The circuit below represents an AC circuit in steady-state in the phasor domain (for the complex numbers, you may assume units are V, A, 2, etc. as appropriate). The voltage source vs is an AC source with w = 5 rad/s. Each box represents the impedance of a single circuit element (a resistor, capacitor or inductor). Given Answers (a) Find the average power Ps supplied by the voltage source. Ps 0.3 W POW İXRMS 0.5/2 A (b) Find the average power P₁ received by the element with impedance Z₁. (c) Find the RMS value of ix.

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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The circuit below represents an AC circuit in steady-state in the phasor domain (for
the complex numbers, you may assume units are V, A, 2, etc. as appropriate). The voltage
source vs is an AC source with w = 5 rad/s. Each box represents the impedance of a single
circuit element (a resistor, capacitor or inductor).
Given Answers
(a) Find the average power P, supplied by the voltage source.
P 0.3 W
P₁OW
İXRMS 0.5√2 A
(b) Find the average power P₁ received by the element with
impedance Z₁.
(c) Find the RMS value of ix.
Im
α
|Vsl = 3 V
||
T-3
Re
VS
ia,
Z₁
Z₂
B.B.
ix
Z₂ = 3
Z₁ = -3j
4. ia
Transcribed Image Text:The circuit below represents an AC circuit in steady-state in the phasor domain (for the complex numbers, you may assume units are V, A, 2, etc. as appropriate). The voltage source vs is an AC source with w = 5 rad/s. Each box represents the impedance of a single circuit element (a resistor, capacitor or inductor). Given Answers (a) Find the average power P, supplied by the voltage source. P 0.3 W P₁OW İXRMS 0.5√2 A (b) Find the average power P₁ received by the element with impedance Z₁. (c) Find the RMS value of ix. Im α |Vsl = 3 V || T-3 Re VS ia, Z₁ Z₂ B.B. ix Z₂ = 3 Z₁ = -3j 4. ia
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