1.5 AC Circuit Analysis Example

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Do curso por Georgia Institute of Technology

Linear Circuits 2: AC Analysis

38 ratings

Georgia Institute of Technology

Linear Circuits 2: AC Analysis

38 ratings

This course explains how to analyze circuits that have alternating current (AC) voltage or current sources. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Na lição

Module 1: AC Circuit Analysis

This module describes how to analyze circuits with sinusoidal inputs

1.0 Introduction to Linear Circuits 2: AC Analysis3:11

1.1 AC Circuits7:50

1.2 Phasors11:06

1.3 Impedance5:29

1.4 Circuit Analysis with AC Impedances10:16

1.5 AC Circuit Analysis Example7:35

1.6 Transfer Functions13:28

1.7 Transfer Function Example9:04

1.8 Module Summary3:50

Conheça os instrutores

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering

[MUSIC]

Welcome back to linear circuits, this is Dr. Ferri.

This lesson is on AC circuit analysis example so

we're just going to demonstrate the impedance method.

And I've been promising all along in this particular module that we

would be able to analyze circuit like this so now is our chance.

Building upon the impedance method, remember, it's a three step process.

First, we redraw the circuit replacing

all the components with their impedances to solve the circuit using this standard

methods that we did with resistance circuits.

And then finally to convert the output phasor back to the sinusoidal form.

Now let's take a look at this circuit with these particular numbers in there.

Step one of our process is to redraw the circuit.

And I'm going to call it Z sub C for now and Z sub L.

Just because it's easier to work with those than to put in the actual numbers.

And my phasor for the input is 4 at an angle of 10 degrees.

And we can define Z sub L is equal to j omega L.

In this particular case, this is omega right here.

And if I substitute in that value for omega and

this value for L, I will get 103.7 j.

And similarly, I can solve for

Z sub C, that's 1 over j omega C.

And with this value of omega,

that value of C, I get -318.3 j.

Okay, so that's step one.

Step two is to solve this using standard methods.

Now, I'm interested, in particular, this voltage right there.

So, if I want to do this voltage, then I can look at it in terms of trying

to do a voltage divider law here, but I have to first combine these two resistors.

So let's let Z sub 1 equal that combination.

Z sub L in parallel with R and

that would be Z sub L product over the sum.

When I plug in those numbers, I'm going to get 42.4 + 81.7 j.

Now most scientific calculators, a lot of scientific calculators these days

handle complex numbers and handle complex arithmetic.

So I'm assuming that you can go ahead and

be able to do this either on your calculator or on an Internet site.

Then if I want to solve for V0 then, I use the voltage divider law, which is Z sub 1,

which is this combination right here, or this right here.

Over the total which is R + Z sub C + Z sub 1 times VS,

which is 4 at an angle of 10 degrees.

When I plug in all my numbers into here,

I get 0.2718 at an angle of 106.9 and

all that gets multiplied by 4 at an angle of 10.

So then I just multiply my magnitudes and add my angles.

And I get that V out in phasor form

is 1.08 at an angle of 116.9.

And the last step was to convert this back to the time domain.

So V out(t) = 1.08, cosine,

the same omega as I had before, 2 pi times 5000 t,

and now the angle's 116.9, and

that's my final solution.

I went ahead and built this circuit, and I wanted to measure it.

In this particular case, I have a V in that looks like this,

and notice right here that the amplitude

is supposed to be 4, so this is 4 divisions right here, that's how it's set.

And the peak to peak is right here, measured at 8.

This is peak to peak is 2 times the amplitude.

And then my V out is right here.

In this particular case, we had solved it to be 1.08.

You can see it's just above one division right here, so that's 1.08 and

it's measured the 2 times this output amplitude is 2.16.

So the measurement right here is exactly what it's supposed to be.

Now let's check our angle right here.

Now right here, the angle is supposed to be 10 at of the original signal,

and 117 of the output signal.

The difference between that, between V in and V out,

Is 106, or 107 rather.

107 degrees between this one and this one.

Because the reason is if I take this at a reference point, there's

no time equals 0 here so it's hard for me to measure 10 degrees on the V input.

It's easier for me to measure the difference between these.

Now, notice that it's 107 degrees and this is positive.

This is a positive angle compared to this one.

I mean this is shifted to the left, V out is shifted to the left by 107 degrees.

So, let's take a look at this.

This is the original signal right here, the input.

This is 360 degrees, that means this is 180 degrees.

So that's 360.

That's 180.

This right here is 90, a 90 degree shift to the left.

And this is a little bit more than a 90, so this is about 107.

So in other words, the measurements here match what I

actually got when I calculated it analytically.

So to summarize here, in this example, we went through the impedance method and

demonstrated the first step, of course,

was to redraw the circuit with its impedance equivalent.

Treat the impedances as if they were resistors and

use standard circuit analysis methods to solve it.

And finally, to convert the output phasor to sinusoidal form.

And what we found is that when we actually did experiments to build that circuit and

test it, that we found that our tested,

our measured values matched those that we found analytically.

All right, thank you

[MUSIC]

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