SSC JE Electrical 2019 with solution SET-1
In an AC network, the load connected is (10 + j10). The phase relation between the voltage applied and the current through the load is:
Total resistance in the given circuit R = (250 + 250)MΩ = 500 MΩ Current I = V/R = 24/(500 × 103) Now the Voltage in the voltmeter = \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3} V = 12 V
Hysteresis Loss = Kh × BM1.67 × f × v watts where Kh = Hysteresis constant depends upon the material Bm = Maximum flux density f = frequency v = Volume of the core Hence the hysteresis loss does not depend upon the ambient temperature.
EXAMIANS