What's the solution? A Fundamentals_of_Electric_Circuits_6th_Ed.pdf - Adobe ReaderFile Edit View Window Help148% -E EFrom Eqs. (3.7.1) to (3.7.4),Оpen98(121 of 991)ToolsFill & SignCommenti = -7.5 A,iz = -2.5 A,iz = 3.93 A,i4 = 2.143 APractice Problem 3.7Use mesh analysis to determine i¡, iz, and iz in Fig. 3.25.Answer: i = 12.379 A, iz = 378.9 mA, iz = 3.284 A.5Ω24 V10 Ω4 A20 Ωİ Nodal and Mesh Analysesby Inspection3.6Figure 3.25This section presents a generalized procedure for nodal or mesh analysis.It is a shortcut approach based on mere inspection of a circuit.When all sources in a circuit are independent current sources, wedo not need to apply KCL to each node to obtain the node-vequations as we did in Section 3.2.mere inspection of the circuit. As an e xample, let us ree xamine thecircuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. Thecircuit has two nonreference nodes and the node equations were de-For Practice Prob. 3.7.oltageWe can obtain the equations byG2V17.99 x 10.24 inV2ww

From the given circuit, I1-I2=4I1=4+I2 .................(a)…

Answered: Use mesh analysis to determine ij, iz,…

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Use mesh analysis to determine ij, iz, and iz in Fig. 3.25.

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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A Fundamentals_of_Electric_Circuits_6th_Ed.pdf - Adobe Reader
File Edit View Window Help
148% -E E
From Eqs. (3.7.1) to (3.7.4),
Оpen
98
(121 of 991)
Tools
Fill & Sign
Comment
i = -7.5 A,
iz = -2.5 A,
iz = 3.93 A,
i4 = 2.143 A
Practice Problem 3.7
Use mesh analysis to determine i¡, iz, and iz in Fig. 3.25.
Answer: i = 12.379 A, iz = 378.9 mA, iz = 3.284 A.
5Ω
24 V
10 Ω
4 A
20 Ω
İ Nodal and Mesh Analyses
by Inspection
3.6
Figure 3.25
This section presents a generalized procedure for nodal or mesh analysis.
It is a shortcut approach based on mere inspection of a circuit.
When all sources in a circuit are independent current sources, we
do not need to apply KCL to each node to obtain the node-v
equations as we did in Section 3.2.
mere inspection of the circuit. As an e xample, let us ree xamine the
circuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The
circuit has two nonreference nodes and the node equations were de-
For Practice Prob. 3.7.
oltage
We can obtain the equations by
G2
V1
7.99 x 10.24 in
V2
ww

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