SSC JE Electrical 2019 with solution SET-1
In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:
Given Data Voltage Va = 200 v Armature Resistance Ra = 0.5Ω Armature Current Ia = 20 A Induced EMF = Ea = ? The Induced EMF of a DC machine working as a Motor is Ea = Va − IaRa Ea = 200 − 20 × 0.5 Ea = 190 V
In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.
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
EXAMIANS