Internal generated voltage E. V5004003002001000.01.02.03.04.05.0 6.07.08.09.0 10.0Field current, ANote: When the field current is zero, E, is about 3 V.Magnetization curve for the DC generator, made at a constant speed of 1800rpmFigure Q4(c) (c)(iv)A separately excited DC generator is rated at 172 kW, 430 V, 400 A and 1800 rpm.Its magnetization curve is shown in Figure Q4(c). The generator's characteristics are:RA= 0.05RF = 2002Radj = 0 to 300(i)(iii)VF = 430 V NF = 1000 turns per poleSketch an equivalent circuit of the generator with proper labels.If Radj = 63 N, and the prime mover is driving at 1500 rpm, calculate thegenerators' no-load terminal voltage.Calculate the terminal voltage if a 360 A load is connected to its terminal.Assuming the generator does not have compensating winding. Its armaturereaction at this load is 450 A. turns. The prime mover is driving at 1500 rpm.For Q4(c)(iii), the terminal voltage is lower than no-load terminal voltagecalculated from Q4(c)(ii). Investigate the adjustment of field current, IF andRadj to raise for the terminal voltage to match the no-load terminal voltage. |

A separately excited DC generator is rated at 172 kW, 430 V, 400 A and 1800 rpm. Its magnetization curve is shown in Figure Q4(c). The generator's characteristics are: RA = 0.05 Ω Radj = 0 to 300 VF = 430 V NF = 1000 turns per pole Sketch an equivalent circuit of the generator with proper labels. (i) RF = 200 Ω

A separately excited DC generator is rated at 172 kW, 430 V, 400 A and 1800 rpm. Its magnetization curve is shown in Figure Q4(c). The generator's characteristics are: RA = 0.05 Ω Radj = 0 to 300 VF = 430 V NF = 1000 turns per pole Sketch an equivalent circuit of the generator with proper labels. (i) RF = 200 Ω

Electric Motor Control

10th Edition

ISBN:9781133702818

Author:Herman

Publisher:Herman

Chapter39: Synchronous Motor Operation

Section: Chapter Questions

Problem 14SQ: Induction motors and welding transformers require magnetizing current, which causes a. lagging power...

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