In a single-phase alternator, a given field current produces an armature current of 250 A on short-circuit and a generated e.m.f. of 1500 V on open-circuit. Calculate the terminal p.d. when a load of 250 A at 6.6 kV and 0.8 p.f. lagging is switched off. Calculate also the regulation of the alternator at the given load. [7,898 V; 19.7%] (Elect. Machines-II, Indore Univ. Dec. 1977) A 500-V, 50-kVA, single-phase alternator has an effective resistance of 0.2 N. A field current of 10 A produces an armature current of 200 A on short-circuit and e.m.f. of 450 V on open circuit. Calculate (a) the synchronous impedance and reactance and (b) the full-load regulation with 0.8 p.f. lagging. [(a) 2.25 N, 2.24 2, (b) 34.4%] (Elect. Technology, Mysore Univ. 1979)

In a single-phase alternator, a given field current produces an armature current of 250 A on short-circuit and a generated e.m.f. of 1500 V on open-circuit. Calculate the terminal p.d. when a load of 250 A at 6.6 kV and 0.8 p.f. lagging is switched off. Calculate also the regulation of the alternator at the given load. [7,898 V; 19.7%] (Elect. Machines-II, Indore Univ. Dec. 1977) A 500-V, 50-kVA, single-phase alternator has an effective resistance of 0.2 N. A field current of 10 A produces an armature current of 200 A on short-circuit and e.m.f. of 450 V on open circuit. Calculate (a) the synchronous impedance and reactance and (b) the full-load regulation with 0.8 p.f. lagging. [(a) 2.25 N, 2.24 2, (b) 34.4%] (Elect. Technology, Mysore Univ. 1979)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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