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Equivalent Circuits: Resistors and
Sources
∗
Don Johnson
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0
†
Abstract
Introduction of equivalent circuits, including the function in v-i relation.
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.
i
vin
+
–
+
R1
R2
v
–
Figure 1
This result generalizes to include sources in a very interesting and useful way. Let's consider our simple
attenuator circuit (shown in the gure (Figure 1)) from the viewpoint of the output terminals. We want to
nd the v-i relation for the output terminal pair, and then nd 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 v and i that describes the kind of element that lurks within
the dashed box. The result is
R2
vin
(1)
v = (R1 k R2 ) i +
R1 + R2
∗ Version
2.25: May 15, 2013 11:10 pm -0500
† http://creativecommons.org/licenses/by/3.0/
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If the source were zero, it could be replaced by a short circuit, which would conrm that the circuit does
indeed function as a parallel combination of resistors. However, the source's presence means that the circuit
is not well modeled as a resistor.
i
+
Req
veq
+
–
v
–
Figure 2:
The Thévenin equivalent circuit.
If we consider the simple circuit of Figure 2, we nd it has the v-i relation at its terminals of
(2)
v = Req i + veq
Comparing the two v-i relations, we nd that
equivalent resistance is Req = R1 k R2 and
they have the same form. In this case the Thévenin
the Thévenin equivalent source has voltage veq =
R2
v
.
Thus,
from
viewpoint
of
the
terminals,
you cannot distinguish the two circuits. Because the
in
R1 +R2
equivalent circuit has fewer elements, it is easier to analyze and understand than any other alternative.
For any circuit containing resistors and sources, the v-i relation will be of the form
(3)
v = Req i + veq
and the Thévenin equivalent circuit for any such circuit is that of Figure 2. This equivalence applies no
matter how many sources or resistors may be present in the circuit. In the example (Example 1) below, we
know the circuit's construction and element values, and derive the equivalent source and resistance. Because
Thévenin's theorem applies in general, we should be able to make measurements or calculations only from
the terminals to determine the equivalent circuit.
To be more specic, consider the equivalent circuit of this gure (Figure 2). Let the terminals be opencircuited, which has the eect of setting the current i to zero. Because no current ows through the resistor,
the voltage across it is zero (remember, Ohm's Law says that v = Ri). Consequently, by applying KVL
we have that the so-called open-circuit voltage voc 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 isc = − Rveqeq . From this property, we can determine the equivalent resistance.
(4)
veq = voc
Req = −
Exercise 1
voc
isc
(5)
(Solution on p. 7.)
Use the open/short-circuit approach to derive the Thévenin equivalent of the circuit shown in
Figure 3.
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i
vin
+
–
+
R1
R2
v
–
Figure 3
Example 1
R2
iin
R1
R3
Figure 4
For the circuit depicted in Figure 4, let's derive its Thévenin equivalent two dierent ways.
Starting with the open/short-circuit approach, let's rst nd the open-circuit voltage voc . We have
a current divider relationship as R1 is in parallel with the series combination of R2 and R3 . Thus,
R3 R1
voc = R1iin
+R2 +R3 . When we short-circuit the terminals, no voltage appears across R3 , and thus
no current ows through it. In short, R3 does not aect the short-circuit current, and can be
R1
. Thus, the Thévenin
eliminated. We again have a current divider relationship: isc = − Ri1in+R
2
R3 (R1 +R2 )
equivalent resistance is R1 +R2 +R3 .
To verify, let's nd 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 R3 is now in parallel with the series combination of
R1 and R2 . Thus, Req = R3 k R1 + R2 , and we obtain the same result.
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i
Sources
and
Resistors
+
v
–
i
i
+
+
Req
veq
+
–
ieq
v
Req
v
–
–
Mayer-Norton Equivalent
Thévenin Equivalent
Figure 5: All circuits containing sources and resistors can be described by simpler equivalent circuits.
Choosing the one to use depends on the application, not on what is actually inside the circuit.
As you might expect, equivalent circuits come in two forms: the voltage-source oriented Thévenin equivalent1 and the current-source oriented Mayer-Norton equivalent (Figure 5). To derive the latter, the v-i
relation for the Thévenin equivalent can be written as
v = Req i + veq
(6)
v
− ieq
Req
(7)
or
i=
where ieq = Rveqeq is the Mayer-Norton equivalent source. The Mayer-Norton equivalent shown in Figure 5
can be easily shown to have this v-i relation. Note that both variations have the same equivalent resistance.
The short-circuit current equals the negative of the Mayer-Norton equivalent source.
Exercise 2
Find the Mayer-Norton equivalent circuit for the circuit below.
1 "Finding Thévenin Equivalent Circuits" <http://cnx.org/content/m0021/latest/>
http://cnx.org/content/m0020/2.25/
(Solution on p. 7.)
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R2
iin
R1
R3
Figure 6
Equivalent circuits can be used in two basic ways. The rst 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 conguration
(making the Thévenin equivalent the best) or a parallel one (making Mayer-Norton the best).
Another application is modeling. When we buy a ashlight 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 specication, they should serve as voltage sources and the Thévenin equivalent serves as
the natural choice. If a load resistance RL is placed across its terminals, the voltage output can be found
RL
using voltage divider: v = RvLeq+R
. If we have a load resistance much larger than the battery's equivalent
eq
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
Req
current divider, and equals i = − RiLeq+R
. For a current that does not vary with the load resistance, this
eq
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.
He was an engineer with France's Postes, Télégraphe et Téléphone. In
1883, he published (twice!) a proof of what is now called the Thévenin equivalent while developing
ways of teaching electrical engineering concepts at the École Polytechnique. He did not realize that
the same result had been published by Hermann Helmholtz2 , the renowned nineteenth century
physicist, thiry years earlier.
Léon Charles Thévenin:
After earning his doctorate in physics in 1920, he turned to communications engineering when he joined Siemens & Halske in 1922. In 1926, he published in a
German technical journal the Mayer-Norton equivalent. During his interesting career, he rose to
lead Siemen's Central Laboratory in 1936, surruptiously leaked to the British all he knew of German
warfare capabilities a month after the Nazis invaded Poland, was arrested by the Gestapo in 1943
for listening to BBC radio broadcasts, spent two years in Nazi concentration camps, and went to
the United States for four years working for the Air Force and Cornell University before returning
to Siemens in 1950. He rose to a position on Siemen's Board of Directors before retiring.
Hans Ferdinand Mayer:
Edward Norton3 was an electrical engineer who worked at Bell Laboratory
from its inception in 1922. In the same month when Mayer's paper appeared, Norton wrote in an
Edward L. Norton:
2 http://www-gap.dcs.st-and.ac.uk/∼history/Mathematicians/Helmholtz.html
3 http://www.ece.rice.edu/∼dhj/norton
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internal technical memorandum a paragraph describing the current-source equivalent. No evidence
suggests Norton knew of Mayer's publication.
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Solutions to Exercises in this Module
Solution to Exercise (p. 2)
2
voc = R1R+R
vin and isc = − vRin1 (resistor R2 is shorted out in this case). Thus, veq =
2
R1 R2
Req = R1 +R2 .
Solution to Exercise (p. 4)
ieq =
R1
R1 +R2 iin
and Req = R3 k R1 + R2 .
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R2
R1 +R2 vin
and