2. Use mesh-current analysis in the circuit below. The component values are R1 = 4 Q, R2 = 5 2, R3 = 8 Q, R4 = 6 Q, R5 = 8 2, and Is = 4A. !! %3D R Determine Mesh i1, i2, and i3 (Hint: look at the example we did so you can know the position of each mesh) The value of current at R4. The value of voltage at R2 and Rs. The value of power at R1 and R3. Ry Ry R4 Exercise 4: What if our circuit contains current source among two mesh i +2(i1 – i3) + 4(i2 – i3) + 10 = 0 i3 2Ω 4Ω Next, we can write a KVL equation for mesh 3: 3iz + 4(i3 – i2) + 2(i3 – i1) = 0 ΙΩ 10 V current source is 5 A. Thus, we have 5 A i2 - i = 5 As usual, we have defined the mesh currents flowing clockwise. We cannot write a KVL equation around mesh 1 because the voltage across the 5-A current source is unknown (and we do not want to increase the number of unknowns in our equations). A solution is to combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined. This yields
2. Use mesh-current analysis in the circuit below. The component values are R1 = 4 Q, R2 = 5 2, R3 = 8 Q, R4 = 6 Q, R5 = 8 2, and Is = 4A. !! %3D R Determine Mesh i1, i2, and i3 (Hint: look at the example we did so you can know the position of each mesh) The value of current at R4. The value of voltage at R2 and Rs. The value of power at R1 and R3. Ry Ry R4 Exercise 4: What if our circuit contains current source among two mesh i +2(i1 – i3) + 4(i2 – i3) + 10 = 0 i3 2Ω 4Ω Next, we can write a KVL equation for mesh 3: 3iz + 4(i3 – i2) + 2(i3 – i1) = 0 ΙΩ 10 V current source is 5 A. Thus, we have 5 A i2 - i = 5 As usual, we have defined the mesh currents flowing clockwise. We cannot write a KVL equation around mesh 1 because the voltage across the 5-A current source is unknown (and we do not want to increase the number of unknowns in our equations). A solution is to combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined. This yields
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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What are the answers to question 2? I put another photo where you can or an explanation to know how to solve the problem.
Transcribed Image Text:2. Use mesh-current analysis in the circuit below. The component values are
R1 = 4 Q, R2 = 5 2, R3 = 8 Q, R4 = 6 Q, R5 = 8 2, and Is = 4A.
!!
%3D
R
Determine
Mesh i1, i2, and i3 (Hint: look at the example we did
so you can know the position of each mesh)
The value of current at R4.
The value of voltage at R2 and Rs.
The value of power at R1 and R3.
Ry
Ry
R4
Transcribed Image Text:Exercise 4: What if our circuit contains
current source among two mesh
i +2(i1 – i3) + 4(i2 – i3) + 10 = 0
i3
2Ω
4Ω
Next, we can write a KVL equation for mesh 3:
3iz + 4(i3 – i2) + 2(i3 – i1) = 0
ΙΩ
10 V
current source is 5 A. Thus, we have
5 A
i2 - i = 5
As usual, we have defined the mesh currents flowing clockwise. We cannot write a
KVL equation around mesh 1 because the voltage across the 5-A current source is
unknown (and we do not want to increase the number of unknowns in our equations).
A solution is to combine meshes 1 and 2 into a supermesh. In other words, we write
a KVL equation around the periphery of meshes 1 and 2 combined. This yields
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