The source voltages vi(t) 10cos(1000t + 30°),v2 (t) = -7sin(1000t-30'). To find the voltage aross the capacitor Vc may found by one or more equations as follows. 2mH Ib Ia 333.333µF O (5+j3) Ia - j3Ib = 1021o -j3Ia - jilb 7Z120 Vc = -j3(Ia - Ib) Ve-104-150 Ve j2 Ve-7260 -3 = 0 口 Ve- 104-150 Ve-7260 = 0 O None of these
The source voltages vi(t) 10cos(1000t + 30°),v2 (t) = -7sin(1000t-30'). To find the voltage aross the capacitor Vc may found by one or more equations as follows. 2mH Ib Ia 333.333µF O (5+j3) Ia - j3Ib = 1021o -j3Ia - jilb 7Z120 Vc = -j3(Ia - Ib) Ve-104-150 Ve j2 Ve-7260 -3 = 0 口 Ve- 104-150 Ve-7260 = 0 O None of these
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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![The source voltages vi(t) = -10cos(1000t + 30 ),v2(t) = -7sin (1000t - 30°). To find the voltage aross the capacitor Vc may
found by one or more equations as follows.
2mH
Ib
Ia
333.333рF
O (5 + j3) Ia – j3Ib = 10210
-j3Ia - jilb =7Z120
Ve = -j3(Ia- Ib)
Ve - 104 - 150 + Ve +
j2
Ve-7260
-3
= 0
Ve- 104-150
5
Ve-7260 = 0
j2
Ve
+
-j3
O None of these](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45ecf938-b980-4d14-ae7c-f2ec31cad7c4%2Ff616d0a7-d22d-49a8-a285-bc588ed19d69%2Fsbr4ab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The source voltages vi(t) = -10cos(1000t + 30 ),v2(t) = -7sin (1000t - 30°). To find the voltage aross the capacitor Vc may
found by one or more equations as follows.
2mH
Ib
Ia
333.333рF
O (5 + j3) Ia – j3Ib = 10210
-j3Ia - jilb =7Z120
Ve = -j3(Ia- Ib)
Ve - 104 - 150 + Ve +
j2
Ve-7260
-3
= 0
Ve- 104-150
5
Ve-7260 = 0
j2
Ve
+
-j3
O None of these
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