#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

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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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
#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:#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
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
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