problems with assistance module 3 – problem 4 filename: pwa_mod03_prob04.ppt next slide go...

Dave ShattuckUniversity of Houston

© Brooks/Cole Publishing Co.

Problems With AssistanceModule 3 – Problem 4

Filename: PWA_Mod03_Prob04.ppt

Next slide

Go straight to the Problem Statement

Go straight to the First Step



Overview of this Problem

In this problem, we will use the following concepts:

• Kirchhoff’s Voltage Law

• Kirchhoff’s Current Law

• Ohm’s Law

• The Mesh-Current Method

Next slide

Go straight to the Problem Statement

Go straight to the First Step



Textbook Coverage

This material is covered in your textbook in the following sections:

• Circuits by Carlson: Sections 4.2 & 4.3• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections

4.1, & 4.5 through 4.7• Basic Engineering Circuit Analysis 6th Ed. by Irwin and

Wu: Section 3.2• Fundamentals of Electric Circuits by Alexander and

Sadiku: Sections 3.4 & 3.5• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections

4-5 & 4-6

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Coverage in this Module

The material for this problem is covered in this module in the following presentations:

DPKC_Mod03_Part03 and DPKC_Mod03_Part04

This same problem is solved with the Node-Voltage Method in

• PWA_Mod03_Prob01

Next slide



iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Problem Statement

Next slide

Use the mesh-current method to solve for the voltage vX.



iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Solution – First Step – Where to Start?

How should we start this problem? What is the first step?

Next slide




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Problem Solution – First Step

How should we start this problem? What is the first step?

a) Write KVL for each mesh

b) Identify the meshes and define the mesh currents

c) Write KCL for each node

d) Combine resistors in parallel or series




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for First Step – Write KVL for each mesh

This is not a good choice for the first step, although we will write KVL equations for most meshes soon.

One purpose of the mesh-current method is to find a systematic way of writing the correct number of equations. It is important, then, to know how many equations we are going to write.

Go back and try again.




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for First Step – Write KCL for each node

This is not a good choice.

The mesh-current method involves writing KVL equations, not KCL equations. While we may write KCL equations as needed for constraint equations, it is not the systematic step that we take in using the mesh-current method. This is not the way to start this method.

Go back and try again.




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for First Step was – Combine resistors in parallel or series

This might be helpful, but is not the best choice for the first step.

Generally, it is a good thing to simplify a circuit, where we can do so. Here, you may have noted that R4 and R3 are in parallel, and can be combined into a single resistor. We will not even lose any dependent source variables. However, the mesh-current method does not require that we simplify the circuits, and sometimes we cannot do so. Therefore, we recommend that you go back and try again.




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for First Step was – Identify the meshes and define the mesh currents

This is the best choice.

The first step is to make sure that we have identified all the meshes and defined the mesh currents.

How many meshes are there in this circuit? Your answer is:

a) 3 meshes

b) 4 meshes

c) 5 meshes

d) 6 meshes

e) 7 meshes




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for the number of essential nodes – 3

This is not correct. Try again.



© Brooks/Cole Publishing Co. Your choice for the number of essential nodes – 4


iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

This is not correct. Try again.



iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for the number of essential nodes – 5


This is correct. Let’s define the mesh currents.



iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for the number of essential nodes – 6This is not correct. Try again.




iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R 4=

33[W

]

iS2=2[S] vX

vX+ -

Your choice for the number of essential nodes – 7This is not correct. Try again.



© Brooks/Cole Publishing Co. Defining the Mesh Currents

The next step is to define the mesh currents.

We have done so here. Now, we are ready to write the Mesh-Current Method Equations. Even before we do, we can predict that we will need to write six equations, one for each mesh (5) and one for the dependent source variable vX.

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB



© Brooks/Cole Publishing Co. Writing the Mesh-Current Equations – 1

The equation for Mesh A is obtained from the current source:

A: 0.5[A]Ai

Next equation

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB




The equation for Mesh B is obtained by writing KVL around mesh B.

B: 5[V] ( )22[ ] 0B Ei i W

Next equation

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB




The equation for Mesh C is:

C: 5[V] 27[ ] ( )33[ ] 0C C Di i i W W

Next equation

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB




The equation for Mesh D is:

D: ( )33[ ] ( )39[ ] 0D C D Ei i i i W W

Next equation

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB




The equation for Mesh E is obtained from the dependent current source, which determines the mesh current.

E: 2[S]E Xi v

Next equation

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB




The equation for the dependent source variable vX is:

: ( )39[ ]X X E Dv v i i W

Make sure that you agree with this equation for the dependent source variable vX . It can be obtained by writing Ohm’s Law for the 39[W] resistor. The branch current through this resistor, going from left to right, is iE-iD.

Next step

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB



© Brooks/Cole Publishing Co.Writing the Node-Voltage Equations – All

A: 0.5[A]

B: 5[V] ( )22[ ] 0

C: 5[V] 27[ ] ( )33[ ] 0

D: ( )33[ ] ( )39[ ] 0

E: 2[S]

: ( )39[ ]

A

B E

C C D

D C D E

E X

X X E D

i

i i

i i i

i i i i

i v

v v i i

W

W W W W

W

The next step is to solve the equations. We can do this by various approaches. We will choose to use MathCAD for this module.

Next step

iS1=0.5[A] R5=

10[W]

R1=22[W]

R2=27[W]

+

-vS=

5[V]

R3=39[W]

R4=33[W]

iS2=2[S] vX

vX+ -

iA

iD

iC

iE

iB



© Brooks/Cole Publishing Co.Solving the Mesh- Current Equations

0.5[A]

0.033[A]

0.189[A]

0.192[A]

0.194[A]

0.097[V]

A

B

C

D

E

X

i

i

i

i

i

v

See Note

A: 0.5[A]

B: 5[V] ( )22[ ] 0

C: 5[V] 27[ ] ( )33[ ] 0

D: ( )33[ ] ( )39[ ] 0

E: 2[S]

: ( )39[ ]

A

B E

C C D

D C D E

E X

X X E D

i

i i

i i i

i i i i

i v

v v i i

W

W W W W

W

The next step is to solve the equations. We can do this by various approaches. We will choose to use MathCAD for this module.

Go back to Overview

slide.


© Brooks/Cole Publishing Co.What if I like the Mesh-Current Method much more than the Node-Voltage Method?

• If you like the Mesh-Current Method more than the Node-Voltage Method, you are in the majority of beginning circuit-analysis students.

• However, if you note that this problem was the same one as was solved in PWA_Prob01 in this module, you will find that here the solution required 6 equations (really only 5, when we ignore the A mesh equation that was not used). When it was done with the Node-Voltage Method, there were only 4 equations, and one of those was not used, and we could solve it easily by hand.

• Even if you prefer one method, learn them both!

Go back to Overview

slide.

problems with assistance module 3 – problem 4 filename: pwa_mod03_prob04.ppt next slide go...

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