Engr228 - Chapter 5, Hayt 8E 1 Chapter 5 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 5 Objectives • State and apply the property of linearity • State and apply the property of superposition • Investigate source transformations • Define and construct Thevenin and Norton equivalent circuits • Investigate maximum power transfer to a load
41
Embed
Chapter 5curt.nelson/engr228/lecture/chap5.pdf · The Superposition Theorem ... Engr228 - Chapter 5, Hayt 8E 12 Source Transformation and Equivalent Sources The sources are …
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Engr228 - Chapter 5, Hayt 8E 1
Chapter 5
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 5 Objectives
• State and apply the property of linearity
• State and apply the property of superposition
• Investigate source transformations
• Define and construct Thevenin and Norton equivalent
circuits
• Investigate maximum power transfer to a load
Engr228 - Chapter 5, Hayt 8E 2
Linear Elements and Circuits
• A linear circuit element has a linear voltage-current
relationship:
– If i(t) produces v(t), then Ki(t) produces Kv(t)
– If i1(t) produces v1(t) and i2(t) produces v2(t), then i1(t) + i2(t) produces
v1(t) + v2(t),
• Resistors and sources are linear elements
– Dependent sources need linear control equations to be linear elements
• A linear circuit is one with only linear elements
The Superposition Concept
For the circuit shown below, the question is:
• How much of v1 is due to source ia, and how much is due to
source ib?
We will use the superposition principle to answer this question.
Engr228 - Chapter 5, Hayt 8E 3
The Superposition Theorem
In a linear network, the voltage across or the current through
any element may be calculated by adding algebraically all the
individual voltages or currents caused by the separate
independent sources acting “alone”, i.e. with
– All other independent voltage sources replaced by short circuits (i.e. set
to a zero value) and
– All other independent current sources replaced by open circuits (also
set to a zero value).
Applying Superposition
• Leave one source ON and turn all other sources OFF:
– Voltage sources: set v=0
These become short circuits.
– Current sources: set i=0
These become open circuits.
• Then, find the response due to
that one source
• Add the responses from the other
sources to find the total response
Engr228 - Chapter 5, Hayt 8E 4
Superposition Example (Part 1 of 4)
Use superposition to solve for the current ix
Superposition Example (Part 2 of 4)
First, turn the current source off:
′ i x =3
6+9= 0.2
Engr228 - Chapter 5, Hayt 8E 5
Superposition Example (Part 3 of 4)
Then, turn the voltage source off:
ix′′ =
6
6 + 9(2) = 0.8
Superposition Example (Part 4 of 4)
Finally, combine the results:
ix = ix′ + ix
′′ = 0.2 + 0.8 =1.0
Engr228 - Chapter 5, Hayt 8E 6
Example: Power Ratings
Determine the maximum positive current to which the source Ix
can be set before any resistor exceeds its power rating.
Answer: Ix < 42.49 mA
Superposition with a Dependent Source
• When applying superposition to circuits with dependent
sources, these dependent sources are never “turned off.”
Engr228 - Chapter 5, Hayt 8E 7
Superposition with a Dependent Source
Current source off
Voltage source off
ix = ix’+ix
’’=2 + (−0.6) = 1.4 A
Superposition Example
Find voltage Vx. First, solve by nodal analysis.
(42-Vx)/6 = (Vx-(-10))/4 + Vx/3
Vx = 54/9 = 6 volts
Engr228 - Chapter 5, Hayt 8E 8
V
Vx V
333.9
42)7/12(6
)7/12(42
)4||3(6
)4||3()42(
=
×+
=×+
=
Example - Continued
V
Vx V
333.3
1042
210
4)3||6(
)3||6()10(
−=
×+
−=×+
−=
Example - Continued
Engr228 - Chapter 5, Hayt 8E 9
V
VxVxVx VV
6333.3333.9
)10()42(
=−=
+=
Example - Continued
Practical Voltage Sources
• Ideal voltage sources: a first approximation model for a
battery.
• Why do real batteries have a current limit and experience
voltage drop as current increases?
• Two car battery models:
Engr228 - Chapter 5, Hayt 8E 10
Practical Source: Effect of Connecting a Load
For the car battery example:
VL = 12 – 0.01 IL
This line represents
all possible RL
Practical Voltage Source
s
sLsc
soc
R
vi
vv
=
=
LssL iRvv −=
OR
s
L
s
sL
R
v
R
vi −=
Engr228 - Chapter 5, Hayt 8E 11
Practical Current Source
sLsc
spLoc
ii
iRv
=
=
p
LsL
R
vii −=
Equivalent Practical Sources
p
LsL
R
vii −=
s
L
s
sL
R
v
R
vi −=
• These two practical sources are equivalent if
RS = RP S
SS
R
vi =
Engr228 - Chapter 5, Hayt 8E 12
Source Transformation and Equivalent Sources
The sources are equivalent if
Rs=Rp and vs=isRs
Source Transformation
• The circuits (a) and (b) are
equivalent at the terminals.
• If given circuit (a), but circuit (b) is
more convenient, switch them.
• This process is called source
transformation.
Engr228 - Chapter 5, Hayt 8E 13
Example: Source Transformation
We can find the current I in the circuit below using a source
transformation, as shown.
I = (45-3)/(5+4.7+3) = 3.307 mA
I = 3.307 mA
Example
Use source transformations to find Ix
Engr228 - Chapter 5, Hayt 8E 14
Example - continued
Example - continued
Engr228 - Chapter 5, Hayt 8E 15
AI X3
1
6
2
321
2==
++=
Example - continued
Textbook Problem 5.24 Hayt 7E
• Using source transformations, determine the power
dissipated by the 5.8 kΩ resistor.
42.97 mW
Engr228 - Chapter 5, Hayt 8E 16
Textbook Problem 5.6 Hayt 8E
(a) Determine the individual contributions of each of the two
current sources to the nodal voltage v1
(b) Determine the power dissipated by the 2Ω resistor