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EGR 2201 Unit 12
Sinusoidal Steady-State Analysis
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

Read Alexander & Sadiku, Chapter 10.
Homework #12 and Lab #12 due next
week.
Quiz next week.
Summary of Chapter 9

We saw that we can apply these
familiar techniques to sinusoidal ac
circuits in the phasor domain:
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


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Ohm’s law (𝐕 = 𝐈𝐙)
Kirchhoff’s laws (KVL and KCL)
Series and parallel combinations
Voltage-divider rule
Current-divider rule
In each case, we must use complex
numbers (phasors) instead of real
numbers.
Steps to Analyze AC Circuits
1.
2.
3.
Transform the circuit from the time
domain to the phasor domain.
Solve the problem using circuit
techniques (Ohm’s law, Kirchhoff’s
laws, voltage-divider rule, etc.)
Transform the resulting phasor to
the time domain.
What’s Next?

Now we’ll see that we can also apply
these other familiar techniques in the
phasor domain:
 Nodal analysis





Mesh analysis
Superposition
Source transformation
Thevenin’s theorem
Norton’s theorem
Review from Unit 3: Steps in
Performing Nodal Analysis on a Circuit
with No Voltage Sources

Given a circuit with n nodes, without
voltage sources, follow these steps:
1.
2.
3.
Select a node as the reference node.
Assign voltages v1, v2, …, vn-1 to the
remaining n-1 nodes. These voltages are
relative to the reference node.
Apply KCL to each of the n-1 nonreference nodes. Use Ohm’s law to express
the branch currents in terms of node
voltages. Then simplify the equations.
Solve the resulting n-1 simultaneous
equations to obtain the unknown node
voltages.
Review from Unit 3: What About
Circuits with Voltage Sources?



As described above, our procedure
applies only to circuits without
voltage sources.
But it’s not hard to extend the
procedure to circuits with voltage
sources.
The way you handle a voltage source
depends on whether the source is
connected to the reference node….
Review from Unit 3: Nodal Analysis with A
Voltage Source That Is Connected to the
Reference Node

A voltage source
connected to the
reference node
is easy to handle,
because it
immediately
reveals the voltage
at one of the
non-reference nodes.
 Example:
In the circuit shown, we can
immediately see that v1 = 10 V.
Review from Unit 3: Nodal Analysis with a
Voltage Source That Is Not Connected to the
Reference Node
A voltage source
not connected to
the reference
node is trickier.
 To handle it,
we treat the
voltage source
and its two
nodes (along with
any elements in parallel with the voltage
source), as a supernode.

Review from Unit 3: How to Handle
a Supernode
We apply KCL
and KVL to the
supernode to
get two
equations.
 Example: In
the circuit shown,
KCL gives
i1 + i4 = i2 + i3


And KVL gives
v2 = 5 + v3
Nodal Analysis in the Phasor
Domain


Everything that we just reviewed about
nodal analysis of DC circuits also
applies to sinusoidal AC circuits in the
phasor domain.
The only difference is that we’ll use
complex numbers throughout.
What’s Next?

We can also apply these other
familiar techniques in the phasor
domain:

Nodal analysis

Mesh analysis




Superposition
Source transformation
Thevenin’s theorem
Norton’s theorem
Review from Unit 4: Steps in
Performing Mesh Analysis on a Circuit
with No Current Sources

Given a circuit with n meshes, without
current sources, follow these steps:
1.
2.
3.
Assign mesh currents i1, i2, …, in to the n
meshes.
Apply KVL to each of the n meshes. Use
Ohm’s law to express the voltages in terms
of mesh currents. Then simplify the
equations.
Solve the resulting n simultaneous
equations to obtain the unknown mesh
currents.
Review from Unit 4: What About
Circuits with Current Sources?



As described above, our procedure
applies only to circuits without
current sources.
But it’s not hard to extend the
procedure to circuits with current
sources.
The way you handle a current source
depends on whether the source is
located in only one mesh or is shared
by two meshes….
Review from Unit 4: Mesh Analysis with
a Current Source Located in Only One
Mesh

A current source
located in only
one mesh
is easy to
handle,
because it
immediately
reveals the mesh current in that mesh.
 Example: In the circuit shown, we can
immediately see that i2 = 5 A.
Review from Unit 4: Mesh Analysis with a
Current Source Shared by Two Meshes
A current source
shared by two
meshes is
trickier.
 To handle it,
we create a
supermesh
by excluding the
current source
and any elements
in series with it.

Review from Unit 4: How to Handle
a Supermesh
We apply KCL and
KVL to the supermesh to get two
equations.
 Example: In the
circuit shown, KCL gives
i2 = i1 +6


And KVL around
the supermesh gives
20 = 6i1 + 10i2 + 4i2
Mesh Analysis in the Phasor
Domain


Everything that we just reviewed about
mesh analysis of DC circuits also applies
to sinusoidal AC circuits in the phasor
domain.
The only difference is that we’ll use
complex numbers throughout.
What’s Next?

We can also apply these other
familiar techniques in the phasor
domain:

Nodal analysis
Mesh analysis

Superposition




Source transformation
Thevenin’s theorem
Norton’s theorem
Review from Unit 5:
Superposition Principle (1 of 2)

The superposition principle says that
to find the total effect of two or more
independent sources in a linear
circuit, you can find the effect of
each source acting alone, and then
combine those effects.
Review from Unit 5:
Superposition Principle (2 of 2)

To find the effect of an independent
source acting alone, you must turn
off all of the other independent
sources.
 To turn off a voltage source,
replace it by a short circuit.
 To turn off a current source,
replace it by an open circuit.
Review from Unit 5: Steps in Using
the Superposition Principle
1. Select one independent source and
turn off the others, as explained
above. Find the desired voltage or
current due to this source using
techniques from earlier chapters.
2. Repeat step 1 for each of the other
independent sources.
3. Add the values obtained from the steps
above, paying careful attention to the
signs (+ or ) of each value.
Superposition in the Phasor
Domain


Everything that we just reviewed about
superposition in DC circuits also applies
to sinusoidal AC circuits in the phasor
domain.
The only difference is that we’ll use
complex numbers throughout.
What’s Next?

We can also apply these other
familiar techniques in the phasor
domain:

Nodal analysis
Mesh analysis
Superposition

Source transformation




Thevenin’s theorem
Norton’s theorem
Review from Unit 5: Source
Transformation



Sometimes when
you're analyzing
a circuit, it can
be useful to
substitute a
voltage source (and a resistor in series with
it) by a current source (and a resistor in
parallel with it), or vice versa.
Either substitution is called a source
transformation.
Two questions:
1. Why would this be useful?
2. Are we allowed to do this?
Review from Unit 5: Example of Why
Source Transformation Can Be Useful


Suppose we wish to
find vo in the circuit
to the right.
This is not difficult,
but it requires a bit
of work.
We may get the
answer more easily if
we can first replace
the current-source-andparallel-resistor with
a voltage-source-and-series-resistor.
Review from Unit 5: Are We
Allowed to Do This?
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Yes, we’re allowed
to make this
substitution,
as long as we
use the right
values for the elements.
The resistor R in series with the voltage
source must be the same size as the one in
parallel with the current source.
The current source’s and voltage source’s
values must satisfy the following:
𝑣𝑠
𝑣𝑠 = 𝑖𝑠 𝑅
or
𝑖𝑠 =
𝑅
Source Transformation in the
Phasor Domain


Everything that we just reviewed about
source transformation in DC circuits
also applies to sinusoidal AC circuits in
the phasor domain.
The only difference is that we’ll use
complex numbers throughout, and
instead of having a source resistance R,
we’ll have a source impedance Z.
What’s Next?

We can also apply these other
familiar techniques in the phasor
domain:

Nodal analysis
Mesh analysis
Superposition
Source transformation

Thevenin’s theorem

Norton’s theorem



Review from Unit 6: Thevenin’s
Theorem


Thevenin’s theorem says that any
linear two-terminal circuit can be
replaced by an equivalent circuit
consisting of an independent voltage
source VTh in series with a resistor
RTh.
There’s a standard procedure for
finding the values of VTh and RTh. But
first, what exactly does this mean
and why is it useful?
Review from Unit 6: What Thevenin’s
Theorem Means

Thevenin’s theorem lets us replace
everything on one side of a pair of
terminals by a very simple equivalent
circuit consisting of just a voltage
source and a resistor.
Original Circuit
Equivalent Circuit
Review from Unit 6: Why It’s
Useful

This greatly simplifies computation
when you wish to find values of
voltage or current for several
different possible values of a load
resistance.
Original Circuit
Equivalent Circuit
Review from Unit 6: Steps in
Finding VTH and RTH
1.
2.
3.
Open the circuit at the two
terminals where you wish to
find the Thevenin-equivalent
circuit. (In the circuit shown,
this means removing the load.)
VTH is the voltage across the
two open terminals. Pay
attention to polarity!
RTH is the resistance looking
into the open terminals with all
independent sources turned off.
(Recall that to turn off a
voltage source we replace it by
a short, and to turn off a
current source we replace it by
an open.)
Thevenin’s Theorem in the Phasor
Domain


Everything that we just reviewed about
Thevenin’s theorem in DC circuits also
applies to sinusoidal AC circuits in the
phasor domain.
The only difference is that we’ll use
complex numbers throughout, and
instead of having a Thevenin resistance
RTh we’ll have a Thevenin impedance
ZTh.
What’s Next?

We can also apply these other
familiar techniques in the phasor
domain:

Nodal analysis
Mesh analysis
Superposition
Source transformation
Thevenin’s theorem

Norton’s theorem




Review from Unit 6: Norton’s
Theorem

Norton’s theorem says that any
linear two-terminal circuit can be
replaced by an equivalent circuit
consisting of an independent current
source IN in parallel with a resistor
RN .
Original Circuit
Norton-Equivalent Circuit
Review from Unit 6: Finding IN and
RN
 The book describes how to find IN and RN.
The procedure is similar to how we find VTh
and RTh.
 But you don’t need to learn this new
procedure. Instead, just find VTh and RTh, and
then apply a source transformation:
𝑅N = 𝑅Th
Thevenin-Equivalent Circuit
𝑉Th
and 𝐼N =
𝑅Th
Norton-Equivalent Circuit
Norton’s Theorem in the Phasor
Domain


Everything that we just reviewed about
Norton’s theorem in DC circuits also
applies to sinusoidal AC circuits in the
phasor domain.
The only difference is that we’ll use
complex numbers throughout, and
instead of having a Norton resistance
RN we’ll have a Norton impedance ZN.