SSC JE Electrical 2019 with solution SET-1
Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:
Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA No. of turns NA = NB = 1000 Flux linkage in a coil with B = Flux linkage in coil A × 80/100 = 0.8 × 5 × 10−5 = 4 × 10−5 wb =0.04 mwb
When variable loss becomes equal to the constant loss, efficiency is maximum. Losses = Pi + Pc Since copper loss is a variable loss therefore Losses = Pi + Pi = 2pi Thus at a maximum efficiency of this transformer total loss = 150 x 2 = 300 W
EXAMIANS