The resistance R3 & R4 are connected in series = (30 + 70) = 100Ω Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel 1/Rp = 1/100 + 1/100 + 1/50 Rp = 100/4 = 25 Ω Resistance 50Ω and 25Ω are connected in series. Rtotal = (50 + 25)Ω Rtotal = 75Ω
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.
The self-inductance is given as L = μN2A/I L ∝ N2 where N is the number of turns of the solenoid A is the area of each turn of the coil l is the length of the solenoid and μ is the permeability constant L1/L2 = N21/N22 0.5/0.5 = N21/N22 N1/N2 = 1
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.
EXAMIANS
