#2 A load consists of a 60-2 resistor in parallel with a 90-uF capacitor. If the load is connected to a voltage source vs(t) = 160 cos 2000t, find the average power delivered to the load. #3.Assuming that vs = 8 cos(2t -40°) V in the circuit of Fig 1 find the average power delivered to Solution #2 20 F4 F5 160Z0 R C 1 1 90 μF =-j5.5556 j@C j90x10x2x103 I=160/Z_par = 2.6667 + 28.8000j Here Z_par (R*Zc)/(R+Zc), where Zc= -j/wC So I abs 28.923 A and phase angle phi = 84.71 degrees So Pavg for resistor = 0.5 * (I_abs)^2*R* Cos(84.71)= 2313.9 W < de 69 8 < DII F6 F7 F8 F9
#2 A load consists of a 60-2 resistor in parallel with a 90-uF capacitor. If the load is connected to a voltage source vs(t) = 160 cos 2000t, find the average power delivered to the load. #3.Assuming that vs = 8 cos(2t -40°) V in the circuit of Fig 1 find the average power delivered to Solution #2 20 F4 F5 160Z0 R C 1 1 90 μF =-j5.5556 j@C j90x10x2x103 I=160/Z_par = 2.6667 + 28.8000j Here Z_par (R*Zc)/(R+Zc), where Zc= -j/wC So I abs 28.923 A and phase angle phi = 84.71 degrees So Pavg for resistor = 0.5 * (I_abs)^2*R* Cos(84.71)= 2313.9 W < de 69 8 < DII F6 F7 F8 F9
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...
Related questions
Question
Power and the average power.
#2 A load consists of a 60-2 resistor in parallel with a 90-uF capacitor. If the load is connected to a
voltage source vs(t) = 160 cos 2000t, find the average power delivered to the load.
I need to show all work please
Transcribed Image Text:#2 A load consists of a 60-2 resistor in parallel with a 90-uF capacitor. If the load is connected to a
voltage source vs(t) = 160 cos 2000t, find the average power delivered to the load.
#3.Assuming that vs = 8 cos(2t -40°) V in the circuit of Fig 1 find the average power delivered to
Transcribed Image Text:Solution #2
20
F4
F5
160Z0
R
C
1
1
90 μF
=-j5.5556
j@C
j90x10x2x103
I=160/Z_par = 2.6667 + 28.8000j
Here Z_par (R*Zc)/(R+Zc), where Zc= -j/wC
So I abs 28.923 A
and phase angle phi = 84.71 degrees
So Pavg for resistor = 0.5 * (I_abs)^2*R* Cos(84.71)= 2313.9 W
<
de
69
8
<
DII
F6
F7
F8
F9
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 3 images
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,