Thevenin Equivalent (1/5)
RTh
a
Resistive circuit
with independent
and dependent
sources
Principles of Electrical
Engineering - I (14:332:221)
a
VTh
+
−
b
b
• Thevenin Equivalent consists of a voltage source in
series with a resistor
Chapter 4 Notes (Part 2)
– VTh corresponds to the open circuit voltage across
terminals a,b
– RTh corresponds to the resistance offered by the
circuit if there is a short-circuit condition across
terminals a,b
14:332:221, Spring 2004
Thevenin Equivalent (3/5)
Thevenin Equivalent (2/5)
5Ω
25 V
+
−
20 Ω
5Ω
25 V
+
−
Open Circuit
3A
3A
a
+
v1
−
Short Circuit
20 Ω
4Ω
va,b
b
4Ω
+
v2
−
• A technique for more easily computing RTh is to
deactivate all independent sources:
Solve for va ,b :
+
⇒ VTh = va ,b = v1
– Replace independent voltage sources with a short
circuit
– Replace independent current sources with an open
circuit
⇒ VTh = 32 V
−
Solve for isc :
a
isc
b
5Ω
8
v
isc = 2 = = 2 A
4 4
V
⇒ RTh = Th = 16 Ω
isc
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8
2Ω
24 V
+
−
v a ,b − 24
2
a
4A
ix
8Ω
b
14:332:221, Spring 2004
RTh = Ra,b = 8 Ω
20 Ω
b
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3 ix
– Deactivate independent sources as described above
– Apply a test voltage (vT) or test current (iT) to the
Thevenin terminals a,b
⇒RTh = vT/iT
KCL at node a :
3i
+
4
Deactivating the independent
sources in Ex. 4.23…
2Ω
iT
ix
a
+ v
T
8Ω
−
b
+ 3i x + 4 = 0
ix =
a
Thevenin Equivalent (5/5)
• If the circuit contains dependent sources
v a ,b
4Ω
3
Thevenin Equivalent (4/5)
x
2
va ,b
KCL at node a : iT = 4ix +
8
vT
2
v
Ohm's Law at 8Ω resistor : i x = T
8
⇒ VTh = v a ,b = 8 V
5
14:332:221, Spring 2004
Solving yields : iT = vT
⇒ RTh =
vT
= 1Ω
iT
6
1
Norton Equivalent
Maximum Power Transfer
RTh
VTh
a
+
−
is
Resistive circuit
with independent
and dependent
sources
RTh
b
 VTh
p = i 2 R L = 
 RTh + R L
Solving
14:332:221, Spring 2004
6Ω
Solve for currents due to each independent source in
isolation
−
i2
4Ω
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−
i′3
3Ω
6Ω
Fig. 4.62
i4
120 V
i′1
+
2Ω
v1
i′2
4Ω
i′4
Voltage source deactivated:
2Ω
3Ω
2

 R L

8
Current source deactivated:
• In circuit analysis involving linear circuit
elements, it is possible to deactivate
independent sources to isolate the contribution a
single source
120 V
RL
Superposition (2/2)
H(A) + H(B) = H(A+B) for all A,B
i3
i
V2
dp
= 0 yields RL = RTh ⇒ pmax = Th
dRL
4 RL
7
• H is a linear operator iff:
i1
+
−
Want to choose RL so as to maximize the amount of power
delivered to the resistive load
Superposition (1/2)
6Ω
VTh
b
– Can be obtained via a source transformation of the
Thevenin equivalent circuit
– Note, is = isc, so the Norton equivalent can be obtained
directly by solving for isc, VTh and then RTh = VTh/isc
14:332:221, Spring 2004
RL
a
b
b
• Norton Equivalent consists of a current source in
parallel with a resistor
+
RTh
a
a
i1′′
12 A
9
v3
2Ω
v1 − 120 v1 v1
+ + =0
5
3 6
i1 = i1′ + i1′′
i2 = i2′ + i2′′ Currents
of Fig. 4.62
i3 = i3′ + i3′′
i4 = i4′ + i4′′
v4
i3′′
3Ω
i2′′
4Ω
i4′′
12 A
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v 3 v3 v 3 − v 4
+ +
=0
3 6
2
v 4 − v3 v 4
+ + 12 = 0
2
4
10
2