Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- Please solve this problem step by step and illustrate your answer properly. Write your answer in a piece of paper please. Box the final answer.arrow_forwardA 2300V, 1000 kVA, 0.8 lagging power factor, two-pole, 60HZ, Wye-connected, synchronous generator has a synchronous reactance of 1.1 ohms and an armature resistance of 0.15 ohms. Its losses due to friction and friction with the air are 24 kW and its core losses are 18 kW. The field circuit has a dc voltage of 200v and a maximum of 10A. The field circuit current is adjustable in the range of 20 to 200 ohms.The magnetization curve of this generator is shown in the figure below.a) How much field current is required for Vt to equal 2300V when the generator is running without load?b) What is the voltage generated intermittently by this machine at nominal conditions?c) How much field current is needed to make the voltage at the terminals Vt=2300V when the generator works in nominal conditions?d) How much power and torque should the prime mover of the generator be able to supply?arrow_forwardD2. When a generator's full-load output voltage is the same as its no-load output voltage it is said to be a) under-compounded b) flat-compounded c) over-compounded d) none of the above D3. The output of an overcompounded DC generator can be lowered by use of a a) rheostat connected in series with the series field b) rheostat connected in parallel with the series field c) rheostat connected in parallel with the shunt field d) none of the above D4. When might it be desirable to use an overcompounded generator? a) never b) always c) to compensate for line loss d) when you need a higher no-load voltage than your full-load voltagearrow_forward
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