Circuit Theorems
Dr. Mustafa Kemal Uyguroğlu
Circuit Theorems Overview
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Introduction
Linearity
Superpositions
Source Transformation
Thévenin and Norton Equivalents
Maximum Power Transfer
INTRODUCTION
A large
complex circuits
Simplify
circuit analysis
Circuit Theorems
‧Thevenin’s theorem
‧Circuit linearity
‧source transformation
‧ Norton theorem
‧ Superposition
‧ max. power transfer
Linearity Property
A linear element or circuit satisfies the properties of
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Additivity: requires that the response to a sum of
inputs is the sum of the responses to each input
applied separately.
If v1 = i1R and v2 = i2R
then applying (i1 + i2)
v = (i1 + i2) R = i1R + i2R = v1 + v2
Linearity Property
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Homogeneity:
If you multiply the input (i.e. current) by some
constant K, then the output response (voltage) is
scaled by the same constant.
If v1 = i1R
then K v1 =K i1R
Linearity Property
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A linear circuit is one whose output is linearly
related (or directly proportional) ito its input.
I0
Suppose vs = 10 V gives i = 2 A. According to
the linearity principle, vs = 5 V will give i = 1 A.
V0
v
Linearity Property - Example
i0
Solve for v0 and i0 as a function of Vs
Linearity Property – Example
(continued)
Linearity Property - Example
Ladder Circuit
3A
5A
+6V-
2A
+3V-
2A
1A
+
+
+
14 V
8V
5V
-
-
-
This shows that assuming I0 = 1 A gives Is = 5 A; the actual source
current of 15 A will give I0 = 3 A as the actual value.
Superposition

The superposition principle states that the
voltage across (or current through) an
element in a linear circuit is the algebraic
sum of the voltages across (or currents
through) that element due to each
independent source acting alone.
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.
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2.
3.
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Turn off voltages sources = short voltage sources; make it
equal to zero voltage
Turn off current sources = open current sources; make it
equal to zero current
Repeat step 1 for each of the other independent sources.
Find the total contribution by adding algebraically all the
contributions due to the independent sources.
Dependent sources are left intact.
Superposition - Problem
2kW
4mA
12V
– +
2mA
1kW
I0
2kW
2mA Source Contribution
2kW
2mA
1kW
I’0
I’0 = -4/3 mA
2kW
4mA Source Contribution
2kW
4mA
1kW
I’’0
I’’0 = 0
2kW
12V Source Contribution
12V
2kW
– +
1kW
I’’’0
I’’’0 = -4 mA
2kW
Final Result
I’0 = -4/3 mA
I’’0 = 0
I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA
Example
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find v using superposition
one independent source at a time,
dependent source remains
KCL: i = i1 + i2
Ohm's law: i = v1 / 1 = v1
KVL: 5 = i (1 + 1) + i2(2)
KVL: 5 = i(1 + 1) + i1(2) + 2v1
10 = i(4) + (i1+i2)(2) + 2v1
10 = v1(4) + v1(2) + 2v1
v1 = 10/8 V
Consider the other independent source
KCL: i = i1 + i2
KVL: i(1 + 1) + i2(2) + 5 = 0
i2(2) + 5 = i1(2) + 2v2
Ohm's law: i(1) = v2
v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2
i2(2) + 5 = i1(2) + 2v2
-2v2 = (i - i2)(2) + 2v2
-2v2 = [v2 + (5+2v2)/2](2) + 2v2
-4v2 = 2v2 + 5 +2v2
-8v2 = 5 => v2 = - 5/8 V
from superposition: v = -5/8 + 10/8
v = 5/8 V
Source Transformation
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A source transformation 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
Source Transformation
vs
vs  is R or is 
R
Source Transformation
V s  Rs I s
Vs
Is 
Rs
Source Transformation
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Equivalent sources can be used to simplify
the analysis of some circuits.
A voltage source in series with a resistor is
transformed into a current source in parallel
with a resistor.
A current source in parallel with a resistor is
transformed into a voltage source in series
with a resistor.
Example 4.6
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Use source transformation to find vo in the
circuit in Fig 4.17.
Example 4.6
Fig 4.18
Example 4.6
we use current division in Fig.4.18(c) to get
2
i
(2)  0.4A
28
and
vo  8i  8(0.4)  3.2V
Example 4.7
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Find vx in Fig.4.20 using source
transformation
Example 4.7
 3  5i  vx  18  0
Applying
KVL around the loop in Fig 4.21(b) gives
(4.7.1)
 3  5i  vx  18  0
Appling KVL to the loop containing only the 3V voltage source, the
resistor, and vx yields
(4.7.2)
 3  1i  vx  0  vx  3  i
Example 4.7
Substituting this into Eq.(4.7.1), we obtain
15  5i  3  0  i  4.5A
Alternatively
 vx  4i  vx  18  0  i  4.5A
thus
vx  3  i  7.5V
Thevenin’s Theorem
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Any circuit with sources (dependent and/or
independent) and resistors can be replaced
by an equivalent circuit containing a single
voltage source and a single resistor.
Thevenin’s theorem implies that we can
replace arbitrarily complicated networks with
simple networks for purposes of analysis.
Implications
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We use Thevenin’s theorem to justify the
concept of input and output resistance for
amplifier circuits.
We model transducers as equivalent sources
and resistances.
We model stereo speakers as an equivalent
resistance.
Independent Sources (Thevenin)
RTh
Voc
Circuit with
independent sources
+
–
Thevenin equivalent
circuit
No Independent Sources
RTh
Circuit without
independent sources
Thevenin equivalent
circuit
Introduction
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Any Thevenin equivalent circuit is in turn
equivalent to a current source in parallel with
a resistor [source transformation].
A current source in parallel with a resistor is
called a Norton equivalent circuit.
Finding a Norton equivalent circuit requires
essentially the same process as finding a
Thevenin equivalent circuit.
Computing Thevenin Equivalent

Basic steps to determining Thevenin
equivalent are
–
–
Find voc
Find RTh
Thevenin/Norton Analysis
1. Pick a good breaking point in the circuit (cannot split a dependent
source and its control variable).
2. Thevenin: Compute the open circuit voltage, VOC.
Norton: Compute the short circuit current, ISC.
For case 3(b) both VOC=0 and ISC=0 [so skip step 2]
Thevenin/Norton Analysis
3. Compute the Thevenin equivalent resistance, RTh
(a) If there are only independent sources, then short circuit
all the voltage sources and open circuit the current sources
(just like superposition).
(b) If there are only dependent sources, then must use a test
voltage or current source in order to calculate
RTh = VTest/Itest
(c) If there are both independent and dependent sources,
then compute RTh from VOC/ISC.
Thevenin/Norton Analysis
4. Thevenin: Replace circuit with VOC in series with RTh
Norton: Replace circuit with ISC in parallel with RTh
Note: for 3(b) the equivalent network is merely RTh , that is, no
voltage (or current) source.
Only steps 2 & 4 differ from Thevenin & Norton!