5/0° A 2Ω j1 Q ww 20-90° V -j2 N 21

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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problem 5

Problem (4)
Find Io
0.25 F
2 H
2Ω
1 H
ell
8 sin (2t + 30°) V
0.5 F
cos 2t A
Problem (5)
Find I
5/0° A
j1 Q
ll
20/-90° V
-j2 2
21
Problem (6)
Find Io
20 µF
50 cos 2000t V
80 2
100 2
40 mH
2 sin 4000t A
60 Ω
24 V
ww
Transcribed Image Text:Problem (4) Find Io 0.25 F 2 H 2Ω 1 H ell 8 sin (2t + 30°) V 0.5 F cos 2t A Problem (5) Find I 5/0° A j1 Q ll 20/-90° V -j2 2 21 Problem (6) Find Io 20 µF 50 cos 2000t V 80 2 100 2 40 mH 2 sin 4000t A 60 Ω 24 V ww
Expert Solution
Step 1

To solve above problem, one should have basic idea about working of AC circuits.

Above given circuit is a R-L-C circuit.

In ac circuits having energy storage elements like L and C their impedance is a function of supply frequency.

Ac circuits cannot be solved directly by applying rules of arithmetic. To solve AC circuit we have to solve them in phasor domain where, reactive components are added together and resistive components together and to add the both phasor sum is required.

Active power is only consumed by resistive loads and reactive power in the circuit comes from inductor and capacitor connected in the circuit.

If reactive power is zero means either the circuit is in resonance no inductors and capacitors present in the circuit.

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