The reactance of a generator is given as 0.25 per-unit based on the generator’s of 18 kV, 500 MVA.  Find its per-unit reactance on a base of 20 kV, 100 MVA. a. 0.0505 b. 0.0605 c. 0.0405 d. 0.0606

Q: What is the load factor? What happens when load factor = 1?

A:

Q: 17 The voltage a cross a load impedance is V = 12270° V and the current through it is I = 32-20° A…

A:

Q: 12.14: the currenit waveforms of a capacitov are shown in Figuse o Determine root The avesage peak…

A: Given Ip=500 At1=100 μst2=300 μst3=500 μsfs=250 Hz

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Q: one (360 degrees), all of the instantaneous power falls under positive loops (no negative loops),…

A: Option A  Resistor is the correct answer.

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A: This is a simple question based on network analysis and simple electrical engineering. The topic is…

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A: It refers to the connection of multiple generators in a way to supply a common load. This…

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.