We have found that the way to think about circuits is to locate and group parallel and series resistor combinations. Those resistors not involved with variables of interest can be collapsed into a single resistance. This result is known as an equivalent circuit: from the viewpoint of a pair of terminals, a group of resistors functions as a single resistor, the resistance of which can usually be found by applying the parallel and series rules.
This result generalizes to include sources in a very interesting and useful way. Let's consider our simple attenuator circuit (shown in the figure) from the viewpoint of the output terminals. We want to find the v-i relation for the output terminal pair, and then find the equivalent circuit for the boxed circuit. To perform this calculation, use the circuit laws and element relations, but do not attach anything to the output terminals. We seek the relation between and that describes the kind of element that lurks within the dashed box. The result is
If we consider the simple circuit of [link], we find it has the v-i relation at its terminals of
For any circuit containing resistors and sources, the v-i relation will be of the form
To be more specific, consider the equivalent circuit of this figure. Let the terminals be open-circuited, which has the effect of setting the current to zero. Because no current flows through the resistor, the voltage across it is zero (remember, Ohm's Law says that ). Consequently, by applying KVL we have that the so-called open-circuit voltage equals the Thévenin equivalent voltage. Now consider the situation when we set the terminal voltage to zero (short-circuit it) and measure the resulting current. Referring to the equivalent circuit, the source voltage now appears entirely across the resistor, leaving the short-circuit current to be . From this property, we can determine the equivalent resistance.
Use the open/short-circuit approach to derive the Thévenin equivalent of the circuit shown in [link].
and (resistor is shorted out in this case). Thus, and .
For the circuit depicted in [link], let's derive its Thévenin equivalent two different ways. Starting with the open/short-circuit approach, let's first find the open-circuit voltage . We have a current divider relationship as is in parallel with the series combination of and . Thus, . When we short-circuit the terminals, no voltage appears across , and thus no current flows through it. In short, does not affect the short-circuit current, and can be eliminated. We again have a current divider relationship: . Thus, the Thévenin equivalent resistance is .
To verify, let's find the equivalent resistance by reaching inside the circuit and setting the current source to zero. Because the current is now zero, we can replace the current source by an open circuit. From the viewpoint of the terminals, resistor is now in parallel with the series combination of and . Thus, , and we obtain the same result.
As you might expect, equivalent circuits come in two forms: the voltage-source oriented Thévenin equivalent and the current-source oriented Mayer-Norton equivalent ([link]). To derive the latter, the v-i relation for the Thévenin equivalent can be written as
Find the Mayer-Norton equivalent circuit for the circuit below.
and .
Equivalent circuits can be used in two basic ways. The first is to simplify the analysis of a complicated circuit by realizing the any portion of a circuit can be described by either a Thévenin or Mayer-Norton equivalent. Which one is used depends on whether what is attached to the terminals is a series configuration (making the Thévenin equivalent the best) or a parallel one (making Mayer-Norton the best).
Another application is modeling. When we buy a flashlight battery, either equivalent circuit can accurately describe it. These models help us understand the limitations of a battery. Since batteries are labeled with a voltage specification, they should serve as voltage sources and the Thévenin equivalent serves as the natural choice. If a load resistance is placed across its terminals, the voltage output can be found using voltage divider: . If we have a load resistance much larger than the battery's equivalent resistance, then, to a good approximation, the battery does serve as a voltage source. If the load resistance is much smaller, we certainly don't have a voltage source (the output voltage depends directly on the load resistance). Consider now the Mayer-Norton equivalent; the current through the load resistance is given by current divider, and equals . For a current that does not vary with the load resistance, this resistance should be much smaller than the equivalent resistance. If the load resistance is comparable to the equivalent resistance, the battery serves neither as a voltage source or a current course. Thus, when you buy a battery, you get a voltage source if its equivalent resistance is much smaller than the equivalent resistance of the circuit to which you attach it. On the other hand, if you attach it to a circuit having a small equivalent resistance, you bought a current source.