Chapter 5: Circuit Theorems
This chapter provides a new powerful technique of solving complicated circuits that are
more conceptual in nature than node/mesh analysis. Conceptually, the method is fairly
straightforward to write down, however, they are complicated to apply in practice. One will
find that the amount of work increases dramatically using these circuit theorems to solve a
circuit. There are four theorems we will look at in this chapter: Source Transformations,
Superposition Principle and Thévenin & Norton Equivalents. There are two useful ideas to
take away from this chapter.
1. Thévenin-Norton source transformations. There is a way to convert a voltage source into
a current source and vice-versa. Source transformations are going to change how you
solve circuits and in some cases, a series of source transformations will have a dramatic
affect in making the circuit calculation quite straight forward. Source transformations will
become an additional technique in your arsenal in solving circuits regardless of the
techniques one uses (SCT, Node, Mesh…).
2. Thévenin-Norton Equivalents. The techniques of Chapter 4 (Mesh and Node) are very
powerful and fairly straight forward to use. When we start studying nonlinear circuits
(circuits with capacitors and inductors), mesh/node techniques will not be applicable.
The way we will solve these nonlinear circuits are similar to solving irreducible
dependent source circuits with KVL and KCL equations. The Thévenin-Norton
Equivalents will help us to simplify circuits as they get more and more challenging.
Picture wise, we will break-up the circuit into two subcircuits: the (i) load circuit and (ii)
subcircuit to be simplified via the Thévenin-Norton Equivalent. That is, we will rewrite it
into an equivalent form that is either a simple series circuit or a simple parallel circuit.
Thévenin – Norton Source Transformation
There are transformations that convert current sources into voltage sources (and vice
versa). That is, there is an equivalent circuit transformation between a series resistor and
voltage source into a parallel resistor and current source:
In order for these two circuits to be equivalent, the two separate subcircuits must have the
same terminal current i and the terminal voltage v. In order to have the same terminal
current and voltage, certain conditions must be met. These are derived by applying KVL to
the series circuit and KCL to the parallel circuit. Here is what we get:
KVL: v  v Th  iRTh 
v
v
v
solving for i
 i  iN 
 Th 

 RTh  RN and v Th  iNRTh
 
KCL: i  iN  v/RN 
RN RTh RTh
necessary conditions for equivalance
That is, a Thévenin-Norton Source Transformation converts a voltage source with a
resistor in series into a current source and the same resistor in parallel:
5-1
Example 5.1
Use source transformations to solve for the current through the 3Ω-resistor.
Solution
There are several ways one can solve for the current in this circuit: simple circuits or
mesh/node. With the help of source transformations, you will see how useful these can be in
reducing the circuit down to one loop before you can change a baby’s diaper.
Important Point: do not apply a source transformation to the 3 element. If a source
transformation changes the 3Ω resistor to a parallel one, that specific information on the
current through that 3Ω resistor is lost. Leave it alone! Let’s convert this circuit using three
different source transformations:
Applying KVL/Mesh to the last loop,
(3  4)  i3  16
 10
3


i3  2.19A
Superposition Theorem
In Chapter 2, we briefly addressed the issue of a linear circuit. We stated then that for a
circuit to be linear it meant that (1) the elements of the circuit are linear themselves and (2)
the total response of an element (the voltages or currents) can be determined by the sum of
the individual responses from each independent source in the circuit. This last statement is
the Superposition theorem. Analogy: Newton’s 2nd law (Fnet = F1 + F2 +…+ Fn)
Superposition Principle
The current iR through an element is equal to the algebraic sum of the currents (i1, i2…in)
produced independently by each individual source. That is,
iR 
i1
current produce by source-1


iN
current produce by source-N


isource
all sources
Solving Strategies
Step 1: Isolate one source and deactivate all other independent sources. To
deactivate an independent current and voltage sources replace a
 voltage source with a short circuit (a short has vshort = 0)
 current source with an open circuit (an open has iopen = 0)
With only one active source in the circuit, calculate the current/voltage of the
element of interest. Repeat this process for each independent source in the circuit.
Step 2: If there are n-sources, then there are n-circuits to solve using superposition.
Example 5.2
Use superposition to find the current ix through the 20kΩ-resistor.
Solution
Superposition states that to calculate the current i20kΩ, this
current is the sum of all of the individual currents produced
by the 12V, 3mA and 9mA-sources:
i20k  i12V  i3mA  i9mA  i1  i2  i3
5-2
Response of the 12V-source
Deactivate both current sources (replace them with opens), and
calculate the current i1 produced by the 12V source using a
KVL/Mesh loop:
36  i1  12  i1   1 mA
3
Response of the 3mA-source
Deactivate the other sources (short 12 V and open 9 mA sources),
and calculate the current i2 produced by the 3mA source using
CDR:
i2 
16
(3mA)  4 mA  i2
3
16 + 20
Response of the 9mA-source
Deactivate the other sources (short 12 V and open 3 mA sources),
and calculate the current i3 produced by the 9mA source using
CDR:
i3 
12
(9A)  3 mA  i3
12 + 24
According to Superposition, the current i20kΩ is the sum of all of these individual currents:
i20k  i1  i2  i3   1  4  3  2 mA  i20k
3
3
Thévenin’s Theorem
History: this theorem was independently derived in 1853 by Helmholtz and in 1883 by Léon
Charles Thévenin. Léon Charles Thévenin was a French telegraph engineer who
extended Ohm's law to the analysis of complex electrical circuits.
The goal of Thévenin’s theorem is to identify and separate a portion of a complex circuit
(called the subcircuit) and replace it with a Thévenin equivalent (series voltage source and
resistor). In doing so, the solving of the circuit problem is greatly simplified. A typical
application of a Thévenin theorem goes something like this:
Step 1. Focus on a load and separate it out from the subcircuit to be replaced with a
Thévenin equivalent.
Step 2. Perform the necessary calculations to rewrite the subcircuit as a Thévenin
equivalent circuit.
Step 3. Reconnect the Thévenin equivalent circuit back onto the load circuit and now solve
the simplified circuit.
To accomplish this, one applies the Thévenin’s Rules.
Thévenin’s Rules (Notation: vTh = vOC, RTh = RN, and iN = iSc)
Method-1: Independent Sources Only (Conceptual Approach)
1. For only Independent Sources in the load circuit
a. Determine voc with all sources activated
5-3
 Apply KVL around the open circuit to determine vOC
 Use any method to solve for vOC
b. Determine RTh.
 Deactivate all sources and find RTh. (this is more conceptual in nature)
c. Redraw the equivalent Thévenin circuit. If the Thévenin resistance is negative,
subtract it from the load resistance.
2. Method-2: Independent & Dependent Sources (Calculational Approach)
a. Find voc with all sources activated
 Apply KVL around the open circuit to determine vOC
 Use any method to solve for vOC
b. Find isc with a short at the terminal points.
 Use any method to solve for iSC
c. Determine RTh by using RTh = vOC/iSC.
d. Redraw the equivalent Thévenin circuit. If the Thévenin resistance is negative,
subtract it from the load resistance.
3. Dependent sources only – the book does not give any problems related to only
dependent sources. I will not cover them; however, if you are transferring to San Jose
State and plan on majoring in EE, at the end of the semester I can teach you these in
about 10-15 minutes.
Independent Sources Only
Because this example circuit only has an independent source, I will not only demonstrate
how a Thévenin equivalent is determine between points a-b, but also show that Method-1
and Method-2 are equivalent.
Suppose the following circuit has its load identified and removed, so that we can determine
the Thévenin equivalent between points a-b.
Both Method-1 and Method-2 determine voc exactly the same.
Step 1: Determine voc
What does the voltage voc mean? It is the effective voltage that the load “sees” between the
points a and b with respect to the rest of the circuit. Identifying nodes and
applying node analysis because one of the nodes is voc, we get
1  1 )  v  ( 1 )  v  ( 1 )  18  0 
Node voc : (12
oc
x

10
10
12
 voc  v Th  12V
 
1
1
1
1
1
Node v x : ( 3  6  10 )  v x  (10 )  voc  ( 6 )  18  0
What a differentiates Method-1 and Method-2 is in the way that the Thévenin resistance is
determined.
Step 2: Determine RTh
Method 1: Deactivating Sources
The fact that there is an independent voltage source in the circuit is the only reason why we
can use Method-1. Method-1 is NOT applicable if there are dependent sources in the
circuit.
We start by deactivating the voltage source by shorting it out
(replace the 18V-source with a wire) and place an Ohmmeter at
the terminal points. Effectively, one way to envision how an
Ohmmeter measures resistance is by sending out a calibrated
5-4
current which is then compared to the returning current, and reads the resistance. The RTh
resistance is
RTh  (10  3 6) 12 
12  (10  3 6)
 6Ω  RTh
12  10  3 6
What does this resistance physically mean? It is the effective resistance that the load “sees”
between the points a and b with respect to the rest of the circuit. The Thévenin equivalent
circuit has replaced a complex circuit with a Thévenin equivalent circuit with values of vTh =
12V and RTh = 6:
Method 2: Determine iSC to calculate RTh = vOC/iSC
Method-2 is applicable for both independent and dependent sources in the circuit.
Place a short across the open and calculate current iSC. There are three loops in the circuit
and will use mesh analysis to determine iSC.
(12  10  6)  i1  (6)  i2  10  iSC  0 
 28 6 10   i1   0 

   

matrix
Loop - i2 : (6  3)  i2  (6)  i1  3  iSC  18
9 3   i2   18 
 
form   6
 10 3 13   i   0 
Loop - iSC : (10  3)  iSC  (10)  i1  (3)  i2  0 

  SC   

Loop - i1:
solution

 iSC  2A
However, I think that the most efficient way is using node analysis since there is only one
node plus a KCL to get iSC. Applying node to the vy, we get
Node v y :
1 )  v  ( 1 )  18  0 
( 31  61  10
 v y  5V
y
6
Now apply KCL at node a:
KCL:
iSC  i18   i10   18
 5  2A  iSC
12 10
Calculating RTh, we get
R Th 
v OC 12V

 6
iSC
2A
This agrees with our previous calculation of RTh and therefore, both methods are equivalent.
Example 5.3
Determine the current through the 14Ω-resistor using a Thévenin equivalent circuit (Methods
1 and 2).
Solution
First, remove the 14Ω-resistor and convert the circuit into a Thévenin equivalent.
5-5
Step 1: Determine voc = vTh.
Focus on the open circuit and see how voc is defined.
Since the 4 resistor is connected to an open loop, there
is no current flowing through it and therefore, has no
voltage drop. Applying a KVL loop around the open circuit
loop, the 20V-source will still contribute to voc and we have
v10  20  voc 
 voc  v10  20
So to determine voc, my real task is first to determine the voltage across the 10 resistor.
Question: what is the most efficient way to determine v10Ω? For Carlos, I’d say a source
transformation on the series 8V and 9Ω, converting them into a parallel current source and
resistor, then add the current sources together. The circuit I get is
Applying CDR to get the current of the 10 resistor and Ohm’s law to get the voltage v10Ω,
we get
CDR : i10  
9
10
Ohm' s
 A   52 A 
v10  R10i2  10  52  4V
law
9  16 9
Using our KVL relation from above, we determine voc:
v oc  v10  20  4  20  24V  v oc  v Th
Step 2: Find RTh.
Method 1: Deactivating Independent Sources
Since the sources are all independent sources, we can use Method-1 and deactivate all of
sources to determine the Thévenin resistance RTh at a-b. It is important to realize that in the
previous voc-circuit, there was no 4Ωresistor; however, in the RTh-circuit the 4Ω resistor
reappears and must be accounted for.
15 10
 10Ω  RTh
15  10
Now that we have both the Thévenin voltage and resistance, I can draw the equivalent
Thévenin circuit, add back in the load 14Ω-resistor, and solve for the current i14Ω.
RTh  4  (6  9) 10  4 
Using Ohm’s law to determine the current i14Ω, we see that the current flows from b to a, not
a to b and therefore, I assign a negative value to this current:
24V
i14 
 1A  i14
6  14
Interpretation of results
1. Effectively, when we calculate the Thévenin equivalent, the 14Ω resistor “sees” an
effective voltage of -24V and an effective resistance of 6Ω from the view point of terminal
5-6
points a-b. That is, regardless of the load (whether it’s a 14Ω resistor or some
complicated load circuit), this load “feels” -24V and 6Ω from the rest of the circuit.
 It does not see the individual 20V, 8V and 2A-sources but a combination of these
that effectively reads the Thévenin voltage of -24V between a-b.
 It does not see some combination of all of the resistors in the circuit, but the effective
Thévenin resistance of 6Ω between a-b.
2. It clearly is a lot more work to calculate the current i14Ω using Thévenin Equivalent
circuits, whereas using Node or Mesh would have given us this current more efficiently.
So what’s all the fuss about learning these Thévenin equivalent techniques then if there
are other techniques more efficient? To put this into prospective, resistive circuits are
only half of this course, with the other have being capacitive and inductive circuits. Node
and Mesh only apply to linear resistive circuits (as well as to ac phasor circuits). When
we start capacitive and inductive circuits (nonlinear circuits), we will not be able to solve
these circuits using Node or Mesh and we will be forced to use Simple Circuit
Techniques (KVL & KCL) along with Thévenin equivalents to guide us. Thévenin
equivalents are a powerful ally in solving these considerably tougher circuits.
Method 2: Finding iSC to determine RTh = vOC/iSC
When a short is placed across a-b to calculate iSC, note that iSC = i4. I expect that you are
most likely thinking of calculating iSC using mesh analysis (3 loops = 2 loop equations plus a
SC). Even though this is a different circuit, I already have reduced the left-hand portion
circuit with a source transformation that combined the 8V-source with the 2A-source. I want
to continue using this simplified circuit and make one more source transformation on the
10/9A-source to convert it into a voltage source. This is what I get:
I immediately see that there is only one node voltage to solve for and a solver equation for
iSC.
Node v :
1  1  1 )  v  ( 1 )  ( 10)  ( 1 )  (20)  0 
( 15
 v  10.4V
4
10 4
15
Solver :
i4 
20  v
 2.4A  iSC
4
Using a combination of voc and iSC, the Thévenin resistance is determined to be
RTh 
vOC 24V

 10
iSC 2.4A
Interpretation of results
5-7
1. Both methods (1 & 2) determine the Thévenin resistance, however, Method-1 is only
valid if there are ONLY independent sources. If there are any dependent sources,
Method-1 is NOT valid and ONLY Method-2 can be used.
2. The combination voc/iSC, is a measure of the effective resistance that the load sees at
terminal points a-b: RTh.
Norton Equivalent Circuits
Independent & Dependent Sources
There are two important changes when dealing with independent and dependent sources in
circuit where Thévenin’s theorem is going to be applied:
1. Without dependent sources, zeroing all the independent sources leaves a resistive
circuit so that the Thévenin resistance can be calculated directly. However, dependent
sources typically alter the Thévenin resistance so those can't be zeroed. That is,
Method-1 is NOT VALID whenever dependent sources are in the circuit and the only way
to determine RTh is by using Medthod-2; determine the ratio vOC/iSC = RTh.
2. Thévenin circuit with dependent sources can have positive and negative Thévenin
resistance. It is not like there is an actual negative resistor that one can actual purchase
at the store or something like that, but has implications that circuits have a “impedance
matching” and can be “resonated.”
Example 5.4
Obtain the Thévenin equivalent for the following circuit.
Solution
The fact that there is a dependent source in the circuit immediately tells us that Method-1
cannot be applied to this circuit. Only Method-2 is valid.
Step 1: Determine voc
Focusing on the open circuit where voc is defined, the open immediately tells us that the 1k
resistor is a “dead resistor” since there is no current flowing through it and has no voltage
drop. So the voltage across the 3k resistor is equivalent to voc:
v oc  v 3k  3iX
Clearly, to determine voc we shift our attention to first determining ix. Thinking in terms of
source transformations first, converting the 2.5mA source into a
voltage source allows us to reduce the circuit to one loop.
KVL: (6  4  3)iX  15  2iX 
 iX  1mA 
 v oc  3i X  3V
5-8
Step 2: Determine isc such that RTh = voc /isc.
Since I’ve already modified the left side of the circuit, I will continue using it. Placing the
short across a-b, isc is the current through the 1kΩ resistor. To determine isc we can either
use (i) mesh analysis which will produce 3 equations and 3 unknowns (isc, ix, iz), (ii) node
analysis which also has 3 equations and 3 unknowns (isc, ix, vy) or (ii) simple circuits that has
2 equations and 2 unknowns (isc, ix). Let’s do simple circuits!
I will use VDR to determine the voltage 3ix (and current ix) and use Ohm’s law to get isc:
VDR: v 3K  v1K  3ix 
3 1
(15  2ix )
3 1  10
solving
1

for ix  i x  3 mA
solving

for i  iSC 
SC
v1k
 1mA  iSC
1 k
Applying RTh = vOC/iSC, we find that
RTh 
v oc
3V

 3kΩ
isc 1mA
Step 3 Redraw the equivalent circuit:
Example 5.5
Determine the current through the 2Ω-resistor at a-b using a Thévenin equivalent circuit.
Solution
Step 1: Determine voc
The simplest way to determine voc is by applying KVL around the top
loop. Why? Because I immediately know the dependent current i since it
is in series with the 4A-source.
i  4A
voc  8  2i  6i  0 
 vOC  24V
Step 2: Determine isc such that RTh = voc/isc.
Placing the short at a-b sets up three loops (i, isc, ix). However, this short really sets up an
unusual situation where the loop current-i (which also has a dependent
source controlled by i) is directly connected to isc via a constant 4Asource. Note that I have completely ignored ix. The trick is to see that
doing a mesh loop for loop-i and applying KCL between a-b, there are 2
equations and 2 unknowns (isc, i) that quickly determines isc:
solving for i
Loop - i :  6i  2i  8 
 i  2A
KCL at a-b: iSC  4  i  6A
Applying RTh = vOC/iSC, we find that
5-9
RTh 
v oc 24 V

 4
isc
6A
Step 3: Redraw the equivalent circuit and solve the circuit problem.
Using Ohm’s law, the current through the 2-resistor is
i2 
v OC
24

 12 A  i2
RTh  2 4  2
Because this current is positive, the current runs positive from terminal a to terminal b.
Interpretation of Negative Thévenin Resistance
Suppose now that the load resistor was replaced with a variable
resistor (i.e., decade resistor box) such that we adjusted the
resistance. How does the circuit behave?
One would think that if one sets Rab = 0Ω, then maximum current would flow through the
load resistance Rab. Let’s look at this:
iload 
v OC
24

 6 A
RTh  Rload 4  0
If I now adjusted the decade resistor box to Rload = 2Ω, the current now reads 12A, double
the value of the short circuit current:
iload 
24
 12 A
4  2
And as you can see, something very special happens as Rload → 4Ω, the current goes to
infinity:
lim iload 
Rload  4 
lim
Rload  4 
24


4  Rload
Physical interpretation
The key phase is “impedance matching.” Negative resistance cannot physically occur in the
case where the circuit is linear and contains only passive components (resistors, capacitors
and inductors). But for active circuits, where dependent sources (amplifiers) are applied, a
virtual negative resistance can be realized. The physical meaning of negative resistance is
that power is absorbed by the circuit with zero phase shift rather than dissipated.
Maximum Power Delivered to a Circuit
The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid
to system design. Simply stated, the maximum amount of power will be dissipated by a
load resistance when that load resistance is equal to the Thévenin resistance of the circuit
supplying the power. If the load resistance is lower or higher than the Thévenin
resistance, its dissipated power will be less than maximum.
This is essentially what is aimed for in radio transmitter design where the antenna or
transmission line “impedance” is matched to final power amplifier “impedance” for
maximum radio frequency power output (impedance matching). Impedance, the
overall opposition to AC and DC current, is very similar to resistance, and must
be equal between source and load for the greatest amount of power to be
transferred to the load. A load impedance that is too high will result in low power
output. A load impedance that is too low will not only result in low power output, but
5-10
possibly overheating of the amplifier due to the power dissipated in its internal Thévenin
impedance.
Mathematically, one can determine the maximum power delivered to the load by calculating
the critical point where dPload/dRload = 0:
v load 
Rload
v2
dP
Rload
d 
2

v Th 
 Pload  load 
 load  v Th

RTh  Rload
Rload
dRload
dRload RTh  Rload 2


0


After doing some algebra, we arrive
Rload
d 

dRload  RTh  Rload 2

 R  R  2R
load
load
  Th
 0 
 RTh  Rload
4

R

R


Th
load

Example 5.6
What is the maximum power that can be delivered to the load resistor RL?
Solution
Step 1: Find vOC
What is the most efficient way to solve for vOC? I first note that the
controlling voltage vy is known since the 2Ω-resistor is in series with the
9A-source: vy = 18V. So I can replace the VCCS as a 36A-source
instead. So if I do the numbers, this is what I get:
4 nodes  2 sources  2 node eqs [ node (v OC ) & SN eq (v1,v 2 )
 SC (v1,v 2 )
 DC (ix )
2 eqs, 2 unknowns (v OC ,v1,v 2 ,ix )
Writing out the equations:
node v OC : 21 v OC  21 v 2  0  9
SN: (1  21 )v 2  3ix  21 v OC  21 v1  21  3ix  27
SC: v 2  v1  2
DC: v1  3ix  2ix
In matrix form and solution:
 21
 1
 2
 0

 0
0
 21
1
2
3
2
1
1
1
0

238
 79.3V
 v OC  
3

0   v OC   9 

190
 


v 
 63.3V
 92   v1   27  solutions 
 1
3



0   v2   2 
184

 
v2 
 61.3V



5   ix   0 
3

i   38  12.7V
 x
3
5-11
Step 2: Find iSC.
Let’s do the numbers:
4 loops  2 sources  2 loops eqs (1 regular (i2 ) & 1 SL (i1,i3 ,ix )
 2 SC
 DC (ix , v y )
 Solver (iSC )
6 eqs, 6 unknowns (i1,i2 ,i3 ,ix ,v y ,iSC )
Writing out the equations,
Loop 2 : 1  2  i2  2i1  3ix
SL:
2i1  2i2  0  2  ix  3ix  2
loop ix
loop 1
2 SC :
i3  i1  9, i3  ix  2v y
DC :
v y  2i2  2i1
Solver :
iSC  i2  i3
The matrix and solutions are
i1  3.58A
 2 3 0 3 0 0   i1   0 




 i
 
i2  2.26A
 2 2 0 5 0 0   2   2 
 1 0 1 0 0 0   i3   9  solutions i3  5.42A

       i  0.129A
 0 0 1 1 2 0   ix   0 
x


 2 2 0 0 1 0  v y
0
v y  2.65V

    



1 1 0 0 1  iSC   0 
 iSC  7.67A
0
Step 3: Find RTH and solve the circuit problem
Applying RTh = vOC/iSC, we find that
RTh 
v oc 79.3V

 10.3 
isc
7.67A
RTh  10.3 
maximum power theorem


RTh RL 3 
vTh  79.3V
Pmax 
vL2 1.52

 0.75W  Pmax
RL
3
5-12