For the circiut shown, R = 600 2, C = 4.0 µF, L = 3.0 H₂= 140 V and f = 100 Hz. The phase anglethe current and voltage for the circuit isbetweena) 71.3º.|AUR AVEAUCb) 65.2°,Meeec) 68.0⁰.RLCd) 76.8°,e) 83.3º.A

Given data, R=600 ΩC=4.0 μFL=3.0 HVm=140 Vf=100 Hz

For the circiut shown, R = 600 Q, C = 4.0 µF, L = 3.0 H. Vm = 140 V and f = 100 Hz. The phase angle between the current and voltage for the circuit is a) 71.3º, |AUR AULAUC b) 65.2°, Leee c) 68.0⁰. R d) 76.8° L C e) 83.3º. A

For the circiut shown, R = 600 Q, C = 4.0 µF, L = 3.0 H. Vm = 140 V and f = 100 Hz. The phase angle between the current and voltage for the circuit is a) 71.3º, |AUR AULAUC b) 65.2°, Leee c) 68.0⁰. R d) 76.8° L C e) 83.3º. A

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 2.7P: Let a 100V sinusoidal source be connected to a series combination of a 3 resistor, an 8 inductor,...

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Concept explainers

Compensation Techniques

The compensation techniques are the methods that are employed in the electrical control systems to enhance the performance of the system. The performance are generally affected by the factors like lag in the compensating devices, slow response of the system of the components like sensors used in the control system. The basic compensation technique involves lag compensation, lead compensation or the combination of the both.

Question

For the circiut shown, R = 600 2, C = 4.0 µF, L = 3.0 H₂
= 140 V and f = 100 Hz. The phase angle
the current and voltage for the circuit is
between
a) 71.3º.
|AUR AVEAUC
b) 65.2°,
Meee
c) 68.0⁰.
R
L
C
d) 76.8°,
e) 83.3º.
A

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