A single phase voltage source "v"is given by v = 141 sin(wt) + 42.3 sin(3wt) + 28.8 sin(5wt). The corresponding current in the load circuit is given by i = 16.5 sin (wt + 54.5°) + 8.43 sin (3wt- 38°) + 4.65 sin (5wt - 34.3°). Determine the following:

A single phase voltage source "v"is given by v = 141 sin(wt) + 42.3 sin(3wt) + 28.8 sin(5wt). The corresponding current in the load circuit is given by i = 16.5 sin (wt + 54.5°) + 8.43 sin (3wt- 38°) + 4.65 sin (5wt - 34.3°). Determine the following:

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.23MCQ: In a balanced system, the phasor sum of the line-to-line voltages and the phasor sum of the...

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Concept explainers

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.

Question

Determine :

(g) The total power of the circuit in W.
(h) The total power factor of the circuit.

A single phase voltage source "v"is given by
v = 141 sin(wt) + 42.3 sin(3wt) + 28.8 sin(5wt).
The corresponding current in the load circuit is given by
i = 16.5 sin (wt + 54.5°) + 8.43 sin (3wt – 38°) + 4.65 sin (5wt – 34.3°).
Determine the following:

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