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A chopper is feeding an RL load as shown in Fig. 2 with V= 220-V, R = 10 Q, and L = 15.5 mH, f= 1-Khz, and E = 110-V and k =

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Given data Vs = 220v R=lor L = 15.5mH = 15.5x10-3 f = 1khz = 10 Hz E-lov k=0.4, T = 2 2 = 1 Given figure = 103 kg IX10 15.5mi-(1-04)810 710 I, = 26 (1-0.4) 10/10 I = Ize is 5x10-3 -110(1-0 15.5x 10-3 = I xe bosst 11x (1-0-0-387) G = I, (0.679) - 3.53e coreut (Joh 6 Average load current (sa) and rm Now find the average load current (Ta). Ia = Iz tij 2 -4.3564 = -0.4744-3.85-(1-04)810 710 I, = 26 (1-0.4) 10/10 I = Ize is 5x10-3 -110(1-0 15.5x 10-3 = I xe bosst 11x (1-0-0-387) G = I, (0.679) - 3.53

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