)]e=−2×10−2x50πcos(10πt)for maximum value of emf cos(10πt)=0e=100100πe=πRegards

Answer 2

Answer:

Current i=5 sin(10πt)

Mutual inductance L=2×10

−2

H=0.02H

Induced emf, E=−M

dt

di

∴ ∣E∣=M×5(10π)cos(10πt)

∣E∣

max

=M×50π=0.02×50π=π volt