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Bangladesh University of Engineering & Technology Assignment on DC Motor Course no(prenominal) : EEE271 Group no.: 01 pick: 0608001 to 0608015 Level: 2, Term: 1 Dept.: I.P.E. accounting entry Date: 26-10-2008 Submitted To: Tanzina Khalaque Dept. of E.E.E., BUET. Question 29.13(d). A five hundredV D.C. shunt push back draws a zephyr veritable of 5A on leisurely- excite. If armature apology is 0.15ohm and electron mountain range resistance is 200ohm, determine the cleverness of the cable car running as a generator delivering a debase watercourse of 40A. solving: It is presumption that, The supply dominance of the shunt push back, V = 500V Armature resistance, Ra = 0.15ohm bowl resistance, Rf = 200ohm On empty-headed load the motor draws line menses, Io = 5A When the configuration whole kit and caboodle as a generator, cogency = ? The correction current, If = = = = 2.5A As the light loaded line current Io consists of armature current Iao and field current If Io = Iao + If Iao = Io - If = (5 - 2.5)A = 2.

5A At light load, introduce world-beater, Po = V Io = 500 * 5 = 2500watt As in the moorage of light load or no load the motor rotates freely and at the eon it re takes no output power. So solely input power ultimately turn into injuryes. in unremitting loss, armature Cu loss (It lurchs with change in Iao) = Iao2 * Ra = 2.52 * 0.15 = 1 watt. unremitting loss (including field Cu loss, no load automatonlike loss, lens nucleus loss) = 2500 1 = 2499 watt When the machine runs as a generator, the load current is at present 40A. I = 40A As it is constant as supply emf and Rf is constant, so at a time the armature current, Ia1 = I + If = 40 + 2.5 = 42.5A Because instantly load current consists of armature current and field current, I = Ia - If So, now pertly armature Cu loss = Ia2 * Ra...If you insufficiency to get a climb essay, order it on our website: Orderessay

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