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j162 j40 2g j40 2 24 2 DO V/30 キーji2 j4 2 =-0 -j10 2= j9 2 10Ω 32 82 ee

j162 j40 2g j40 2 24 2 DO V/30 キーji2 j4 2 =-0 -j10 2= j9 2 10Ω 32 82 ee

Introductory Circuit Analysis (13th Edition)
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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j16 2 j40 2 ll j40 2 24 Ω 100 V/30 ( --jl 2 j4 2 =-j82 j10 2 -j10 2= j9 2 10Ω 3Ω 82 le
Transcribed Image Text:j16 2 j40 2 ll j40 2 24 Ω 100 V/30 ( --jl 2 j4 2 =-j82 j10 2 -j10 2= j9 2 10Ω 3Ω 82 le
QUESTION 3 Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the voltage across the 18 ohm capacitor. O 14V RMS O 8.5V RMS O 34V RMS O 23V RMS QUESTION 4 Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the current through the j9 ohm inductor. O 2.8A RMS O 11.1A RMS O 850mA RMS O 6.3A RMS
Transcribed Image Text:QUESTION 3 Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the voltage across the 18 ohm capacitor. O 14V RMS O 8.5V RMS O 34V RMS O 23V RMS QUESTION 4 Using Figure 23-24 on page 688 of your textbook, calculate the magnitude of the current through the j9 ohm inductor. O 2.8A RMS O 11.1A RMS O 850mA RMS O 6.3A RMS
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