Circuit Theorems - PowerPoint PPT Presentation

About This Presentation
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:129
Avg rating:3.0/5.0
Slides: 39
Provided by: MustafaKe8
Category:
Tags: circuit | theorems

less

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
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.
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!
Write a Comment
User Comments (0)
About PowerShow.com