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
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
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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Question

Transcribed Image Text: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

Transcribed Image Text:QUESTION 4
Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the current through the -j10 ohm capacitor.
7A RMS
4.4A RMS
18.9A RMS
12A RMS
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