in part (b,across the-j11-N capacitance in part (c).(C)j16 2-jll 2el24 Qj40 Ω:-j1 Qj4Ω -3Ω-j8 Q100 V/30° (j10 Q-j102j9 210 ΩFigure 23-24ion 23-5 Source ConversionConvert the 120-V source and QUESTION 4Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the current through the -j10 ohm capacitor.7A RMS4.4A RMS18.9A RMS12A RMS

Assume 100[30°] is a RMS Value.

Answered: across the j16 2 j40 2 ணை j40 2 24 2…

Engineering

Electrical Engineering

across the j16 2 j40 2 ணை j40 2 24 2 100 V/30° ( -jl Q j4 2 =-j82 -j10 2= j10 2 j9 2 10 Ω 32 8 Ω ll ll

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter7: Alternating Current, Power Distribution, And Voltage Systems

Section: Chapter Questions

Problem 7RQ

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Graph theory

It is that part of network analysis where the configuration is called graph or structure of the network system when all of the elements (e.g., R, L, and C) in a network system are substituted with lines having circles/rings or dots across both endpoints. The study of the solution of the networks and circuits with the help of graphs is known as graph theory. The objective is to know the mathematics behind the network analysis.

Question

in part (b,
across the-j11-N capacitance in part (c).
(C)
j16 2
-jll 2
el
24 Q
j40 Ω
:-j1 Q
j4Ω -3Ω
-j8 Q
100 V/30° (
j10 Q
-j102
j9 2
10 Ω
Figure 23-24
ion 23-5 Source Conversion
Convert the 120-V source and

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