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In the network shown in Fig. 4-24, find the current in the 10-2 resistor 10 1 ref 5 Ω 2 0 2 A 4 A 2 Fig. 4-24 The nodal equations in matrix form are written by inspection. 1 1 1 2 10 5 1 1 Va -6 5 -0.20 0.70 0.30 -0.20 2 -6 = 1.18 V -0.20 0.70 Then, I V,/10 0.118 A ILIPS 01 SLC Saudi Lighting Date: .. ZA 2 4 hare I 2A I3 =4A IIM+4 0 -247 A ILO7A

In the network shown in Fig. 4-24, find the current in the 10-2 resistor 10 1 ref 5 Ω 2 0 2 A 4 A 2 Fig. 4-24 The nodal equations in matrix form are written by inspection. 1 1 1 2 10 5 1 1 Va -6 5 -0.20 0.70 0.30 -0.20 2 -6 = 1.18 V -0.20 0.70 Then, I V,/10 0.118 A ILIPS 01 SLC Saudi Lighting Date: .. ZA 2 4 hare I 2A I3 =4A IIM+4 0 -247 A ILO7A

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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In the network shown in Fig. 4-24, find the current in the 10-2 resistor 10 1 ref 5 Ω 2 0 2 A 4 A 2 Fig. 4-24 The nodal equations in matrix form are written by inspection. 1 1 1 2 10 5 1 1 Va -6 5 -0.20 0.70 0.30 -0.20 2 -6 = 1.18 V -0.20 0.70 Then, I V,/10 0.118 A
Transcribed Image Text:In the network shown in Fig. 4-24, find the current in the 10-2 resistor 10 1 ref 5 Ω 2 0 2 A 4 A 2 Fig. 4-24 The nodal equations in matrix form are written by inspection. 1 1 1 2 10 5 1 1 Va -6 5 -0.20 0.70 0.30 -0.20 2 -6 = 1.18 V -0.20 0.70 Then, I V,/10 0.118 A
ILIPS 01 SLC Saudi Lighting Date: .. ZA 2 4 hare I 2A I3 =4A IIM+4 0 -247 A ILO7A
Transcribed Image Text:ILIPS 01 SLC Saudi Lighting Date: .. ZA 2 4 hare I 2A I3 =4A IIM+4 0 -247 A ILO7A
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