EGR 2201 Unit 12Sinusoidal Steady-State Analysis

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: Ohm’s law () 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. Transform the circuit from the time domain to the phasor domain.

2. Solve the problem using circuit techniques (Ohm’s law, Kirchhoff’s laws, voltage-divider rule, etc.)

3. 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. 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.

2. Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express the branch currents in terms of node voltages. Then simplify the equations.

3. Solve the resulting n-1 simultaneous equations to obtain the unknown node voltages.

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: What About Circuits with Voltage Sources?

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 itimmediately 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 sourcenot connected to the reference node is trickier.

To handle it, we treat the voltage source and its twonodes (along withany elements in parallel with the voltage source), as a supernode.

Review from Unit 3: How to Handle a Supernode

We apply KCLand KVL to thesupernode toget two equations.

Example: In the circuit shown,KCL givesi1 + 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 analysisMesh 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. Assign mesh currents i1, i2, …, in to the n

meshes.2. 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.

3. Solve the resulting n simultaneous equations to obtain the unknown mesh currents.

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: What About Circuits with Current Sources?

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 itimmediately 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 sourceshared by twomeshes is trickier.

To handle it, we create asupermesh 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 super-mesh to get twoequations.

Example: In thecircuit shown, KCL gives i2 = i1 +6

And KVL aroundthe 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 analysisSuperposition 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 SuperpositionSource 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 bitof work.

We may get the answer more easily ifwe can first replace the current-source-and-parallel-resistor with a voltage-source-and-series-resistor.

Review from Unit 5: Are We Allowed to Do This?

Yes, we’re allowedto 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:

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 transformationThevenin’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. 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.)

2. VTH is the voltage across the two open terminals. Pay attention to polarity!

3. 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 theoremNorton’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:

Norton-Equivalent CircuitThevenin-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.