Chapter 1 Thévenin’s and Norton's Theorems

Thévenin’s Theorem Review

In this chapter, we will cover the following topics:

• Thévenin’s Theorem

• Finding Thévenin’s equivalents

• Example of finding a Thévenin’s equivalent

Thévenin’s Theorem Defined

Thévenin’s Theorem is another equivalent circuit. Thévenin’s Theorem can be stated as follows:

Any circuit made up of resistors and sources, viewed from two terminals of that circuit, is equivalent to a voltage source in series with a resistance.

Figure 1.1 Thévenin Equivalent. Courtesy of David Shattuck, used with personal permission.

The voltage source is equal to the open-circuit voltage for the
two-terminal circuit, and the
resistance is equal to the
equivalent resistance of the circuit. We can write this in equation form as

We have introduced a term called the open-circuit voltage. This is the voltage for the circuit that we are finding the equivalent of, with nothing connected to the circuit. Connecting nothing means an open circuit. This voltage is shown here, with the voltage v_OC being defined at the terminals of the yellow circuit, with nothing being added.

The polarities of the source with respect to the terminals is important. If the reference polarity for the open-circuit voltage is as given here (voltage drop from A to B), then the reference polarity for the voltage source must be as given here at the right of Figure 1.2 (positive terminal closer to A, and negative terminal closer to B).

Open Circuit Voltage Definition.png — Figure 1.2 Definition of the Open Circuit Voltage. Courtesy of David Shattuck, used with personal permission.

We have introduced a term called the equivalent resistance. This is the resistance for the circuit that we are finding the equivalent of, with the independent sources set equal to zero. Any dependent sources are left in place.

As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalent circuits.

Short Circuit Current Definition.png — Figure 1.3 Definition of the Short Circuit Current. Courtesy of David Shattuck, used with personal permission.

voc isc req definitions.png — Equation 1.2

A useful concept is the concept of short-circuit current. This is the current that flows through a wire, or short circuit, connected to the terminals of the circuit. This current is shown here as i_SC.

The polarities of the source with respect to the terminals is important. If the reference polarity for the short-circuit current is as given here (flowing through the wire from A to B), then the reference polarity for the voltage source must be as given here (positive terminal closer to A, and negative terminal closer to B), where we are relating these three quantities with the equation,

FOhm's Law.png — Equation 1.3 This is not Ohm's Law, but it does look just like Ohm's Law.

When we look at the circuit on the right of Figure 1.3, we can see that the short-circuit current is equal to v_TH/R_TH, which is also v_OC/R_EQ. Thus, we obtain the important expression for i_SC, shown here in Equation 1.3.

Note: You will note be credited with getting the correct answer unless your exact wording is the same as mine. This is not important, since credit is being given for participation here. We will discuss the correct answer in class later.

Any Two of the Three Quantities

For the Thévenin equivalent, the open-circuit voltage is equal to the short-circuit current times the equivalent resistance. In general we can find the Thévenin equivalent of a circuit by finding
any two of the following three things in the equation

FOhm's Law.png — Equation 1.3 (repeated)

The three things are

1) the open circuit voltage, v_OC,

2) the short-circuit current, i_SC, and

3) the equivalent resistance, R_EQ.

Once we find any two, we can find the third by using this equation.

One more time, the reference polarities of our voltages and currents matter. If we pick v_OC at A with respect to B, then we need to pick i_SC going from A to B. If not, we need to change the sign in this equation.

Plus sign voc isc relationship.png — Figure 1.4 Relationship of v_OC and i_SC. Courtesy of David Shattuck, used with personal permission.

As an example, if we pick v_OC and i_SC with the reference polarities given here, we need to change the sign in the equation as shown.

Minus sign voc isc relationship.png — Figure 1.5 Alternative relationship of v_OC and i_SC. Courtesy of David Shattuck, used with personal permission.

FOhm's Law with minus sign.png — Equation 1.4

This is a consequence of the sign in Ohm’s Law. For a further explanation, see the next diagram. Note that in this diagram, where the short circuit current i_SC is defined going from B to A through the short circuit, that current is flowing through the resistance R_EQ in active sign relationship with respect to the voltage v_OC which is set by the polarity of the voltage source, in turn set by the choice of definition of the open circuit voltage at A with respect to B.

Be very careful here! We have labeled the voltage across the resistance R_EQ as v_OC. This is true only for this special case. This v_OC is not the voltage at A with respect to B in this circuit. In this circuit, that voltage is zero due to the short. Due to the short, the voltage across R_EQ is v_OC.

Example showing ASC voc isc.png — Figure 1.6 Alternative relationship of v_OC and i_SC. Courtesy of David Shattuck, used with personal permission.

When we have dependent sources in the circuit in yellow, it will make some calculations more difficult, but does not change the validity of the theorem.

Summary to this Point

1. We can find the Thévenin equivalent of any circuit made up of voltage sources, current sources, and resistors. The sources can be any combination of dependent and independent sources.

2. We can find the values of the Thévenin equivalent by finding the open-circuit voltage and short-circuit current. The reference polarities of these quantities are important.

3. To find the equivalent resistance, we need to set the independent sources equal to zero. However, the dependent sources will remain. This requires some care. We will discuss finding the equivalent resistance with dependent sources in a later part of this section.

4. As with all equivalent circuits, the Thévenin equivalent is equivalent only with respect to the things connected to it.

Example Problem 1.1

We wish to find the Thévenin equivalent of the circuit below, as seen from terminals A and B.

Thev Example Problem 1.png — Figure 1.7 Circuit for Example 1.1. Note that there is an unstated assumption here; we assume that we will later connect something to these two terminals. Having found the Thévenin equivalent, we will be able to solve that circuit more easily by using that equivalent. Courtesy of David Shattuck, used with personal permission.

We wish to find the open-circuit voltage v_OC with the polarity defined in the circuit given below. We have also defined the node voltage v_C, which we will use to find v_OC.

In general, remember, we need to find two out of three of the quantities v_OC, i_SC, and R_EQ. In this problem we will find two, and then find the third just as a check. In general, finding the third quantity is not required.

Thev Example Problem 2.png — Figure 1.8 Open-Circuit Voltage Circuit for Example 1.1. Courtesy of David Shattuck, used with personal permission.

Writing KCL at the node encircled with a dashed red line, we have

Thev Example Equation 1.png — Equation 1.5

Thev Example Equation 2.png — Equation 1.6

Plugging in values, we get

Then, using VDR, we can find

Thev Example Equation 3.png — Equation 1.7

Next, we will find the equivalent resistance, R_EQ. The first step in this solution is to set the independent sources equal to zero. We then have the circuit below.

Thev Example Problem 3.png — Figure 1.9 Circuit for Equivalent Resistance. Note that the voltage source becomes a short circuit, and the current source becomes an open circuit. These represent zero-valued sources. Courtesy of David Shattuck, used with personal permission.

To find the equivalent resistance, R_EQ, we simply combine resistances in parallel and in series. The resistance between terminals A and B, which we are calling R_EQ, is found be recognizing that R₁ and R₃ are in parallel. That parallel combination is in series with R₂. That series combination is in parallel with R₄. We have

Thev Example Equation 4.png — Equation 1.8

To complete this problem, we would typically redraw the circuit, showing the complete Thévenin’s equivalent, along with terminals A and B. This has been done here. This shows the proper polarity for the voltage source.

Thev Example Problem 4.png — Figure 1.10 Solution for Example Problem 1.1. Because we solved for the voltage at A with respect to B, the value is attached to the voltage source with the plus sign closer to A, and the minus sign closer to B. Courtesy of David Shattuck, used with personal permission.

Let’s check this solution, by finding the short-circuit current in the original circuit, and compare it to the short-circuit current in the Thévenin’s equivalent. We will start with the Thévenin’s equivalent shown here. We have

Thev Example Problem 5.png — Figure 1.11 Testing the Solution for Example Problem 1.1. Courtesy of David Shattuck, used with personal permission.

Thev Example Equation 5.png — Equation 1.9

Let’s find the short-circuit current in the original circuit.

Thev Example Problem 6.png — Figure 1.12 Testing the Solution for Example Problem 1.1, in a Different Way. Courtesy of David Shattuck, used with personal permission.

We have KCL at the dashed closed surface,

Thev Example Equation 6.png — Equation 1.10

Note that resistor R₄ is neglected, since it has no voltage across it, and therefore no current through it. With this result, we can find the short-circuit current in the original circuit,

Thev Example Equation 7.png — Equation 1.11

This is the same result that we found using the Thévenin’s equivalent earlier. The solution checks. This is important. This shows that we could indeed have found any two of three of the quantities: open-circuit voltage, short-circuit current, and equivalent resistance, as we can see from

Thev Example Equation 8.png — Equation 1.12

Summary

What is the deal here? Is this worth all this trouble? This is a good question. The deal here is that Thévenin’s Theorem is a very big deal. It is difficult to convey the full power of it at this stage in your education. However, you may be able to imagine that it is very useful to be able to take a very complicated circuit, and replace it with a pretty simple circuit. In many cases, it is very definitely worth all this trouble. There is one example you may have seen in electronics laboratories. There, the signal generator outputs are typically labeled 50[ ]. This means that the Thévenin’s equivalent resistance, for the complicated circuit inside the generator, is 50[ ], as viewed from the output terminals. Knowing this makes using the generator easier. We view the generator as just an adjustable voltage source in series with a 50[ ] resistor.

Norton's Theorem

In this part, we will cover the following topics:

• Norton’s Theorem Defined
• Finding Norton’s equivalents
• Example of finding a Norton’s equivalent

Norton’s Theorem Defined

Norton’s Theorem is another equivalent circuit. Norton’s Theorem can be stated as follows: Any circuit made up of resistors and sources, viewed from two terminals of that circuit, is equivalent to a current source in parallel with a resistance.

The current source is equal to the short-circuit current for the two-terminal circuit, and the resistance is equal to the equivalent resistance of the circuit.

Norton Equiv Diagram 1.png — Figure 1.13 Norton's Theorem. Courtesy of David Shattuck, used with personal permission.

This holds for when

Norton Equiv Equation 1.png — Equation 1.13

It is probably obvious to you, if you studied the last parts of this chapter, that if Thévenin’s Theorem is valid, then Norton’s Theorem is valid, because Norton’s Theorem is simply a source transformation of Thévenin’s Theorem. Note that the resistance value is the same in both cases, that is, R_TH = R_N = R_EQ. We remember that

Norton Equiv Equation 2.png — Equation 1.14

The polarities for sources are almost always important. Here the polarity of the current source depends on the reference polarity of the short-circuit current.

Norton Equiv Diagram 2.png — Figure 1.14 Polarity of Current Source. Courtesy of David Shattuck, used with personal permission.

As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalent circuits. When we have dependent sources in the circuit shown here, it will make some calculations more difficult, but does not change the validity of the theorem. The open-circuit voltage that results from the Norton equivalent is equal to the product of the Norton current source and the Norton resistance. This leads to the same equation that we used previously, for the Thévenin equivalent,

Norton Equiv Equation 3.png — Equation 1.15

Norton Equiv Diagram 3.png — Figure 1.15 Equation for the Open-Circuit Voltage. Courtesy of David Shattuck, used with personal permission.

We have shown that for the Norton equivalent, the open-circuit voltage is equal to the short-circuit current times the equivalent resistance. This is fundamental and important. However, it is not Ohm’s Law. Remember that i_SC = i_N, and R_EQ = R_N. This was given earlier in Equation 1.13. In general we can find the Norton equivalent of a circuit by finding any two of the following three things:

1) the open circuit voltage, v_OC,
2) the short-circuit current, i_SC, and
3) the equivalent resistance, R_EQ.

As before, in Thevenin's theorem, we have to be careful about our choices of reference polarities for v_OC and i_SC. This is explained in the following figures.

Norton Equiv Diagram 4.png — Figure 1.16 Sign of the Equation in Relation to Reference Polarities. Courtesy of David Shattuck, used with personal permission.

The alternative option is shown in the next figure.

Norton Equiv Diagram 5.png — Figure 1.17 Alternative Sign in Relation to Reference Polarities. Courtesy of David Shattuck, used with personal permission.

As an example, if we pick v_OC and i_SC with the reference polarities given here, we need to change the sign in the equation as shown. This is a consequence of Ohm’s Law, which for resistor R_N requires a minus sign, since the voltage and current are in the active sign relationship in the next figure, for the resistance R_N.

Norton Equiv Diagram 6.png — Figure 1.18 Explanation for Sign Used. Courtesy of David Shattuck, used with personal permission.

Summary

1. We can find the Norton equivalent of any circuit made up of voltage sources, current sources, and resistors. The sources can be any combination of dependent and independent sources.

2. We can find the values of the Norton equivalent by finding the open-circuit voltage and short-circuit current. The reference polarities of these quantities are important.

3. To find the equivalent resistance, we need to set the independent sources equal to zero. However, the dependent sources will remain. This requires some care. We will discuss finding the equivalent resistance with dependent sources in the next part of this chapter.

4. As with all equivalent circuits, the Norton equivalent is equivalent only with respect to the things connected to it.

What is the deal here? Is this worth all this trouble? This is a good question. The deal here is that Norton’s Theorem is a very big deal. It is difficult to convey the full power of it at this stage in your education. However, you may be able to imagine that it is very useful to be able to take a very complicated circuit, and replace it with a pretty simple circuit. In many cases, it is very definitely worth all this trouble. Norton’s Theorem is often treated as being less important than Thévenin’s Theorem. Actually, they are of equal importance, since they say essentially the same thing. One is no more important than the other.

Example Problem 1.2

We wish to find the Norton equivalent of the circuit below, as seen from terminals A and B. Note that there is an unstated assumption here; we assume that we will later connect something to these two terminals. Having found the Norton equivalent, we will be able to solve that circuit more easily by using that equivalent.

Norton Example Diagram 1.png — Figure 1.19 Circuit for Example Problem 1.2. Courtesy of David Shattuck, used with personal permission.

We wish to find the open-circuit voltage v_OC with the polarity defined in the circuit given below. In general, remember, we need to find two out of three of the quantities v_OC, i_SC, and R_EQ. In this problem we will find two, and then find the third just as a check. In general, finding the third quantity is not required.

Norton Example Diagram 2.png — Figure 1.20 Defining v_OC for this Circuit. Courtesy of David Shattuck, used with personal permission.

Some students may be tempted to remove resistor R₃ from this circuit. We should not do this. In future problems, if we are asked to find “the equivalent circuit seen by resistor R₃”, then we assume that the resistor “does not see itself”, and remove it. In this problem, we are not given this instruction. Leave the resistor in place, even though the open-circuit voltage is across it.

We wish to find the voltage v_OC. Writing KCL at the node encircled with a dashed red line, we have

Norton Example Equation 1.png — Equation 1.16

Substituting in values, we have

Norton Example Equation 2.png — Equation 1.17

Next, we will find the equivalent resistance, R_EQ. The first step in this solution is to set the independent sources equal to zero. We then have the circuit below.

Norton Example Diagram 3.png — Figure 1.21 Find the Equivalent Resistance, R_EQ. Courtesy of David Shattuck, used with personal permission.

To find the equivalent resistance, R_EQ, we simply combine resistances in parallel and in series. The resistance between terminals A and B, which we are calling R_EQ, is found by recognizing that R₅ and R₄ are in series. That series combination is in parallel with R₂. That parallel combination is in parallel with R₃. We have

Norton Example Equation 3.png — Equation 1.18

Some students may have difficulty getting this expression. Remember that we are finding the resistance seen at these two terminals, A and B. The idea is that we would have this resistance if you connected a source to these two terminals.

To complete this problem, we would typically redraw the circuit, showing the complete Norton’s equivalent, along with terminals A and B. This has been done here. To get this, we need to use our equation to get the Norton current,

Norton Example Equation 4.png — Equation 1.19

Norton Example Diagram 4.png — Figure 1.22 Solution for Example Problem 1.2. Courtesy of David Shattuck, used with personal permission.

Let us check this solution, by finding the short-circuit current in the original circuit, and compare it to the short-circuit current in the Norton’s equivalent. We redraw the original circuit, with the short circuit current shown. We wish to find this short circuit current, i_SC.

Norton Example Diagram 5.png — Figure 1.23 Testing the Solution, Using i_SC. Courtesy of David Shattuck, used with personal permission.

We start by noting that that there is no current through resistor R₃, since there is no voltage across it. Another way of saying this is that the resistor R₃ is in parallel with a short circuit. The parallel combination of the resistor and the short circuit, will be a short circuit. The same exact argument can be made for the series combination of R₅ and R₄. This series combination is in parallel with a short circuit. Thus, we can simplify this circuit to the circuit in the next figure.

Norton Example Diagram 6.png — Figure 1.24 Simplified Circuit for i_SC. Courtesy of David Shattuck, used with personal permission.

Here, we have removed resistors R₃, R₄ and R₅ since they do not affect the short circuit current, i_SC. When we look at this circuit, we note that the voltage source v_S is directly across the resistor R₂, and so we can write directly,

Norton Example Equation 5.png — Equation 1.20

This short-circuit current is the same result that we found in the Norton’s Equivalent earlier. In retrospect, it is now clear that we did not take the best possible approach to this solution. If we had solved for the short-circuit current, and the equivalent resistance, we would have gotten the solution more quickly and more easily. One of our goals is to be so good at circuit analysis that we can see ahead of time which approach will be the best for a given problem.

Thévenin’s and Norton’s Theorems including Dependent Sources

In this part, we will cover the following topics:

Dependent Sources and Equivalent Resistance

The Test-Source Method

Example of finding an equivalent resistance with a dependent source present

Thévenin’s and Norton’s Theorems Reviewed

Thévenin’s Theorem and Norton’s Theorem can be stated as follows:

Any circuit made up of resistors and sources, viewed from two terminals of that circuit, is equivalent to a voltage source in series with a resistance, or to a current source in parallel with a resistance.

The voltage source is equal to the open-circuit voltage for the two-terminal circuit, the current source is equal to the short-circuit current for that circuit, and the resistance is equal to the equivalent resistance of that circuit.

Figure 1.25 The Definition of Equivalents. Courtesy of David Shattuck, used with personal permission.

Equivalent Resistance Reviewed

When we find the equivalent resistance for a Thévenin’s equivalent or a Norton’s equivalent, we set the independent sources equal to zero, and find the equivalent resistance of what remains.

When a dependent source is present, trying to find the equivalent resistance results in a situation we have not dealt with yet. What do we mean by the equivalent resistance of a dependent source?

The answer must be stated carefully. If the ratio of voltage to current for something is a constant, then that something can be said to have an equivalent resistance, since it is behaving as a resistance

Equivalent Resistance of a Source

So, what we mean by the equivalent resistance of a dependent source is that in this case the ratio of voltage to current is a constant. Then the source can be said to have an equivalent resistance, since it is behaving as a resistance. The equivalent resistance of a dependent source depends on what voltage or current it depends on, and where that voltage or current is in the circuit. It is not easy to predict the answer.

No Equivalent Resistance for an Independent Source

The equivalent resistance of a dependent source, in this case, is the ratio of voltage to current, which is a constant. Then the source can be said to have an equivalent resistance, since it is behaving as a resistance. This will only be meaningful for a dependent source. It is not meaningful to talk about the equivalent resistance of an independent source. The ratio of voltage to current will not be constant for an independent source.

Figure 1.25 Repeated Here for Convenience. Courtesy of David Shattuck, used with personal permission.

Simple Example with a Dependent Source

We will try to explain this by starting with a simple example. We wish to find the equivalent resistance of the circuit below, as seen at terminals A and B.

This will mean that the ratio of the voltage across the circuit, labeled V_Q, to the ratio of the current through the circuit, labeled i_Q, must be a constant. Let’s find that constant by finding the ratio.

Figure 1.26 Simple Example for Equivalent Resistance. Courtesy of David Shattuck, used with personal permission.

We wish to find the equivalent resistance of the circuit above, as seen at terminals A and B. Let’s find the ratio of the voltage across the circuit, labeled v_Q, to the ratio of the current through the circuit, labeled i_Q. This must be a constant. Let’s look first at the circuit equivalent on the right. We note that from Ohm’s Law applied to R_X, we can say

Dependent Source Example Equation 1.png — Equation 1.21

Next, we apply KCL at the A node to write that

Dependent Source Example Equation 2.png — Equation 1.22

We take the ratio of them, and plug in the expressions that we found for each. When we do this, we get

Dependent Source Example Equation 3.png — Equation 1.23

Note that ratio is a constant. The ratio has units of resistance, which is what we expect when we take a ratio of a voltage to a current.

The dependent source is in parallel with the resistor R_X. Since the parallel combination is 25[ ], the dependent source must be behaving as if it were a 33.33[ ] resistor. However, this value depends on R_X; in fact, it is R_X /3.

2nd Simple Example with a Dependent Source

We wish to find the equivalent resistance of a second circuit, given below, as seen at terminals A and B.

Let’s find the ratio of the voltage across the circuit, labeled v_Q, to the ratio of the current through the circuit, labeled i_Q. This must be a constant. We note that from Ohm’s Law applied to R_X, we can say that

Dependent Source Example Equation 4.png — Equation 1.24

Figure 1.27 Circuit for Second Simple Example. Courtesy of David Shattuck, used with personal permission.

Next, we apply KCL at the A node to write that

Dependent Source Example Equation 5.png — Equation 1.25

Note the change in polarity for the source, from the previous example.

We take the ratio of them, and plug in the expressions that we found for each. When we do this, we get

Dependent Source Example Equation 6.png — Equation 1.26

Note that ratio has changed when we simply changed the polarity of the dependent source. The magnitude is not the only thing that changed; the equivalent resistance is now negative!

The dependent source is in parallel with the resistor R_X. Since the parallel combination is -50[ ], the dependent source must be behaving as if it were a -33.33[ ] resistor. This value depends on R_X; in fact, it is -R_X /3.

Note 1 -- Equivalent Resistance can be Negative

When we find the equivalent resistance for a Thévenin’s equivalent or a Norton’s equivalent, we set the independent sources equal to zero, and find the equivalent resistance of what remains. We can see that the equivalent resistance can be negative. This is one reason why we have been so careful about polarities all along. We need to get the polarities right to be able to get our signs right.

Figure 1.25 Repeated for Convenience. Courtesy of David Shattuck, used with personal permission.

Note 2 -- Test Source Method

In the simple examples that we just did, we were effectively applying a source to the terminals of the circuit. This results in a circuit like others that we have solved before, and we can find the ratio of voltage to current. This is usually easier to think about for most students. It is as if we were applying a source just to test the circuit; we call this method the Test-Source Method.

Test-Source Method – Defined

To get the equivalent resistance of a circuit, as seen by two terminals of that circuit, we follow these steps.

Set all independent sources equal to zero.

Find the equivalent resistance. Do one of the two following bulleted items.

If there are no dependent sources, find this equivalent resistance using the equivalent resistance rules that have been used before. These include series combinations, parallel combinations, and delta-to-wye equivalents.

If there are dependent sources present, apply a test source to the two terminals. It can be either a voltage source or a current source. If you apply a voltage source, find the current through that voltage source. If you apply a current source, find the voltage across that current source. Then, find the ratio of the voltage to the current, which will be the equivalent resistance.

Figure 1.28 Test Source Method. Courtesy of David Shattuck, used with personal permission.

Test-Source Method – Notes

Do not forget step one. We always set the independent sources to zero. Some call this "killing" the independent sources. While this sounds violent, it is important not to forget this step.

In step two, you have two choices. Pick the easier one, in general. You do not need to do both. You could actually pick the second option every time, but the first option is easier. Use it if you can.

When you apply this voltage and this current, v_T and i_T, apply them in the active sign relationship for the source. This gives the proper sign; that is, it allows you to write the equation that looks like Ohm's Law with a positive sign. This actually is Ohm's Law, but it is applied to the circuit made up of resistors and dependent sources, where this voltage and current are in the passive sign relationship.

Notes

The Test-Source Method usually requires some practice before it becomes natural for students. It is important to work several problems to get this practice in.

There is a tendency to assume that one could just ignore the Test-Source Method, and just find the open-circuit voltage and short-circuit current whenever a dependent source is present. However, sometimes this does not work. In particular, when the open-circuit voltage and short-circuit current are zero, we must use the Test-Source Method. Learn how to use it.

Example Problem 1.3

Figure 1.29 Circuit for Example Problem 1.3. Courtesy of David Shattuck, used with personal permission.

We wish to find the Thévenin equivalent of the circuit above, as seen from terminals A and B. We will start by find the open-circuit voltage v_OC at those two terminals, as defined below.

Test Source Example Diagram 1.png — Figure 1.30 Defining the Open-Circuit Voltage v_OC. Courtesy of David Shattuck, used with personal permission.

To find v_OC, we will first find v_D, by writing KCL at the top center node. We have

Test Source Example Equation 1.png — Equation1.27

Note that we recognize that the current through R₂ must be zero since R₂ is in series with an open circuit. We can substitute in the value for i_S, 25[mS]v_C. We note that since the current through R₂ is zero, the voltage across it is zero, so v_C is zero. So, we write

Test Source Example Equation 2.png — Equation 1.28

Next, we substitute in values and solve for v_D. We write

Test Source Example Equation 3.png — Equation 1.29

Now, we can take KVL around the loop, and we write

Test Source Example Equation 4.png — Equation 1.30

The Thévenin voltage is equal to this open-circuit voltage, so the Thévenin voltage must be zero. The short-circuit current will also be zero. To get the resistance, we need to use the Test-Source Method.

Test Source Example Diagram 2.png — Figure 1.31 Applying a Test Source. Courtesy of David Shattuck, used with personal permission.

We have applied a test current source to the two terminals. We have also labeled a voltage across this current source, v_T. This voltage has been defined in the active sign relationship for the current source. As noted earlier, this will give us the passive sign relationship for v_T and i_T for the circuit that we are finding the equivalent resistance of. Thus, we will have

Test Source Example Equation 5.png — Equation 1.31

We have applied a test current source to the two terminals. We don’t need to do this, but doing so makes it clear that we are now just solving another circuit, like the many that we have solved before. We have even given the source a value, in this case, 1[A]. This is just a convenience. Many people choose to leave this as an arbitrary source. We choose to use a value, an easy value like 1[A], to allow us to find an actual value for v_T. We have applied a test current source to the two terminals. A test voltage source would have been just as good. We chose a current source because we thought it might make the solution a little easier, since we can find v_C so easily now. But it really does not matter. Don’t worry about which one to choose. Let us solve.

Let us solve for v_T. We note that we can write an expression for v_C using Ohm’s Law, and get

Test Source Example Equation 6.png — Equation 1.32

This voltage may seem very large. Don’t let this bother you. We do not actually have this voltage; it is just for calculating the resistance. Next, let’s write KCL for the top center node. We get

Test Source Example Equation 7.png — Equation 1.33

Solving for v_D yields

Test Source Example Equation 8.png — Equation 1.34

Taking KVL, we get

Test Source Example Equation 9.png — Equation 1.35

So, we can find the equivalent resistance by finding

Test Source Example Equation 10.png — Equation 1.36

So, the Thévenin equivalent is given in the circuit below. Note that the Thévenin voltage is zero, and so we don’t even show the voltage source at all. The Thévenin resistance is shown, and in this case, it is the Thévenin equivalent.

Test Source Example Diagram 3.png — Figure 1.32 Solution for Example Problem 1.3. Notice that the solution is only a resistance, since the source is zero. Courtesy of David Shattuck, used with personal permission.

Summary

Is the Test-Source Method really that important? This is a good question. Basically, the answer is yes. There are many cases where we have dependent sources present, and wish to use a Thévenin equivalent. While in some cases we can get the Thévenin resistance from the open-circuit voltage and the short-circuit current, there are others where we cannot. The Test-Source Method is also quicker in some cases. Some students go to great lengths to avoid learning this method. This seems like a waste of energy. Just learn it and use it.

All figures and all equations in this chapter were developed by David Shattuck, University of Houston, Houston, TX, USA.

Circuit Analysis II

Table of Contents for Circuit Analysis II