EEE 201
Circuits I
Part 4 : Thevenin’s Theorem
4.1 Thevenin’s Theorem
4.2 Case 1: Independent Sources Only
4.3 Case 2: Independent & Dependent
Sources
4.4 Case 3: Dependent Sources Only
4.5 More Examples of Cases 1, 2, 3
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EEE 201: Thevenin's
Theorem
9.1 Thevenin’s Theorem (named for Charles Leon Thévenin (1857-1926)
i
a
+
v i
–
b
Network A
Linear
or
Nonlinear
Network B
Conditions:
1. Network A and Network B are connected at
only a and b.
2. No dependent source in B is controlled by a
voltage or current in A, and vice versa.
Thevenin’s Theorem shows that:
i a
R
Th
VTh
+
–
Network A
+
v i
–
b
(Unchanged)
Network B
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EEE 201: Thevenin's
Theorem
RTh
VTh
i
+
–
Network A
a
+
v
–
b
i
(Unchanged)
Network B
RTh is the equivalent resistance between a-b with (1) B
disconnected, (2) the independent sources in A set to 0
(“killed”), and (3) the dependent sources in A unchanged.
RTh is also known as the “lookback resistance.”
VTh is the open-circuit voltage of Network A, obtained with
Network B disconnected.
By using equivalent sources, we can get the “Norton
Equivalent”:
RTh
VTh +

–
VTh / RTh
Thevenin
Equivalent
RTh
Norton
Equivalent
Note that the open-circuit voltage is VTh and VTh / RTh is the
short-circuit current (RTh is the same for both the Thevenin &
Norton Equivalents).
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EEE 201: Thevenin's
Theorem
Summary:
RTh
Voc
+
–

Thevenin
Equivalent
Isc
RTh
Norton
Equivalent
Voc = RTh Isc
Memorize
This!
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EEE 201: Thevenin's
Theorem
9.2 Case 1: Independent Sources Only
(No dependent sources are present)
Example 1
Find the Thevenin Equivalent for ‘A.’ (What part of the
circuit we choose to include in ‘A’, with the remainder
lumped into ‘B’, is arbitrary.)
3W
12 V
+
–
8W
a
load
1A
6W
b
A
B
Solution:
For RTh :
3W
8W
6W
a
i = 0
RTh
Sources
set to 0!
b
RTh =
8 + 3 // 6 = 10 W
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EEE 201: Thevenin's
Theorem
Example 1 (cont.):
For VTh (Voc):
3W
12 V
+
–
8W
a
+
because we are
finding the openVoc circuit voltage!
1A
6W
No current flows
through the 8 W
–
b
voc
12 V 3 W
12 V
+
–
voc is also the
node voltage!
1A
6W
b
KCL: (Voc– 12) / 3 + Voc / 6 – 1 = 0
 Voc = 10 V
Thus the Thevenin and Norton equivalents are:
10 W
10 V
a
a

+
–
1A
b
10 W
b
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EEE 201: Thevenin's
Theorem
Example 2
Find i
a
4W
i
12 W
3A
6W
+
–
6W
12 V
b
Solution:
To find
RTh :
4W
a
i=0
12 W
6W
6W
b
RTh =
4 + 6 // 12 = 8 W
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EEE 201: Thevenin's
Theorem
Solution:
To find
VTh :
v
a
4W
+
12 W
6W
3A
+
–
12 V
voc
6W
–
b
KCL:
v / 6 + (v – 12) / 12 = 3
 v = 16 V
so that Voc = 4 • 3 + 16 = 28 V = VTh
The Thevenin equivalent can now be used to find
the current that flows when the 6 W resistor is put
back:
a
8W
i
+
28 V –
i = 28 / ( 8+6 )
6W
= 2A
b
EEE 201: Thevenin's
Theorem
201
Example 3
Find P4W (power absorbed by 4 W resistor)
c 4W
a
12 W
6W
3A
+
–
6W
12 V
b
Solution:
Note that this circuit is the same as the one in the previous
example. However, the Thevenin equivalent we want now
will be different. Must re-draw to emphasize what we
want:
4W
c
a
12 W
6W
3A
+
–
6W
12 V
b
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EEE 201: Thevenin's
Theorem
RTh
c
(cont.)
a
RTh =
RTh :
6//12 + 6
12 W
6W
6W
= 10 W
+ Voc –
Voc :
c
a
Voc = vc - va
12 W
3A
6W
+
–
6W
6/18•12 – 3 • 6
= – 14 V
12 V
The Thevenin equivalent can now be used to find the
current that flows when the 4 W resistor is put back:
4W
10 W
–14 V
+
–
c
v4 = 4/14 • (–14)
a
+ v4 –
= –4V
P4W = v42 / 4
= 4W
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EEE 201: Thevenin's
Theorem
• Stop here
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EEE 201: Thevenin's
Theorem
9.3 Case 2: Independent & Dependent Sources
Example 4
Find the Thevenin
equivalent
at a-b.
20 V +
–
100 W
a
ib
9 ib
10 W
Solution:
100 W
VTh :
20 V
b
+
–
voc
ib
9 ib
10 W
a
+
Voc = VTh
–
b
KCL at node a: (Voc– 20) / 100 – 9 ib + Voc / 10 = 0
Constraint:
ib = ( 20 – Voc) / 100
Solving gives: Voc = 10 V
RTh :
100 W
ib
9 ib
Independent source set to 0!
Now what ??!!
10 W
RTh
Dependent source is not!
Can proceed three ways!!
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EEE 201: Thevenin's
Theorem
Three ways to find RTh
Method 1. Use RTh = voc / isc (keeping the independent
source in the circuit).
Recall the Thevenin model:
RTh
a
Voc
RTh
Voc
+
–
a
Isc
+
–
b
b
RTh = Voc / Isc, so find Isc and use it with Voc to find RTh.
Thus, in our example,
100 W
20 V
+
–
ib
a
9 ib
10 W
Isc
b
ib = 20 / 100 = 0.2 A Tricky! (The 10 W is shorted,
so 20 V appears across the 100 W!)
isc = ib + 9 ib = 10 ib = 2 A
 RTh = Voc / Isc
= 10 / 2
= 5W
EEE 201: Thevenin's
Theorem
206
Three ways to find RTh (cont)
Method 2. Apply a current source at a-b, compute
the corresponding voltage, and take the ratio R = V / I.
a
100 W
ib
9 ib
Independent source set to 0!
KCL at node a:
10 W
+
1A
v
Dependent source is not! b Note –
reference
direction for
the 1A
source!
v / 100 – 9 ib + v / 10 – 1 = 0
constraint:
ib = – v / 100
solving gives:
v = 5V
 RTh = v / 1
= 5/1
= 5W
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EEE 201: Thevenin's
Theorem
Three ways to find RTh (cont)
Method 3. Apply a voltage source at a-b, compute
the corresponding current, and take the ratio R = V / I.
i
100 W
ib
9 ib
10 W
a
+
–
b
Independent source set to 0! Dependent source not!
1V
Again,
note
reference
directions!
ib = – 1 / 100
9 ib = – 9 / 100
i = – ib – 9 ib + 1 / 10
KCL
 i = 2 / 10 A
 RTh = 1 / i
= 1 / ( 2 / 10 )
= 5W
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EEE 201: Thevenin's
Theorem
Example 5
15 W
Find the Thevenin
Equivalent as seen
by R3.
2Vx +
–
+
2A
5W
R3
Vx
–
Solution
15 W
First find the shortcircuit current:
2Vx +
–
Isc
2A
5W
Isc = 2 A
(tricky!)
Next, find the opencircuit voltage:
...
15 W
+
Voc = 15 V
2Voc
(Use KCL at node vx)
The Thevenin resistance
is then Voc/Isc, or 7.5 W
and the Thevenin
Equivalent circuit is as
shown.
+
–
2A
5W
Vx = Voc
–
7.5 W
15 V
+
–
R3
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EEE 201: Thevenin's
Theorem
9.4 Case 3: Dependent Sources Only
No independent sources, some dependent sources.
Example 6
Find the Thevenin equivalent at a-b.4 W
Solution:
+ v1 –
a
6W
6W
No independent sources,
+
– 1.5 v1
thus voc and isc are
undetermined. Therefore
RTh = voc / isc is indeterminate!
b
Try applying a 1 V source at a,b:
4W
v + v1 – i a
1.5 v1
+
–
6W
6W
+
–
1V
b
KCL:
(v – 1.5 v1) / 4 + v / 6 + (v – 1)/ 6 = 0
constraint: v1 = v – 1
Solving gives: v1 = – 2 V,
so that
i = – v1 / 6 = 0.333 A
and RTh = 1 / i = 3 W
a
3W
b
EEE 201: Thevenin's
Theorem
210
Example 7
Find the Thevenin equivalent at a-b
Solution:
voc and isc undetermined
(no independent sources) so
RTh = voc / isc is indeterminate.
Try applying a 1 A source at a,b:
2 va
4W
4W
a
+
2W
vab
–
b
The node voltage va is
what we want:
va = vab!
a
va
2 vab
1A
2W
b
KCL:
2 va + va / 2 – 1 = 0
Solving gives: va = 2 / 5
and so RTh = (2 / 5) / 1 = 0.4 W.
Thevenin equivalent:
a
0.4 W
b
EEE 201: Thevenin's
Theorem
211
Case Summary
Case 1. Independent
sources only. (No
dependent sources
are present.)
Case 2. Independent
and dependent
sources.
1. With the sources in, calculate voc.
2. Set the sources to 0 and calculate
RTh , by combining resistances if
possible. Or, go back to step 1,
calculate isc and calculate the ratio voc
/ isc for RTh .
With the sources in, calculate voc and
isc. Then calculate the ratio voc / isc for
RTh .
or
With the independent sources set to 0,
apply a voltage source (current source)
and calculate the corresponding source
current (source voltage), and then
calculate RTh as the ratio of the source
voltage to the source current.
Case 3. No
independent
sources, some
dependent sources.
voc and isc are undetermined.
Apply a voltage (current) source and
calculate the current (voltage) and
calculate their ratio for RTh .
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EEE 201: Thevenin's
Theorem
9.5 More Examples
Example 8
[Case 2]
Find the value of RL that will cause the current iL to be 2 A.
4i
a
+ –
10 A
2W
iL
i
RL
4i
b
Solution
a
+ –
10 A
2W
First find the open-circuit voltage:
+
Voc
i
–
Voc = – 4(10) + 2(10) = – 20 V
b
4i
Next, find the short-circuit current: i
must be 0 because for the current 10 A
through the 2W resistor, i = 4i/2,
which is satisfied only by i = 0. Thus
iSC = 10 A.
The Thevenin resistance is then
Voc/Isc, or – 2 W. (Strange!) and the
Norton Equivalent circuit is as
shown. Using current division, we
find that for iL to be 2 A, RL must be
– 8 W. Strange!
+ –
2W
10 A
EEE 201: Thevenin's
Theorem
i
iSC
iL
RL
–2 W
213
[ Case 2 ]
Example 9
Find the maximum power that can be absorbed by the 3W
resistor.
4W
10 V
+
–
Ix
3W
20 W
5W
3Ix
Solution
First we find the Thevenin Equivalent circuit as seen by the
3-W resistor. One way to do this is to find the open-circuit
voltage and the short-circuit current at that branch. The
results are (please verify on your own!):
4W
10 V
+
–
Ix
20 W
+
Voc= 20 V
5W
3Ix
5W
3Ix
–
10 V
+
–
Ix
20 W
Isc= 5 A
4W
With Voc= 20 Vand Isc= 5 A, RTh = 20 / 5 =  4 W
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EEE 201: Thevenin's
Theorem
Solution (cont.)
As an aside we note that another way to find the Thevenin
Equivalent resistance as seen by the 3W resistor is to insert a
current source, compute the resulting voltage across the
source, and compute the ratio. The results are as shown
(please verify on your own). Thus, RTh = 20/3 / 1 = 20/3 W,
as expected.
4W
10 V
+
–
Ix
20 W
+
–4V
1A
5W
3Ix
–
Still another way to find the Thevenin Equivalent resistance
as seen by the 3-W resistor is to insert a voltage source,
compute the resulting current through the source, and
compute the ratio. The results are as shown.
Thus, RTh = 1 / (3/20) = 20/3 W, as expected.
4W
10 V
+
–
Ix
20 W
– 1/4 A
+ 1V
–
5W
3Ix
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EEE 201: Thevenin's
Theorem
Solution (cont.)
With the use of the Thevenin Equivalent, the equivalent circuit
for calculating the maximum power absorbed by the 3W
resistor is:
–4W
+
20 V
+
–
3W
V3
–
The maximum power that can be absorbed by the 3-W resistor
is calculated as follows:
V3 
3  20 

20  3 
3
3
–4W
V32
p3 
3


 3  20  

 
20
 3 
 3 

3



3
2
 1.43 W
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EEE 201: Thevenin's
Theorem