Chapter 4
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 4 Objectives
• Understand and be able to use the node-voltage method to
solve a circuit.
• Understand and be able to use the mesh-current method to
solve a circuit.
• Be able to determine which technique is best for a particular
circuit.
• Understand source transformations and be able to use them to
solve a circuit.
• Understand the concept of Thevenin and Norton equivalent
circuits and be able to construct one.
• Know the condition for maximum power transfer to a resistive
load and be able to calculate the value of the load resistor that
satisfies this condition.
Engr228 - Chapter 4, Nilsson 9E
1
Circuit Analysis
• As circuits get more complicated, we need an organized
method of applying KVL, KCL, and Ohm’s law.
• Nodal analysis assigns voltages to each node, and then we
apply Kirchhoff's Current Law to solve for the node voltages.
• Mesh analysis assigns currents to each mesh, and then we
apply Kirchhoff’s Voltage Law to solve for the mesh currents.
Engr228 - Chapter 4, Nilsson 9E
2
Review - Nodes, Paths, Loops, Branches
• These two networks are equivalent.
• There are three nodes and five branches:
– Node: a point at which two or more elements have a common connection
point.
– Branch: a single path in a network composed of one simple element and
the node at each end of that element.
• A path is a sequence of nodes.
• A loop is a closed (circular) path.
Review - Kirchhoff’s Current Law
• Kirchhoff’s Current Law (KCL) states that the algebraic sum
of all currents entering a node is zero.
iA + iB + (−iC) + (−iD) = 0
Engr228 - Chapter 4, Nilsson 9E
3
Review - Kirchhoff’s Voltage Law
• Kirchhoff’s Voltage Law (KVL) states that the algebraic sum
of the voltages around any closed path is zero.
-v1 + v2 + -v3 = 0
Node Example
• Node = every point along the same wire
6K
V
10V
4K
3 nodes
Engr228 - Chapter 4, Nilsson 9E
4
Nodes
How many nodes are there in the circuit(s) below?
Notes on Writing Nodal Equations
• All terms in the equations are in units of current.
• Everyone has their own style of writing nodal equations
– The important thing is that you remain consistent.
• Probably the easiest method if you are just getting started is to
remember that:
Current entering a node = current leaving the node
• Current directions can be assigned arbitrarily, unless they are
previously specified.
Engr228 - Chapter 4, Nilsson 9E
5
The Nodal Analysis Method
• Assign voltages to every node relative to a reference node.
Choosing the Reference Node
• By convention, the bottom node is often the reference node.
• If a ground connection is shown, then that becomes the
reference node.
• Otherwise, choose a node with many connections.
• The reference node is most often assigned a value of 0.0 volts.
Engr228 - Chapter 4, Nilsson 9E
6
Apply KCL to Find Voltages
• Assume reference voltage = 0.0 volts
• Assign current names and directions
• Apply KCL to node v1 ( Σ out = Σ in)
• Apply Ohm’s law to each resistor:
v1 v1 − v 2
+
= 3.1
2
5
Apply KCL to Find Voltages
• Apply KCL to node v2 ( Σ out = Σ in)
• Apply Ohm’s law to each resistor:
v1 − v2 v2 − 0
=
+ (−1.4)
5
1
We now have two equations for the two unknowns v1
and v2 and we can solve them simultaneously:
v1 = 5V and v2 = 2V
Engr228 - Chapter 4, Nilsson 9E
7
Example: Nodal Analysis
Find the current i in the circuit below.
Answer: i = 0 (since v1=v2=20 V)
Nodal Analysis: Dependent Source Example
Determine the power supplied
by the dependent source.
Key step: eliminate i1 from the
equations using v1=2i1
15 =
v1 − v2 v1 − 0
+
1
2
v1 − v2
v −0
+ 3i1 = 2
1
3
i1 =
v1 − 0
2
Answer: 4.5 kW being generated
Engr228 - Chapter 4, Nilsson 9E
8
Example #1
V1
V2
0V
V1 = 3
V1−V 2
V2−0
+2=
5
1
13
V2=
6
Example #2
• How many nodes are in this circuit?
• How many nodal equations must you write to solve for the
unknown voltages?
V2
V1
V3
0V
Engr228 - Chapter 4, Nilsson 9E
9
Example #2 – node V1
V2
V1
V3
0V
V1−V 3
V1−V 2
+3+
=0
4
3
96 + 3V 1 − 3V 3 + 36 + 4V 1 − 4V 2 = 0
8+
At node V1
7V 1 − 4V 2 − 3V 3 = −132
Example #2 – node V2
V1
V2
V3
0V
At node V2
Engr228 - Chapter 4, Nilsson 9E
V 2 − V1
V 2 −V 3 V 2 − 0
−3+
+
=0
3
2
1
2V 2 − 2V 1 − 18 + 3V 2 − 3V 3 + 6V 2 = 0
− 2V 1 + 11V 2 − 3V 3 = 18
10
Example #2 – node V3
V2
V1
V3
0V
At node V3
V 3 − V 2 V 3 − V1
V3−0
+
− 25 +
=0
2
4
5
10V 3 − 10V 2 + 5V 3 − 5V 1 − 500 + 4V 3 = 0
− 5V 1 − 10V 2 + 19V 3 = 500
All 3 Equations
7V 1 − 4V 2 − 3V 3 = −132
− 2V 1 + 11V 2 − 3V 3 = 18
− 5V 1 − 10V 2 + 19V 3 = 500
Answer:
Engr228 - Chapter 4, Nilsson 9E
V1 = 0.956V
V2 = 10.576V
V3 = 32.132V
11
Voltage Sources and the Supernode
If there is a DC voltage source between two non-reference
nodes, you can get into trouble when trying to use KCL
between the two nodes because the current through the voltage
source may not be known, and an equation cannot be written
for it. Therefore, we use a supernode instead.
The Supernode Analysis Technique
• Apply KCL at Node v1.
• Apply KCL at the supernode.
• Add the equation for the
voltage source inside the
supernode.
v1 = 1.0714V
v2 = 10.5V
v3 = 32.5V
Engr228 - Chapter 4, Nilsson 9E
v1 − v3 v1 − v2
+
= −3 − 8
4
3
v1 − v2 v1 − v3
v v
+
= −3 + 2 + 3 − 25
3
4
1 5
v3 − v2 = 22
12
Modified Example
V2
V1
V3
0V
Modified Example – node V1
V2
V1
V3
0V
At node V1
V1−V 3
V1−V 2
+3+
=0
4
3
96 + 3V 1 − 3V 3 + 36 + 4V 1 − 4V 2 = 0
8+
7V 1 − 4V 2 − 3V 3 = −132
Engr228 - Chapter 4, Nilsson 9E
13
Modified Example – node V2, supernode
supernode
V2
V1
V3
0V
At node V2
V 2 − V1
V 3 − V1
V3−0 V2−0
−3+
− 25 +
+
=0
3
4
5
1
20V 2 − 20V 1 − 180 + 15V 3 − 15V 1 − 1500 + 12V 3 + 60V 2 = 0
− 35V 1 + 80V 2 + 27V 3 = 1680
Supernode
V 2 −V 3 = 1
Solving Simultaneous Equations
7V 1 − 4V 2 − 3V 3 = −132
− 35V 1 + 80V 2 + 27V 3 = 1680
V 2 −V 3 = 1
V1 = -4.952 V
V2 = 14.333 V
V3 = 13.333 V
Engr228 - Chapter 4, Nilsson 9E
14
Mesh Analysis: Nodal Alternative
•
•
•
•
A mesh is a loop that does not contain any other loops within it.
In mesh analysis, we assign mesh currents and solve using KVL.
All terms in the equations are in units of voltage.
Remember – voltage drops in the direction of current flow except
for sources that are generating power.
• The circuit below has four meshes:
Mesh Example
Engr228 - Chapter 4, Nilsson 9E
15
The Mesh Analysis Method
Mesh currents
Branch currents
Mesh: Apply KVL
Apply KVL to mesh 1
( Σ drops = 0 ):
-42 + 6i1 +3(i1-i2) = 0
Apply KVL to mesh 2
( Σ drops = 0 ):
3(i2-i1) + 4i2 -10 = 0
i1 = 6A
i2 = 4A
Engr228 - Chapter 4, Nilsson 9E
16
Example: Mesh Analysis
Determine the power supplied by the 2 V source.
Applying KVL to the meshes:
−5 + 4i1 + 2(i1 − i2) − 2 = 0
+2 + 2(i2 − i1) + 5i2 + 1 = 0
i1 = 1.132 A i2 = −0.1053 A
Answer: -2.474 W (the source is generating power)
A Three Mesh Example
Follow each
mesh clockwise
Simplify
Solve the equations:
i1 = 3 A, i2 = 2 A, and i3 = 3 A
Engr228 - Chapter 4, Nilsson 9E
17
Example
• Use mesh analysis to determine Vx
I2
I1
I3
Example - continued
− 7 + 1( I 1 − I 2) + 6 + 2( I 1 − I 3) = 0
I2
I1
3I1 − I 2 − 2 I 3 = 1
Equation I
1( I 2 − I 1) + 2 I 2 + 3( I 2 − I 3) = 0
I3
− I1 + 6 I 2 − 3I 3 = 0 Equation II
2( I 3 − I1) − 6 + 3( I 3 − I 2) + I 3 = 0
− 2 I 1 − 3I 2 + 6 I 3 = 6
Equation III
I1 = 3A, I2 = 2A, I3 = 3A
Vx = 3(I3-I2) = 3V
Engr228 - Chapter 4, Nilsson 9E
18
Current Sources and the Supermesh
• If a current source is present in the network and shared
between two meshes, then you must use a supermesh formed
from the two meshes that have the shared current source.
We can eliminate the need for
introducing a voltage variable
by applying KVL to the
supermesh formed by joining
meshes 1 and 3.
Supermesh Example
• Use mesh analysis to evaluate Vx
I2
I1
I3
Engr228 - Chapter 4, Nilsson 9E
19
Supermesh Example - continued
I2
I1
I3
Loop 2: 1( I 2 − I 1) + 2 I 2 + 3( I 2 − I 3) = 0
− I 1 + 6 I 2 − 3I 3 = 0
Equation I
Supermesh Example - continued
Supermesh
I2
I1
I3
I1 = 9A
I2 = 2.5A
I3 = 2A
Vx = 3(I3-I2) = -1.5V
Engr228 - Chapter 4, Nilsson 9E
−7 + 1( I1 − I 2) + 3( I 3 − I 2) + I 3 = 0
I1 − 4 I 2 + 4 I 3 = 7
Equation II
I1 − I 3 = 7
Equation III
20
Node or Mesh: How to Choose?
• Use the one with fewer equations, or
• Use the method you like best, or
• Use both (as a check).
Example
Does nodal analysis or loop analysis have fewer equations?
7V
V1
V2
V3
0V
Engr228 - Chapter 4, Nilsson 9E
21
Dependent Source Example
Find i1
Answer: i1 = - 250 mA.
Dependent Source Example
Find Vx
I2
I1
I3
Engr228 - Chapter 4, Nilsson 9E
22
Dependent Source Example - continued
I1 = 15 A Equation I
I2
1( I 2 − I 1) + 2 I 2 + 3( I 2 − I 3) = 0
− I1 + 6 I 2 − 3I 3 = 0 Equation II
I1
I3
1
I 3 − I1 = Vx Equation III
9
Vx = 3( I 3 − I 2) Equation IV
I1=15A, I2=11A, I3=17A
Vx = 3(17-11) = 18V
Textbook Problem 4.52 Hayt 7E
• Obtain a value for the current labeled i10 in the circuit below
I10 = -4mA
Engr228 - Chapter 4, Nilsson 9E
23
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 4, Nilsson 9E
24
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 4, Nilsson 9E
25
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 =
Engr228 - Chapter 4, Nilsson 9E
3
= 0.2
6+9
26
Superposition Example (Part 3 of 4)
Then, turn the voltage source off:
ix′ =
6
(2) = 0.8
6+9
Superposition Example (Part 4 of 4)
Finally, combine the results:
ix = ix′ + ix′ = 0.2 + 0.8 = 1.0
Engr228 - Chapter 4, Nilsson 9E
27
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 4, Nilsson 9E
28
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 4, Nilsson 9E
29
Example - Continued
(3 || 4)
(12 / 7)
× 42 =
× 42
6 + (3 || 4)
6 + (12 / 7)
Vx( 42V ) =
= 9.333V
Example - Continued
Vx(10V ) = −
(6 || 3)
2
×10 = −
×10
(6 || 3) + 4
2+4
= −3.333V
Engr228 - Chapter 4, Nilsson 9E
30
Example - Continued
Vx = Vx( 42V ) + Vx(10V )
= 9.333 − 3.333 = 6V
Source Transformation and Equivalent Sources
The sources are equivalent if
Rs=Rp and vs=isRs
Engr228 - Chapter 4, Nilsson 9E
31
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.
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
Engr228 - Chapter 4, Nilsson 9E
32
Example
Use source transformations to find Ix
Example - continued
Engr228 - Chapter 4, Nilsson 9E
33
Example - continued
Example - continued
IX =
Engr228 - Chapter 4, Nilsson 9E
2
2 1
= = A
1+ 2 + 3 6 3
34
Textbook Problem 5.24 Hayt 7E
• Using source transformations, determine the power
dissipated by the 5.8 kΩ resistor.
42.97 mW
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
v17A = 6.462V, v14A = -2.154V, v1tot = 4.31V, P2Ω = 3.41W
Engr228 - Chapter 4, Nilsson 9E
35
Textbook Problem 5.17 Hayt 8E
Determine the current labeled i after first transforming the
circuit such that it contains only resistors and voltage
sources.
i = -577µA
Textbook Problem 5.18 Hayt 8E
Using source transformations, reduce the circuit to a voltage
source in series with a resistor, both of which are in series
with the 6 MΩ resistor.
Vs = -6.284V Rs = 825.4 kΩ
Engr228 - Chapter 4, Nilsson 9E
36
Textbook Problem 5.19 Hayt 8E
Find the power generated by the 7V source.
P7v = 17.27W (generating)
Thévenin Equivalent Circuits
Thévenin’s theorem: a linear network can be replaced by its
Thévenin equivalent circuit, as shown below:
Engr228 - Chapter 4, Nilsson 9E
37
Thévenin Equivalent using Source Transformations
• We can repeatedly
apply source
transformations on
network A to find its
Thévenin equivalent
circuit.
• This method has
limitations - not all
circuits can be source
transformed.
Finding the Thévenin Equivalent
• Disconnect the load.
• Find the open circuit voltage voc.
• Find the equivalent resistance Req of the network with all
independent sources turned off.
Then:
VTH = voc and
RTH = Req
Engr228 - Chapter 4, Nilsson 9E
38
Thévenin Example
Example
Find Thevenin’s equivalent circuit and the current
passing thru RL given that RL = 1Ω
Engr228 - Chapter 4, Nilsson 9E
39
Example - continued
Find VTH
10V
6V
6V
0V
0V
0V
VTH =
3
×10 = 6V
2+3
Example - continued
Find RTH
RTH = 10 + 2 || 3 + 2
Short voltage source
2×3
+2
2+3
= 13.2Ω
= 10 +
RTH
Engr228 - Chapter 4, Nilsson 9E
40
Example - continued
Thevenin’s equivalent circuit
The current thru RL = 1Ω is
6
= 0.423 A
13.2 + 1
Example
Find Thevenin’s equivalent circuit
Engr228 - Chapter 4, Nilsson 9E
41
Example - continued
Find VTH
5V
3V
3V
0V
0V
0V
VTH = 1× 3 = 3V
Example - continued
Find RTH
Open circuit
current source
RTH = 10 + 3 + 2
= 15Ω
RTH
Engr228 - Chapter 4, Nilsson 9E
42
Example - continued
Thevenin’s equivalent circuit
Example: Bridge Circuit
Find Thevenin’s equivalent circuit as seen by RL
Engr228 - Chapter 4, Nilsson 9E
43
Example - continued
Find VTH
10V
8V
2V
0V
VTH = 8-2 = 6V
Example - continued
Find RTH
RTH
Engr228 - Chapter 4, Nilsson 9E
44
Example - continued
RTH = 2 K || 8 K + 4 K || 1K
= 1.6 K + 0.8 K = 2.4 K
Example - continued
Thevenin’s equivalent circuit
Engr228 - Chapter 4, Nilsson 9E
45
Thevenin’s Equivalent Circuit - Recap
Norton Equivalent Circuits
Norton’s theorem: a linear network can be replaced by its Norton
equivalent circuit, as shown below:
Engr228 - Chapter 4, Nilsson 9E
46
Finding the Norton Equivalent
• Replace the load with a short circuit.
• Find the short circuit current isc.
• Find the equivalent resistance Req of the network with all
independent sources turned off.
Then:
IN = isc and RN = Req
Source Transformation: Norton and Thévenin
• The Thévenin and Norton equivalents are source
transformations of each other.
RTH=RN =Req and vTH=iNReq
Engr228 - Chapter 4, Nilsson 9E
47
Example: Norton and Thévenin
Find the Thévenin and Norton equivalents for the network faced
by the 1-kΩ resistor.
The load
resistor
This is the circuit we will simplify
Example: Norton and Thévenin
Norton
Thévenin
Source
Transformation
Engr228 - Chapter 4, Nilsson 9E
48
Thévenin Example: Handling Dependent Sources
One method to find the Thévenin equivalent of a circuit with a
dependent source: find VTH and IN and solve for RTH=VTH / IN
Thévenin Example: Handling Dependent Sources
Finding the ratio VTH / IN fails when both quantities are zero!
Solution: apply a test source.
Engr228 - Chapter 4, Nilsson 9E
49
Thévenin Example: Handling Dependent Sources
Solve: vtest =0.6 V, and so RTH = 0.6 Ω
v test v test − (1.5i)
+
=1
2
3
i = −1
Example
• Find Norton’s equivalent circuit and the current
through RL if RL = 1Ω
Engr228 - Chapter 4, Nilsson 9E
50
Example - continued
Find IN
Find R total:
Find I total:
Current divider:
2 + 3 || (10 + 2) = 2 +
I=
3 × 12
= 4.4Ω
3 + 12
V 10
=
= 2.27 A
R 4.4
I SC =
3
× 2.27 = 0.45 A
3 + 12
Example - continued
Find Rn
RTH = 10 + 2 || 3 + 2
2×3
+2
2+3
= 13.2Ω
= 10 +
Engr228 - Chapter 4, Nilsson 9E
51
Norton’s Equivalent Circuit
The current thru RL = 1Ω is
13.2
× 0.45 = 0.418 A
13.2 + 1
Relationship Between Thevenin and Norton
RTH = RN
VTH = I N × RTH
I
ISC
Slope = - 1/RTH
VOC
Engr228 - Chapter 4, Nilsson 9E
V
52
Recap: Thevenin and Norton
Thevenin’s equivalent circuit
Norton’s equivalent circuit
Same R value
RTH = RN
6 = 0.45 ×13.2
VTH = I N × RTH
Textbook Problem 5.50 Hayt 7E
Find the Thevenin equivalent of the circuit below.
RTH = 8.523 kΩ
VTH = 83.5 V
Engr228 - Chapter 4, Nilsson 9E
53
Equivalent Circuits with Dependent Sources
We cannot find RTH in circuits with dependent sources using
the total resistance method.
But we can use
RTH =
VOC
I SC
Example
Find Thevenin and Norton’s equivalent circuits
Engr228 - Chapter 4, Nilsson 9E
54
Example - continued
Find Voc
I2
I1
KVL
loop1
− 1 + 250 I1 + 4000( I1 − I 2) = 0
4250 I1 − 4000 I 2 = 1
KVL
loop2
4000( I 2 − I 1) + 2000 I 2 + 80 I 2 − 100Vx = 0
Vx = 4000( I 1 − I 2)
− 404000I 1 + 406080 I 2 = 0
Example - continued
I2
I1
Solve equations
I1 = 3.697mA
I2 = 3.678mA
VOC = 80 I 2 − 100Vx = 80 I 2 − 400000( I 1 − I 2)
= 80(3.678mA) − 400000(3.697 − 3.678)
= −7.3V
Engr228 - Chapter 4, Nilsson 9E
55
Example - continued
Find Isc
I2
I1
KVL
loop1
KVL
loop2
I3
− 1 + 250 I 1 + 4000 ( I 1 − I 2 ) = 0
4250 I 1 − 4000 I 2 = 1
4000 ( I 2 − I 1) + 2000 I 2 + 80 ( I 2 − I 3) − 100Vx = 0
Vx = 4000 ( I 1 − I 2 )
− 404000 I 1 + 406080 I 2 − 80 I 3 = 0
KVL
loop3
80 ( I 3 − I 2) + 100Vx = 0
400000 I 1 − 400080 I 2 + 80 I 3 = 0
Example - continued
Find Isc
I1
I2
I3
I1 = 0.632mA
I2 = 0.421mA
I3 = -1.052 A
Isc = I3 = -1.052 A
Engr228 - Chapter 4, Nilsson 9E
56
Example - continued
RTH =
VOC
− 7.28
=
= 6.94Ω
I SC − 1.052
Thevenin’s equivalent circuit
Norton’s equivalent circuit
Maximum Power Transfer
Thevenin’s or Norton’s equivalent circuit, which has an RTH
connected to it, delivers a maximum power to the load RL for
which RTH = RL
Engr228 - Chapter 4, Nilsson 9E
57
Maximum Power Theorem Proof
P = I 2 RL
I=
and
2
VTH
RTH + RL
2
 VTH 
VTH RL


P
=
⋅
R
=
Plug it in
L
R +R 
( RTH + RL ) 2
L 
 TH
2
2
dP ( RTH + RL ) 2 VTH − VTH RL ⋅ 2( RTH + RL )
=
=0
dRL
( RTH + RL ) 4
Maximum Power Theorem Proof - continued
2
2
dP ( RTH + RL ) 2 VTH − VTH RL ⋅ 2( RTH + RL )
=
=0
dRL
( RTH + RL ) 4
2
2
( RTH + RL ) 2VTH = VTH RL ⋅ 2( RTH + RL )
( RTH + RL ) = 2 RL
RTH = RL
For maximum power transfer
Engr228 - Chapter 4, Nilsson 9E
58
Example
Evaluate RL for maximum power transfer and find the power.
Example - continued
Thevenin’s equivalent circuit
RL should be set to 13.2Ω to get max. power transfer
Max. power is
Engr228 - Chapter 4, Nilsson 9E
V 2 ( 6 / 2) 2
=
= 0.68W
R
13.2
59
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:
Practical Source: Effect of Connecting a Load
For the car battery example:
VL = 12 – 0.01 IL
This line represents
all possible RL
Engr228 - Chapter 4, Nilsson 9E
60
Practical Voltage Source
vL = vs − Rs iL
OR
iL =
vs v L
−
Rs Rs
voc = v s
i Lsc =
vs
Rs
Practical Current Source
iL = is −
vL
Rp
v Loc = R p is
iLsc = is
Engr228 - Chapter 4, Nilsson 9E
61
Equivalent Practical Sources
iL =
vs v L
−
Rs Rs
iL = is −
vL
Rp
• These two practical sources are equivalent if
iS =
vS
RS
RS = RP
Chapter 4 Summary
• Illustrated the node-voltage method to solve a circuit.
• Illustrated the mesh-current method to solve a circuit.
• Practiced choosing which technique is better for a particular
circuit.
• Explained source transformations and how to use them to
solve a circuit.
• Illustrated the techniques of constructing Thevenin and Norton
equivalent circuits.
• Explained the principle of maximum power transfer to a
resistive load and showed how to calculate the value of the
load resistor that satisfies this condition.
Engr228 - Chapter 4, Nilsson 9E
62