iök Ri 25k FVe ta Ial.330k IL R MA R3 R9 20k Rs Ic SISK 10k

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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can you let me know if my mesh equations are correct?

2 Io
must ditimine all mesh ciments
from HW#5 putblim 3
find
fail
I-3mA foV
KVE IR-IF
Ri
25k FV ta Ial3304 ILO Zvx Rr 25R
Ial330k IE
VIR
Rz10K
3MA
6 mA
R9
20k
R3
Rs 1UK
310K Re 30k
Rs
Id
2 rugu msh ? In,īe d IL Id
VK- IRI= 3mA(25k) =75V
ZV 30kCIbId.
0=25Ta-25I-30(Ia- Ib)-IS(Ic-Id) - 29 Ic -20IA
0=30(16- Ia)-ZVx- 16I2-IS(Id-Ic)
Ia-Ic 3mA
Id-Ib-6mA
Transcribed Image Text:2 Io must ditimine all mesh ciments from HW#5 putblim 3 find fail I-3mA foV KVE IR-IF Ri 25k FV ta Ial3304 ILO Zvx Rr 25R Ial330k IE VIR Rz10K 3MA 6 mA R9 20k R3 Rs 1UK 310K Re 30k Rs Id 2 rugu msh ? In,īe d IL Id VK- IRI= 3mA(25k) =75V ZV 30kCIbId. 0=25Ta-25I-30(Ia- Ib)-IS(Ic-Id) - 29 Ic -20IA 0=30(16- Ia)-ZVx- 16I2-IS(Id-Ic) Ia-Ic 3mA Id-Ib-6mA
Expert Solution
Step 1

One method used for analysis of electrical circuits is the mesh analysis. This method includes assuming a distinct current within the mesh and the polarities of drops in the individual elements is determined by the direction of current so assumed for the particular loop. The branch currents are then determined by solving the set of equations obtained by applying KVL in each closed loop.

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