A load with an active power of P = 6000Watts is connected to the network witha load with a voltage of V=220<0°Volts.The load is a capacitive load, and theReactive power will be taken as Qc5105 VAR.Compensation is planned for the saidload. As a result of the compensation,the power factorFind the value of the inductance (L)that must be connected to the circuitso that cose = 0.99 (capacitive).NOTE: The frequency will be 50 Hz.V=220/0° [V]Z=

Given data, Active power P = 6000 W Voltage V=220∠00 V Reactive power, Qc=5105 VAR

Answered: A load with an active power of P = 6000…

A load with an active power of P = 6000 Watts is connected to the network with a load with a voltage of V=220<0° Volts. The load is a capacitive load, and the Reactive power will be taken as Qc =

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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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

A load with an active power of P = 6000
Watts is connected to the network with
a load with a voltage of V=220<0°
Volts.
The load is a capacitive load, and the
Reactive power will be taken as Qc
5105 VAR.
Compensation is planned for the said
load. As a result of the compensation,
the power factor
Find the value of the inductance (L)
that must be connected to the circuit
so that cose = 0.99 (capacitive).
NOTE: The frequency will be 50 Hz.
V=220/0° [V]
Z
=

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