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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
simplify a circuit;
•  Understand the concept of Thevenin and Norton equivalent
circuits and be able to derive 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.
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.
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 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
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
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.
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.
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
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
v1 − v2 v1 − 0
15 =
+
1
2
v1 − v2
v −0
+ 3i1 = 2
1
3
v1 − 0
i1 =
2
Answer: 4.5 kW being generated
Example #2
•  How many nodes are in this circuit?
•  How many nodal equations must you write to solve for the
unknown voltages?
4Ω
-3A
V2
3Ω
V1
2Ω
1Ω
-8A
0V
V3
5Ω
-25A
Example #2 – node V1
4Ω
-3A
3Ω
V1
V2
1Ω
-8A
2Ω
V3
5Ω
-25A
0V
At node V1
V1 − V 3
V1 −V 2
8+
+3+
=0
4
3
96 + 3V 1 − 3V 3 + 36 + 4V 1 − 4V 2 = 0
7V 1 − 4V 2 − 3V 3 = −132
Example #2 – node V2
4Ω
-3A
V1
3Ω
V2
1Ω
-8A
2Ω
V3
5Ω
-25A
0V
At node V2
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
Example #2 – node V3
4Ω
-3A
V1
3Ω
V2
1Ω
-8A
2Ω
V3
5Ω
-25A
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:
V1 = 0.956V
V2 = 10.576V
V3 = 32.132V
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 create a supernode.
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
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
Supernode Example
4Ω
-3A
V1
3Ω
V2
1Ω
-8A
0V
V3
1V
5Ω
-25A
Supernode Example – Node V1
4Ω
-3A
3Ω
V1
V2
1Ω
-8A
V3
1V
5Ω
-25A
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
7V 1 − 4V 2 − 3V 3 = −132
8+
Supernode Example – Nodes V2 and V3
4Ω
-3A
3Ω
V1
supernode
V2
1Ω
-8A
V3
1V
5Ω
-25A
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
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
Simple resistive circuit showing three paths, which represent
three mesh currents. Note that IR3 = I1 – I2
The Mesh Analysis Method
Mesh currents
Branch currents
Mesh: Apply KVL
Apply KVL to mesh 1
( Σ voltage drops = 0 ):
-42 + 6i1 +3(i1-i2) = 0
i1 = 6A
i2 = 4A
Apply KVL to mesh 2
( Σ voltage drops = 0 ):
3(i2-i1) + 4i2 -10 = 0
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
Example
Use mesh analysis to determine Vx
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
6V
2Ω
I3
1Ω
Example - continued
1Ω
7V
I2
2Ω
− 7 + 1( I1 − I 2) + 6 + 2( I1 − I 3) = 0
3I 1 − I 2 − 2 I 3 = 1
Equation I
+ Vx -
I1
3Ω
6V
2Ω
I3
1Ω
1( I 2 − I1) + 2 I 2 + 3( I 2 − I 3) = 0
− 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
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.
Supermesh Example
Use mesh analysis to evaluate Vx
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
2Ω
I3
1Ω
Supermesh Example - continued
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
2Ω
I3
1Ω
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
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
2Ω
I1 = 9A
I2 = 2.5A
I3 = 2A
Vx = 3(I3-I2) = -1.5V
I3
1Ω
−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
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.
Dependent Source Example
Find i1
Answer: i1 = - 250 mA.
Dependent Source Example
Find Vx
1Ω
15A
I2
2Ω
+ Vx -
I1
3Ω
1/9 Vx
I3
2Ω
1Ω
Dependent Source Example - continued
I1 = 15 A Equation I
1Ω
15A
I1
I2
2Ω
+ Vx -
− I1 + 6 I 2 − 3I 3 = 0 Equation II
3Ω
1/9 Vx
I3
2Ω
1( I 2 − I1) + 2 I 2 + 3( I 2 − I 3) = 0
1Ω
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
Textbook Problem 4.56 Nilsson 10th
Find the power absorbed by the 20V source in the circuit below.
Power20V = 480 mW absorbed
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.
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
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:
3
iʹ′x =
= 0.2
6+9
Superposition Example (Part 3 of 4)
Then, turn the voltage source off:
6
ʹ′
ix =
(2) = 0.8
6+9
Superposition Example (Part 4 of 4)
Finally, combine the results:
ix = ixʹ′ + ixʹ′ = 0.2 + 0.8 =1.0
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
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
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 5.19 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:
Thévenin Equivalent using Source Transformations
•  We can repeatedly apply
source transformations
on network A to find its
Thévenin equivalent
circuit.
•  This method has
limitations – due to
circuit topology, 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.
–  Set voltage sources to zero volts → short circuit
–  Set current sources to zero amps → open circuit
Then:
VTH = voc and
RTH = Req
Thévenin Example
Example
Find Thévenin’s equivalent circuit and the current
passing thru RL given that RL = 1Ω
2Ω
10V
10Ω
3Ω
RL
2Ω
Example - continued
Find VTH
10V
6V
6V
2Ω
10Ω
3Ω
10V
2Ω
0V
0V
VTH
3
=
×10 = 6V
2+3
0V
Example - continued
2Ω
10Ω
Find RTH
3Ω
10V
2Ω
Short voltage source
2Ω
10Ω
3Ω
RTH = 10 + 2 || 3 + 2
2×3
= 10 +
+2
2+3
= 13.2Ω
RTH
2Ω
Example - continued
Thévenin’s equivalent circuit
13.2Ω
6V
The current thru RL = 1Ω is
RL
6
= 0.423 A
13.2 + 1
Example: Bridge Circuit
Find Thévenin’s equivalent circuit as seen by RL
R1=2K
R3=4K
RL=1K
10V
+
R2=8K
R4=1K
Example - continued
Find VTH
10V
R1=2K
10V
R3=4K
8V
2V
R4=1K
R2=8K
0V
VTH = 8-2 = 6V
Example - continued
Find RTH
R1=2K
R3=4K
RTH
R2=8K
R1=2K
R3=4K
R2=8K
R4=1K
R4=1K
R1=2K
R3=4K
R2=8K
R4=1K
Example - continued
R1=2K
R3=4K
R2=8K
R4=1K
RTH = 2 K || 8 K + 4 K || 1K
= 1.6 K + 0.8 K = 2.4 K
Example - continued
Thévenin’s equivalent circuit
2.4K
6V
RL
Norton Equivalent Circuits
Norton’s theorem: a linear network can be replaced by its Norton
equivalent circuit, as shown below:
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 (same as Thévenin)
–  Set voltage sources to zero volts → short circuit;
–  Set current sources to zero amps → open circuit.
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
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
Thévenin Example: Handling Dependent Sources
The normal technique for finding Thévenin or Norton
equivalent circuits can not usually be used if a dependent
source is present. In this case, we can find both VTH and IN and
solve for RTH=VTH / IN
Thévenin Example: Handling Dependent Sources
Another situation that rarely arises, is if both VTH and IN are zero,
or just IN is zero. In this situation, we can apply a test source to
the output of the network and measure the resulting short-circuit
(IN) current, or open-circuit voltage (VTH). RTH is then calculated
as VTH /IN
Thévenin Example: Handling Dependent Sources
Solve: vtest =0.6 V, so RTH = 0.6 Ω
v test v test − (1.5i)
+
=1
2
3
i = −1
€
Recap: Thévenin and Norton Thévenin’s equivalent circuit
Norton’s equivalent circuit
13.2
RL
6V
RTH = RN
VTH = I N × RTH
0.45
13.2
RL
Same R value
6 = 0.45 ×13.2
Textbook Problem 5.50 Hayt 7E
Find the Thévenin equivalent of the circuit below.
RTH = 8.523 kΩ
VTH = 83.5 V
Maximum Power Transfer
Thévenin’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
Maximum Power Theorem Proof
RTH
RL
VTH
2
VTH
I=
RTH + RL
and
P = I RL
2
2
⎛ VTH ⎞
VTH RL
⎟⎟ ⋅ RL =
Plug it in P = ⎜⎜
2
R
+
R
(
R
+
R
)
L ⎠
TH
L
⎝ TH
2
2
dP ( RTH + RL ) 2 VTH − VTH RL ⋅ 2( RTH + RL )
=
=0
4
dRL
( RTH + RL )
Maximum Power Theorem Proof - continued
2
2
dP ( RTH + RL ) 2 VTH − VTH RL ⋅ 2( RTH + RL )
=
=0
4
dRL
( RTH + RL )
2
2
2
( RTH + RL ) VTH = VTH RL ⋅ 2( RTH + RL )
( RTH + RL ) = 2 RL
RTH = RL
For maximum power transfer
Example
Evaluate RL for maximum power transfer and find the power.
2Ω
10V
10Ω
3Ω
RL
2Ω
Example - continued
Thévenin’s equivalent circuit
13.2
RL
6V
RL should be set to 13.2Ω to get maximum power transfer.
2
Max. power is
2
V
(6 / 2)
=
= 0.68W
R
13.2
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
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
simplify 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.