problems with assistance module 3 – problem 2 filename: pwa_mod03_prob02.ppt this problem is...

Dave ShattuckUniversity of Houston

© Brooks/Cole Publishing Co.

Problems With AssistanceModule 3 – Problem 2

Filename: PWA_Mod03_Prob02.ppt

This problem is adapted from:

Problem 4.6, page 183 in Circuits by A. Bruce Carlson

Brooks/Cole Thomson Learning

2000

ISBN: 0-534-37097-7

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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 Node-Voltage Method

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Go straight to the Problem Statement

Go straight to the First Step



Textbook Coverage

The material for this problem is covered in your textbook in the following sections:

• Circuits by Carlson: Sections 4.1 & 4.3• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections

4.2 through 4.4• Basic Engineering Circuit Analysis 6th Ed. by Irwin and

Wu: Section 3.1• Fundamentals of Electric Circuits by Alexander and

Sadiku: Sections 3.2 & 3.3• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections

4-2 through 4-4

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Coverage in this Module

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

• DPKC_Mod03_Part01 and DPKC_Mod03_Part02

A similar problem is worked in:

• PWA_Mod03_Prob01

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© Brooks/Cole Publishing Co. Problem Statement

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Find v2, v4, and the power supplied by each source.

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co. Solution – First Step – Where to Start?

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

Next slide


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co.Problem Solution – First Step

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

a) Write KCL for each node

b) Identify the essential nodes

c) Write KVL for each loop

d) Pick the reference node

e) Combine resistors in parallel or series


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co.Your choice for First Step – Write KCL for each node


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

This is not the best choice for the first step, although we will write KCL equations for most nodes soon.

It is generally worth while to spend some time looking at the problem and choosing an approach before beginning to write equations. Note that we have six variables defined already, but will not need that many.

Go back and try again.


© Brooks/Cole Publishing Co.Your choice for First Step – Write KVL for each loop

This is not the best choice for the first step.

It is generally worth while to spend some time looking at the problem and choosing an approach before beginning to write equations. Note that we have six variables defined already, but will not need that many.

Go back and try again.Find v2, v4, and the power supplied by each source.

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co.Your choice for First Step was – Pick the reference node


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

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

The node-voltage method indeed requires that we pick, and label, the reference node. However, it is usually wise to be sure that we know where all the essential nodes are, and how many connections they have, before making this choice. Thus, while it may not be necessary in simple problems like this, we recommend that you go back and try again.



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

series


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

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, we cannot do so since, there are no resistors in series or parallel. Therefore, we recommend that you go back and try again.

Note to advanced students: We could use delta-to-wye or wye-to-delta transformations, but we are going to take a different approach here.



Your choice for First Step was – Identify the essential nodes


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

This is the best choice. By making sure that we have identified the essential nodes, we can determine how many equations will be needed in the node-voltage method.

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

a)3 essential nodes

b)4 essential nodes

c)5 essential nodes


© Brooks/Cole Publishing Co. Your choice for the number of essential nodes – 4This is not correct. Remember

that essential nodes must have at least 3 connections. In addition, remember that two nodes connected by a wire were really only one node. Try again.


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co. Your choice for the number of essential nodes – 3

This is correct. The essential nodes are marked with red in this schematic. There is a non-essential node, which is marked with green.

With only 3 essential nodes, the node-voltage method is a good choice, since we will have only 2 simultaneous equations. The next step is to pick one of them as the reference node. Which one should we pick?


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co. Your choice for the number of essential nodes – 5

This is not correct. Try again.

Remember that two nodes connected by a wire were really only one node.


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6


© Brooks/Cole Publishing Co. Choosing the Reference Node

The next step is to pick one of them as the reference node. We have chosen the node at the bottom as the reference node. This is considered to be the best choice, since it has 4 connections to it. The equations will probably be easier to write with this as reference node. In addition, the two node voltages that result are the voltages we were asked to find. Next, we define the node-voltages.


+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6



Defining the Node-Voltages

The next step is to define the node-voltages.

We have done so here. Now, we are ready to write the Node-Voltage Method Equations. Even before we do, we can predict that we will need to write two equations, one for each non-reference essential node.

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4



© Brooks/Cole Publishing Co. Writing the Node-Voltage Equations – 1

The equation for Node 2 is given here. Note that we used an expression for the current in R1 to express the current in the voltage source. The resistor R1 and the voltage source are in series.

2 2 4 2

2 2 4 2

Node 2: 0, or10[k ] 18[k ] 15[k ]

60[V]Node 2: 0

10[k ] 18[k ] 15[k ]

Sv v v v v

v v v v

W W W

W W W

Next equation

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4



© Brooks/Cole Publishing Co. Writing the Node-Voltage Equations – 2

The equation for Node 4 is given here.

4 4 2Node 4: 3[mA] 012[k ] 18[k ]

v v v

W W

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4


Next step


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

2 2 4 2

4 4 2

60[V]Node 2: 0

10[k ] 18[k ] 15[k ]

Node 4: 3[mA] 012[k ] 18[k ]

v v v v

v v v

W W W

W W

The next step is to solve the equations. Let’s solve.

Next step

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4



© Brooks/Cole Publishing Co.Solving the Node-Voltage Equations

When we solve, we find that

v2 = 36[V], and

v4 = 36[V].

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4


Next step

We have used MathCAD to solve the two simultaneous equations. This is shown in a MathCAD file called

PWA_Mod03_Prob02_Soln.mcd

which should be available in this module.



Using the Node-Voltages to Solve for Desired Quantities – Part 1We found that

v2 = 36[V], and

v4 = 36[V].

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4


We can use this to find the power supplied by the current source directly. Note that v4 is the voltage across the current source. Note also that v4 and iS are in the active convention for this source. Therefore, we can write:

pdel,iS = v4 iS = 36[V]3[mA] = 108[mW]

Next step



Using the Node-Voltages to Solve for Desired Quantities – Part 2We found that

v2 = 36[V], and

v4 = 36[V].

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4


Next, we want to find the power supplied by the voltage source. For this, we need to find the current through the voltage source, which has already been labeled as i1. We write the expression for this just as we had when writing the KCL expression for node 2.

21

1

60[V]

10[k ]

60[V] 36[V]2.4[mA]

10[k ]

vi

i

W

W

Next step



Using the Node-Voltages to Solve for Desired Quantities – Part 3

See Note

We found that

v2 = 36[V], and

v4 = 36[V].

+

-

vS=60[V]

R1=10[kW]

R2=15[kW]

R3=18[kW]

R4=12[kW]

iS=3[mA]

v2

+

-

v4

+

-

i1 i3i2 i4

Figure P4.6

2 4


Now, we can find the power supplied by the voltage source. Note that vS and i1 are in defined in the active convention for the voltage source. Thus, we can write:

, 1 60[V]2.4[mA] 144[mW]Sdel v Sp v i


© Brooks/Cole Publishing Co.What happened? The two node-voltages were the same!

• It is true that in this problem, the two node-voltages were the same. This occurred because Carlson chose the values of vS, iS, R1, R2 and R4 to make this happen. You can prove to yourself that R3 makes no difference in this case by varying its value, and solving again. For all nonzero values of R3, the solution will be the same.

• This raises yet another important point. The Node-Voltage Method gives us general equations which apply for the way the circuit is laid out, called the topology. Once you have the equations, you could set v2=v4, and solve for values of vS, iS, R1, R2 and R4 to make this happen. The node-voltage technique, once in hand, has many uses.

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