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
During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
As you can see from the below figure in load Z1 is connected with the only current coil. In Load Z2 both current from the current coil (CC) and voltage from voltage coil (PC) are present (Power = V × I). Hence the Wattmeter will read power consumed by Z2.
EXAMIANS