Title: Circuit Theorems
1Circuit Theorems
 Dr. Mustafa Kemal Uyguroglu
2Circuit Theorems Overview
 Introduction
 Linearity
 Superpositions
 Source Transformation
 Thévenin and Norton Equivalents
 Maximum Power Transfer
3INTRODUCTION
A large complex circuits
Simplify circuit analysis
Circuit Theorems
?Thevenins theorem ? Norton theorem ?Circuit
linearity ? Superposition ?source
transformation ? max. power transfer
4Linearity Property
 A linear element or circuit satisfies the
properties of  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
5Linearity Property

 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
6Linearity Property
 A linear circuit is one whose output is linearly
related (or directly proportional) to its input.
Suppose vs 10 V gives i 2 A. According to the
linearity principle, vs 5 V will give i 1 A.
7Linearity Property  Example
Solve for v0 and i0 as a function of Vs
8Linearity Property Example (continued)
9Linearity Property  Example
Ladder Circuit
3 A
5 A
1 A
3 V 
6 V 
2 A
2 A
5 V 
8V 
14 V 
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.
10Superposition
 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.
11Steps to apply superposition principle
 Turn off all independent sources except one
source. Find the output (voltage or current) due
to that active source using nodal or mesh
analysis.  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.
12Superposition  Problem
132mA Source Contribution
I0 4/3 mA
144mA Source Contribution
I0 0
1512V Source Contribution
I0 4 mA
16Final Result
I0 4/3 mA I0 0 I0 4 mA I0 I0
I0 I0 16/3 mA
17Example
 find v using superposition
18one 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) (i1i2)(2) 2v1 10
v1(4) v1(2) 2v1 v1 10/8 V
19Consider the other independent source
KCL i i1 i2 KVL i(1 1) i2(2) 5
0i2(2) 5 i1(2) 2v2Ohm's law i(1)
v2v2(2) i2(2) 5 0 gt i2 (52v2)/2i2(2)
5 i1(2) 2v22v2 (i  i2)(2) 2v22v2
v2 (52v2)/2(2) 2v24v2 2v2 5
2v28v2 5 gt v2  5/8 V from
superposition v 5/8 10/8 v 5/8 V
20Source Transformation
 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
21Source Transformation
22Source Transformation
23Source Transformation
 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.
24Example 4.6
 Use source transformation to find vo in the
circuit in Fig 4.17.
25Example 4.6
Fig 4.18
26Example 4.6
 we use current division in Fig.4.18(c) to get

 and
27Example 4.7
 Find vx in Fig.4.20 using source transformation
28Example 4.7
 Applying KVL around the loop in Fig 4.21(b) gives

(4.7.1)  Appling KVL to the loop containing only the 3V
voltage source, the resistor, and vx yields

(4.7.2)
29Example 4.7
 Substituting this into Eq.(4.7.1), we obtain
 Alternatively

 thus
30Thevenins Theorem
 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.  Thevenins theorem implies that we can replace
arbitrarily complicated networks with simple
networks for purposes of analysis.
31Implications
 We use Thevenins 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.
32Independent Sources (Thevenin)
33No Independent Sources
34Introduction
 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.
35Computing Thevenin Equivalent
 Basic steps to determining Thevenin equivalent
are  Find voc
 Find RTh
36Thevenin/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 VOC0 and ISC0 so skip step
2 
37Thevenin/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.
38Thevenin/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!
