# Circuit Theorems - PowerPoint PPT Presentation

1 / 38
Title:

## Circuit Theorems

Description:

### Circuit Theorems Dr. Mustafa Kemal Uyguro lu Circuit Theorems Overview Introduction Linearity Superpositions Source Transformation Th venin and Norton Equivalents ... – PowerPoint PPT presentation

Number of Views:104
Avg rating:3.0/5.0
Slides: 39
Provided by: MustafaKe8
Category:
Tags:
Transcript and Presenter's Notes

Title: Circuit Theorems

1
Circuit Theorems
• Dr. Mustafa Kemal Uyguroglu

2
Circuit Theorems Overview
• Introduction
• Linearity
• Superpositions
• Source Transformation
• Thévenin and Norton Equivalents
• Maximum Power Transfer

3
INTRODUCTION
A large complex circuits
Simplify circuit analysis
Circuit Theorems
?Thevenins theorem ? Norton theorem ?Circuit
linearity ? Superposition ?source
transformation ? max. power transfer
4
Linearity 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

5
Linearity 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

6
Linearity 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.
7
Linearity Property - Example
Solve for v0 and i0 as a function of Vs
8
Linearity Property Example (continued)
9
Linearity Property - Example
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.
10
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.

11
Steps 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.

12
Superposition - Problem
13
2mA Source Contribution
I0 -4/3 mA
14
4mA Source Contribution
I0 0
15
12V Source Contribution
I0 -4 mA
16
Final Result
I0 -4/3 mA I0 0 I0 -4 mA I0 I0
I0 I0 -16/3 mA
17
Example
• find v using superposition

18
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) (i1i2)(2) 2v1 10
v1(4) v1(2) 2v1 v1 10/8 V

19
Consider 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) 2v2-2v2 (i - i2)(2) 2v2-2v2
v2 (52v2)/2(2) 2v2-4v2 2v2 5
2v2-8v2 5 gt v2 - 5/8 V from
superposition v -5/8 10/8 v 5/8 V
20
Source 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

21
Source Transformation
22
Source Transformation
23
Source 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.

24
Example 4.6
• Use source transformation to find vo in the
circuit in Fig 4.17.

25
Example 4.6
Fig 4.18
26
Example 4.6
• we use current division in Fig.4.18(c) to get
• and

27
Example 4.7
• Find vx in Fig.4.20 using source transformation

28
Example 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)

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

30
Thevenins 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.

31
Implications
• 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.

32
Independent Sources (Thevenin)
33
No Independent Sources
34
Introduction
• 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.

35
Computing Thevenin Equivalent
• Basic steps to determining Thevenin equivalent
are
• Find voc
• Find RTh

36
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 VOC0 and ISC0 so skip step
2

37
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.

38
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!