Question

When a 250 turn coil carries 2 A current a flux of 0.3 mWb is established in it. If the

current is reduced to zero in 2 ms, the voltage induced in neighboring coil is 63.75 V. If

coupling coefficient is 0.85 between the coils, find the self inductances of coils, mutual

inductance and number of turns in second coil.​

Answer 1

Answer:

Here

N1= 250

I1= 2A

Flux =0.3mWb = 0.3×10-³

e= 63.75V

t= 2×10-³

k=0.85

Explanation:

L1= (N1×flux)/I1 = (250×0.3×10-³)/2 = 0.0375H

e= M di/dt

di= 2

dt =2ms = 2×10-³

therefore, M= (e×dt)/di

M= (63.75×2×10-³)/2= 0.06375H

M=(N2×K×flux)/I1

N2=(M×I1)/ k×flux

therefore,

N2 = (0.06375×2) / 0.85×0.3×10-³

=500 turns

M²=K²×L1×L2

L2 = M²/ K²×L1

L2 = (0.06375)²/(0.85)²×(0.0375)

= 0.15 H