Thevenin’s, Norton’s, and Maximum Power Transfer Theorems
Network Analysis Techniques
Suppose a 12 volt leadacid battery has an internal resistance of 20 milliohms (20 mΩ):

If a shortcircuit were placed across the terminals of this large battery, the fault current would be quite large: 600 amps!
Now suppose three of these batteries were connected directly in parallel with one another:

Reduce this network of parallelconnected batteries into either a Thévenin or a Norton equivalent circuit, and then recalculate the fault current available at the terminals of the threebattery “bank” in the event of a direct shortcircuit.
What would happen if a wire having no resistance at all (0 Ω) were connected directly across the terminals of a 6volt battery? How much current would result, according to Ohm’s Law?

Suppose we were to shortcircuit a 6volt battery in the manner just described and measure 8 amps of current. Why don’t the calculated figures from the previous paragraph agree with the actual measurement?
Don’t just sit there! Build something!! 
Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuitbuilding exercises, follow these steps:
 Carefully measure and record all component values prior to circuit construction.
 Draw the schematic diagram for the circuit to be analyzed.
 Carefully build this circuit on a breadboard or other convenient medium.
 Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements onebyone on the diagram.
 Mathematically analyze the circuit, solving for all values of voltage, current, etc.
 Carefully measure those quantities, to verify the accuracy of your analysis.
 If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully recalculate the values and remeasure.
Avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistors between 1 kΩ and 100 kΩ, unless, of course, the purpose of the circuit is to illustrate the effects of meter loading!
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another timesaving technique is to reuse the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
Sometimes you will see amplifier circuits expressed as collections of impedances and dependent sources:

With this model, the amplifier appears as a load (Z_{in}) to whatever signal source its input is connected to, boosts that input voltage by the gain factor (A_{V}), then outputs the boosted signal through a series output impedance (Z_{out}) to whatever load is connected to the output terminals:

Explain why all these impedances (shown as resistors) are significant to us as we seek to apply amplifier circuits to practical applications. Which of these impedances do you suppose are typically easier for us to change, if they require changing at all?
The voltage divider network employed to create a DC bias voltage for many transistor amplifier circuits has its own effect on amplifier input impedance. Without considering the presence of the transistor or the emitter resistance, calculate the impedance as “seen” from the input terminal resulting from the two resistors R_{1} and R_{2} in the following commoncollector amplifier circuit:

Remember, what you are doing here is actually determining the Thévenin/Norton equivalent resistance as seen from the input terminal by an AC signal. The input coupling capacitor reactance is generally small enough to be safely ignored.
Next, calculate the input impedance of the same circuit, this time considering the presence of the transistor and emitter resistor, assuming a current gain (β or h_{fe}) of 60, and the following formula for impedance at the base resulting from β and R_{E}:


Develop an equation from the steps you take in calculating this impedance value.
Determining the output impedance of a commonemitter amplifier is impossible unless we know how to model the transistor in terms of components whose behavior is simple to express.

When in its active mode, a transistor operates like a current regulator. This is similar enough to the behavior of a current source that we may use a source to model the transistor’s behavior for the sake of this impedance determination:

Now, apply the same steps you would use in determining the Thévenin or Norton equivalent impedance to the output of this amplifier circuit, and this will yield the amplifier’s output impedance. Draw an equivalent circuit for the amplifier during this Thévenizing/Nortonizing process to show how the output impedance is determined.
An electric arc welder is a lowvoltage, highcurrent power source designed to supply enough electric current to sustain an arc capable of welding metal with its high temperature:

It is possible to derive a Norton equivalent circuit for an arc welder based on empirical measurements of voltage and current. Take for example these measurements, under loaded and noload conditions:


Based on these measurements, draw a Norton equivalent circuit for the arc welder.
Resistive voltage dividers are very useful and popular circuits. However, it should be realized that their output voltages ßag” under load:

Just how much a voltage divider’s output will sag under a given load may be a very important question in some applications. Take for instance the following application where we are using a resistive voltage divider to supply an engine sensor with reduced voltage (8 volts) from the 12 volt battery potential in the automobile:

If the sensor draws no current (I_{sensor} = 0 mA), then the voltage across the sensor supply terminals will be 8 volts. However, if we were asked to predict the voltage across the sensor supply terminals for a variety of different sensor current conditions, we would be faced with a much more complex problem:
Sensor current (I_{sensor}) Sensor supply voltage
0 mA 8 volts
1 mA
2 mA
3 mA
4 mA
5 mA
One technique we could use to simplify this problem is to reduce the voltage divider resistor network into a Thévenin equivalent circuit. With the threeresistor divider reduced to a single resistor in series with an equivalent voltage source, the calculations for sensor supply voltage become much simpler.
Show how this could be done, then complete the table of sensor supply voltages shown above.
A voltage source is a source of electricity that (ideally) outputs a constant voltage. That is, a perfect voltage source will hold its output voltage constant regardless of the load imposed upon it:

In real life, there is no such thing as a perfect voltage source, but sources having extremely low internal resistance come close.
Another type of electricity source is the current source, which (ideally) outputs a constant current regardless of the load imposed upon it. A common symbol for a current source is a circle with an arrow inside (always pointing in the direction of conventional flow, not electron flow!). Another symbol is two intersecting circles, with an arrow nearby pointing in the direction of conventional flow:

Predict how an ideal current source would behave for the following two load scenarios:

Ideal voltage sources and ideal current sources, while both being sources of electrical power, behave very differently from one another:

Explain how each type of electrical source would behave if connected to a variableresistance load. As this variable resistance were increased and decreased, how would each type of source respond?
A very common sort of graph used in electronics work is the load line, showing all possibilities of load voltage and load current that a particular power source is able to supply to a load:

Note how the load line shows the voltage “sag” of the power source in relation to the amount of current drawn by the load. At high currents, the output voltage will be very low (upperleft end of load line). At low currents, the output voltage will be near its maximum (lowerright end of load line). If all internal components of the power source are linear in nature, the load line will always be perfectly straight.
Plot the load line for a power source having an internal voltage (V_{internal}) of 11 volts and an internal resistance (R_{internal}) of 1.2 kΩ. Superimpose your load line onto the load line graph shown above. Hint: it only takes two points to define a line!
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire) connections. Inside this box, you were told, was a voltage source (an ideal voltage source connected in series with a resistance):

How would you experimentally determine the voltage of the ideal voltage source inside this box, and how would you experimentally determine the resistance of the series resistor? By “experimentally,” I mean determine voltage and resistance using actual test equipment rather than assuming certain component values (remember, this “black box” is sealed, so you cannot look inside!).
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire) connections. Inside this box, you were told, was a current source (an ideal current source connected in parallel with a resistance):

How would you experimentally determine the current of the ideal current source inside this box, and how would you experimentally determine the resistance of the parallel resistor? By “experimentally,” I mean determine current and resistance using actual test equipment rather than assuming certain component values (remember, this “black box” is sealed, so you cannot look inside!).
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire) connections. Inside this box, you were told, was a voltage source connected in series with a resistance.

Your task was to experimentally determine the values of the voltage source and the resistor inside the box, and you did just that. From your experimental data you then sketched a circuit with the following component values:

However, you later discovered that you had been tricked. Instead of containing a single voltage source and a single resistance, the circuit inside the box actually looked like this:

Demonstrate that these two different circuits are indistinguishable from the perspective of the two metal terminals, and explain what general principle this equivalence represents.
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire) connections. Inside this box, you were told, was a current source connected in parallel with a resistance.

Your task was to experimentally determine the values of the current source and the resistor inside the box, and you did just that. From your experimental data you then sketched a circuit with the following component values:

However, you later discovered that you had been tricked. Instead of containing a current source and a resistor, the circuit inside the box was actually a voltage source connected in series with a resistor:

Demonstrate that these two different circuits are indistinguishable from the perspective of the two metal terminals, and explain what general principle this equivalence represents.
Calculate the voltage dropped across the load resistor, and the current through the load resistor, for the following load resistance values in this circuit:

R_{load} E_{load} I_{load}
1 k Ω
2 k Ω
5 k Ω
8 k Ω
10 k Ω
Do the “boxed” components in this circuit behave more like a constant voltage source, or a constant current source? Explain your answer.
Calculate the voltage dropped across the load resistor, and the current through the load resistor, for the following load resistance values in this circuit:

R_{load} E_{load} I_{load}
1 k Ω
2 k Ω
5 k Ω
8 k Ω
10 k Ω
Do the “boxed” components in this circuit behave more like a constant voltage source, or a constant current source? Explain your answer.
In the following circuit, an adjustable voltage source is connected in series with a resistive load and another voltage source:

Determine what will happen to the current in this circuit if the adjustable voltage source is increased.
In this next circuit, an adjustable voltage source is connected in series with a resistive load and a current source:

Now determine what will happen to the current in this second circuit if the adjustable voltage source is increased.
One way to define electrical resistance is by comparing the change in applied voltage (∆V) to the change in resultant current (∆I). This is mathematically expressed by the following ratio:

From the perspective of the adjustable voltage source (V_{adjust}), and as defined by the above equation, which of these two circuits has the greatest resistance? What does this result suggest about the equivalent resistance of a constantvoltage source versus the equivalent resistance of a constantcurrent source?
A practical current source may be built using a battery and a special semiconductor component known as a currentlimiting diode:

The currentlimiting diode acts as a variable resistance, to regulate current through it at a constant value: if current increases, its resistance increases to reduce the current back to where it should be; if current decreases, its resistance decreases to increase current up to where it should be.
Determine the amount of voltage output by an opencircuited (ideal) current source. Contrast this with the voltage output by the practical current source shown in the diagram. Finally, draw an equivalent circuit showing an ideal current source somehow connected to a resistance in such a way that its opencircuited output voltage is identical to the practical current source.
Electrochemical batteries are supposed to act as constant voltage sources, outputting an unchanging voltage for a wide range of load currents. The output voltage of real batteries, though, always “sags” to some degree under the influence of a load.
Explain why this is so, in terms of modeling the battery as an ideal voltage source combined with a resistance. How do you suggest the internal resistance of a chemical battery be experimentally measured?
Load lines are special types of graphs used in electronics to characterize the output voltage and current behavior of different power sources:

If we know that all the internal components of a power source are inherently linear, we know that the load line plot will indeed by a straight line. And, if we know the plot will be a straight line, all we need in order to plot a complete load line are two data points.
Usually, the easiest data points to gather for a circuit  whether it be a real circuit or an hypothetical circuit existing on paper only  is the opencircuit condition and the shortcircuit condition. In other words, we see how much voltage the source will output with no load connected (I_{load} = 0 milliamps) and then we see how much current the source will output into a direct short (V_{load} = 0 volts):

Suppose we have two differentlyconstructed power sources, yet both of these sources share the same opencircuit voltage (V_{OC}) and the same shortcircuit current (I_{SC}). Assuming the internal components of both power sources are linear in nature, explain how we would know without doubt that the two power sources were electrically equivalent to one another. In other words, explain how we would know just from the limited data of V_{OC} and I_{SC} that these two power sources will behave exactly the same when connected to the same load resistance, whatever that load resistance may be.

Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire) connections. Inside this box, you were told, was a voltage source connected in series with a resistance.

Your task was to experimentally determine the values of the voltage source and the resistor inside the box, and you did just that. From your experimental data you then sketched a circuit with the following component values:

However, you later discovered that you had been tricked. Instead of containing a single voltage source and a single resistance, the circuit inside the box actually looked like this:

Demonstrate that these two different circuits are indistinguishable from the perspective of the two metal terminals, and explain what general principle this equivalence represents.
Examine this circuit, consisting of an ideal voltage source and several resistors:

First, calculate the voltage seen at the load terminals with a voltmeter directly connected across them (an opencircuit condition):

Next, calculate the current seen at the load terminals with an ammeter directly connected across them (a shortcircuit condition):

Based on these open and shortcircuit calculations, draw a new circuit consisting of a single voltage source and a single (series) resistor that will respond in the exact same manner. In other words, design an equivalent circuit for the circuit shown here, using the minimum number of possible components.
An electric arc welder is a lowvoltage, highcurrent power source used to generate hot arcs capable of melting metal. Note the voltage and current measurements taken for this particular welder:

Determine two Thévenin equivalent circuits for the arc welder. The first circuit will simply be an AC voltage source and an internal impedance. The second circuit will be a voltage source and internal impedance connected through an ideal transformer with a stepdown ratio of 8 to 1:

An electric arc welder is a lowvoltage, highcurrent power source used to generate hot arcs capable of melting metal. Note the voltage and current measurements taken for this particular welder:

Determine two Thévenin equivalent circuits for the arc welder. The first circuit will simply be an AC voltage source and an internal impedance. The second circuit will be a voltage source and internal impedance connected through an ideal transformer with a stepdown ratio of 8 to 1:

Use Thévenin’s Theorem to determine a simple equivalent circuit for the 10 volt source, the 5 kΩ resistor, and the 2.2 kΩ resistor; then calculate the voltage across the 1 kΩ load:

This exercise may seem pointless, as it is easy enough to obtain the answer simply by seriesparallel analysis of this circuit. However, there is definite value in determining a Thévenin equivalent circuit for the voltage source, 5 kΩ resistor, and 2.2 kΩ resistor if the load voltage for several different values of load resistance needs to be predicted. Explain why Thévenin’s Theorem becomes the more efficient way to predict load voltage if multiple load resistor values are considered.
An ideal (perfect) current source is an abstraction with no accurate realization in life. However, we may approximate the behavior of an ideal current source with a highvoltage source and large series resistance:

Such a Thevenin equivalent circuit, however imperfect, will maintain a fairly constant current through a wide range of load resistance values.
Similarly, an ideal (perfect) voltage source is an abstraction with no accurate realization in life. Thankfully, though, it is not difficult to build voltage sources that are relatively close to perfect: circuits with very low internal resistance such that the output voltage sags only a little under highcurrent conditions.
But suppose we lived in a world where things were the opposite: where closetoideal current sources were simpler and more plentiful than closetoideal voltage sources. Draw a Norton equivalent circuit showing how to approximate an ideal voltage source using an ideal (perfect) current source and a shunt resistance.
One day an electronics student decides to build her own variablevoltage power source using a 6volt battery and a 10 kΩ potentiometer:

She tests her circuit by connecting a voltmeter to the output terminals and verifying that the voltage does indeed increase and decrease as the potentiometer knob is turned.
Later that day, her instructor assigns a quick lab exercise: measure the current through a parallel resistor circuit with an applied voltage of 3 volts, as shown in the following schematic diagram.

Calculating current in this circuit is a trivial exercise, she thinks to herself: 3 V ÷ 500 Ω = 6 mA. This will be a great opportunity to use the new power source circuit, as 3 volts is well within the voltage adjustment range!
She first sets up her circuit to output 3 volts precisely (turning the 10 kΩ potentiometer to the 50% position), measuring with her voltmeter as she did when initially testing the circuit. Then she connects the output leads to the two parallel resistors through her multimeter (configured as an ammeter), like this:

However, when she reads her ammeter display, the current only measures 1 mA, not 6 mA as she predicted. This is a very large discrepancy between her prediction and the measured value for current!
Use Thévenin’s Theorem to explain what went wrong in this experiment. Why didn’t her circuit behave as she predicted it would?
Observe the following circuit:

Note that it is not reducible to a single resistance and power source. In other words, it is not a seriesparallel combination circuit. And, while it is a bridge circuit, you are not able to simply analyze the resistor ratios because it is obviously not in a state of balance!
If you were asked to calculate voltage or current for any component in this circuit, it would be a difficult task . . . unless you know either Thévenin’s or Norton’s theorems, that is! Apply either one of these theorems to the determination of voltage across the 2.2 kΩ resistor.
Hint: consider the 2.2 kΩ resistor as the load in a Thévenin or Norton equivalent circuit.
Suppose we were planning to use a photovoltaic panel to generate electricity and electrolyze water into hydrogen and oxygen gas:

Our goal is to electrolyze as much water as possible, and this means we must maximize the electrolysis cell’s power dissipation. Explain how we could experimentally determine the optimum internal resistance of the electrolysis cell, prior to actually building it, using nothing but the solar panel, a rheostat, and a DMM (digital multimeter).
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