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(1) Use the current-division principle to calculate i1 and iz in Figure 5-1. R = 102 R2 =52 3 A Fig. 5-1 (2) Use the voltage-division principle to calculate v1, V2 and v3 in Fig.5-2. R = 52 + 25 V R = 132 Fig. 5-2

(1) Use the current-division principle to calculate i1 and iz in Figure 5-1. R = 102 R2 =52 3 A Fig. 5-1 (2) Use the voltage-division principle to calculate v1, V2 and v3 in Fig.5-2. R = 52 + 25 V R = 132 Fig. 5-2

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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侍分 (1) Use the current-division principle to calculate ii and iz in Figure 5-1. iz 3 A 10 2 Fig. 5-1 (2) Use the voltage-division principle to calculate v1, V2 and v3 in Fig.5-2. R = 52 25 V R = 132 Fig. 5-2
Transcribed Image Text:侍分 (1) Use the current-division principle to calculate ii and iz in Figure 5-1. iz 3 A 10 2 Fig. 5-1 (2) Use the voltage-division principle to calculate v1, V2 and v3 in Fig.5-2. R = 52 25 V R = 132 Fig. 5-2
(1) Find the equivalent resistance R, in Fig. 7-1. eq 30 30N 122 242 42 Fig. 7-1 (2) Find the equivalent capacitance Ce in Fig.7-2. 2 µF Ceq 1 μF 2 uF Fig. 7-2
Transcribed Image Text:(1) Find the equivalent resistance R, in Fig. 7-1. eq 30 30N 122 242 42 Fig. 7-1 (2) Find the equivalent capacitance Ce in Fig.7-2. 2 µF Ceq 1 μF 2 uF Fig. 7-2
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