ECE 1311: Electric Circuits
Chapter 4: Circuit Theorems
Circuit Theorem Overview
 Linearity Property
 Superposition
 Source Transformation
 Thevenin’s Theorem
 Norton’s Theorem
 Maximum Power Transfer
Linearity Property (1)
 It is the property of an element describing a linear relationship
between cause and effect.
 A linear circuit is one whose output is linearly related (or directly
proportional) to its input.
v=iR
→
kv=kiR
 In this class, we will only consider circuits in which the voltage
and the currents are linearly related to the independent sources.
 Hence, the current through AND the voltage across each circuit
element is linearly proportional to the independent source
amplitude.
Example 4.2:
By assume Io = 1 A, use linearity to find the actual value of Io in
the circuit shown below.
Answer: Io = 3A
Linearity Property (2)
Answer: Vs/2
Linearity Property (3)
Superposition Theorem (1)
 It states that the voltage across (or current through) an
element in a linear circuit is the algebraic sum of the
voltage across (or currents through) that element due to
EACH independent source acting alone
 The principle of superposition helps us to analyze a
linear circuit with more than one independent source by
calculating the contribution of each independent source
separately.
Superposition Theorem (2)
 Steps to apply superposition principle
1. Turn off all independent sources except one
source. Find the output (voltage or current)
due to that active source using nodal or
mesh analysis.
2. Repeat step 1 for each of the other
independent sources.
3. Find the total contribution by adding
algebraically all the contributions due to the
independent sources.
Example 4.3:
 Use the superposition theorem to find v in the circuit
shown below.
3A is discarded by opencircuit
6V is discarded by
short-circuit
*Refer to in-class illustration, text book, answer v = 10V
Practice 4.1
 Find I in the circuit using superposition
Practice 4.2
 Use superposition to determine vx
Practice 4.3
Superposition
Source Transformation (1)
 Recall: Resistor in series, parallel, delta-wye
 It is the process of replacing a voltage source vS in series
with a resistor R by a current source iS in parallel with a
resistor R, or vice versa.
 The two circuits are equivalent if they have the same
current-voltage relationship at their terminals
Source Transformation (2)
(a) Independent source transform
(b) Dependent source transform
• The arrow of the
current source is directed
toward the positive
terminal of the voltage
source.
• The source
transformation is
not possible when
R = 0 for voltage
source and R = ∞
for current source.
Source Transformation (3)
DOES NOT WORK IF R = 0!!!
Example
 Find io in the circuit shown below using source
transformation.
Answer: io = 1.78A
Recap:
Recap:
Practice 4.4
Thevenin’s Theorem
It states that a linear two-terminal circuit (Fig. a) can
be replaced by an equivalent circuit (Fig. b)
consisting of a voltage source VTH in series with a
resistor RTH,
where
• VTH is the open-circuit voltage at the
terminals.
• RTH is the input or equivalent resistance at
the terminals when the independent
sources are turned off.
Norton’s Theorem
It states that a linear two-terminal circuit can be replaced
by an equivalent circuit of a current source IN in parallel
with a resistor RN,
Where
• IN is the short circuit current through
the terminals.
• RN is the input or equivalent resistance
at the terminals when the independent
sources are turned off.
The Thevenin’s and Norton equivalent circuits are related by a
source transformation.
Finding Thevenin’s and Norton
Equivalent
Thevenin’s and Norton Equivalent
Resistance (1)
Thevenin’s and Norton Equivalent
Resistance (2)
Thevenin’s and Norton Equivalent
summary
Practice 4.5
Practice 4.6
 Using Thevenin’s theorem, find the equivalent circuit to
the left of the terminals in the circuit shown below.
6W
6W
Hence find i.
4W
RTh
(a)
6W
2A
6W
2A
(b)
Answer VTH = 6V, RTH = 3W, i = 1.5A
4W
+
VTh

Practice 4.7
 Find the Norton equivalent circuit of the circuit shown
2vx
below.
i
+ 
+
vx

6W
2W
ix
+
vx

1V
+

(a)
2vx
+ 
6W
2W
10 A
(b)
Answer: RN = 1W, IN = 10A.
+
vx

Isc
Practice 4.8
Maximum Power
If the entire circuit is replaced by its Thevenin
equivalent except for the load, the power
delivered to the load is:
2
 VTh 
 RL
P  i 2 RL  
 RTh  RL 
For maximum power dissipated in RL, Pmax,
for a given RTH,
and VTH,
2
RL  RTH

Pmax
V
 Th
4 RL
The power transfer profile with different RL
Practice 4.10
Determine the value of RL that will draw the
maximum power from
the rest of the circuit shown below. Calculate the
maximum power.
+
vx

4W
v0
2W
+
vx
4W

Fig. a
2W
i
1W
1W
+

+

1V
+

9V
io
+

3vx
(a)
Answer, RL = 4.22W, Pm = 2.901W
(b)
3vx
+
VTh

=> To determine RTH
Fig. b
=> To determine VTH