A) , = 48 / (4+0.8571) = 9.8824 A → 12 = (6×9.8824) / (6+1) = 8.4706 A V (C÷R) 13 = 9.8824 – 8.4706 = 1.4118 A → (KCR) 4 = l3/ 2 = 0.7059 A V (C÷R) %3D %3D - %3D %3D O Z05 O A 1118 A (KCR) |
A) , = 48 / (4+0.8571) = 9.8824 A → 12 = (6×9.8824) / (6+1) = 8.4706 A V (C÷R) 13 = 9.8824 – 8.4706 = 1.4118 A → (KCR) 4 = l3/ 2 = 0.7059 A V (C÷R) %3D %3D - %3D %3D O Z05 O A 1118 A (KCR) |
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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It's not a graded question, it's a quiz question just can't work out where the answers come from
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UNIVERSITY OF
Email: pmbryant@liv.ac.uk
E LIVERPOOL
A) = 48/ (4+0.8571) = 9.8824 A →
12 = (6x9.8824) / (6+1) = 8.4706 AJ (C÷R)
2 = 9.8824 –8.4706 = 1.4118 A → (KCR)
4 = 13/ 2 = 0.7059 A V (C÷R)
Is = 0.7059 A → l6 = 1.4118 A + (KCR) I,
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B) I3 = 16 V / (5+0.8+1) = 2.353 A →
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= 2.353 A
9,
4 = 1.1765 A
%3D
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12 = 2.353 – 0.4706 = 1.8824 A ↑ (KCR)
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C) Is = 24 V /3.4872 = 6.8823 A +
4 = (5.8x6.8823)/ (5.8+2) = 5.1176 (C÷R)
I3 = 15 - 14 = 1.7647 A (KCR)
, = (1×1.7647) / (1 + 4) = 0.3529 A (C÷R)
12 = 13 – 1, = 1.4118 A V l6 = 12 + 1, = 1.7647 A →
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ENotes
17:35
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Transcribed Image Text:ト 4》
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UNIVERSITY OF
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it, b) zero or short circuit c) open circuit
to diode. Other ccts obey Ohms law (i.e.
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= 2 V, Vout = 10.48; Vin = 4 V, Vout= 20.96;
L 42
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Notes
17:34
07/11/2021
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