AC CalculationsW12-D2

What’s Happening? I gave a test with a problem you should have been

able to think your way through. After the exam I wanted to explain AC circuits to you

but you just kept talking among yourselves. You lost about 20-30 minutes of explanation.

Today I will cover the material in a very cursory way. You can learn it from the book since listening didn’t seem to occur.

To be sure you DO read the text, there is a WebAssign that will be due next week. Do the problems.

Today we will have a brief discussion of AC circuits and how they work. I will assume you read the beginning of the chapter.

What’s Happening??

Exam III – Done AC Discussion – Probably NO experiment

You will probably do the experiment on Friday. What the heck is RMS??

WebAssign Read the first three sections of chapter 23 and do the

problems assigned. They will be due on Wednesday.

I am not sure when I will grade the papers. Probably over the weekend unless something else pops up.

Attitude There is a population in the class who doesn’t

like physics and seems to blame me. Not liking physics is NOT an excuse to do poorly but this group seems to be doing exactly this.

There is another group in this class that works hard and tries to understand the material. They are doing fairly well.

The first group tends to be hostile, the other wants to learn and make the best of the situation.

Last semester nobody failed this course. This is NOT true for this class.

Learning I do not teach you. You learn by

Doing the Units to the best of your ability. Clear and not sloppy

Not Copying from your neighbors but written AFTER and independently of your team’s completion of the work.

Show some individual thinking. Think your way through problems rather than complain

about them. This is college and not High School

You are adults! Act that way.

I know that you think I am a poor instructor. This may be true but learning is totally YOUR responsibility. In this format, it is your responsibility learn.

Think about it …. it isn’t too late do well.

Read that stuff and learn it … it WILL be on the final examination.

Monday Wednesday

Friday Saturday

November

15 – Short Review &Unit 15 AC

17 – EXAM #3Unit 15 AC

19Unit 15 AC

22 Unit 16-Optics

24 QUIZUnit 16-Optics

26 Holiday

December

29Unit 17-18 Optics

1Unit 18 Optics

3 QUIZUnit 18 Optics

6 LAST DAYEXAM #4

11 FINAL

Approximate Calendar

2

The Sine

Average Value of Sine = <sine>=0

POWER = I2R so average I2 is important

Average value of Sin22 2

2 2

2

2

2 2 20

22 00

sin cos 1

sin cos 1

1sin

21

(sin ) sin 0.7072

Power=I

0.7072

rms

R I RSin t

I RPower I R

Your outlet delivers 120 Volts rms,@ 60 Hz.,. What is the peak voltage?

A. 120 VB. 240VC. 170VD. 340V 1

2

1.41 120 170

2 376sec

170sin(376 )

peak

peak

VVrms

V V

f

V t

Amplitude

Sine & Cosine

LAST TIME

f

tt

tt

2

)sin()2

cos(

)cos()2

sin(

Draw these graphs and convince yourself that this makes sense!

I was going to discuss this mayerial.

AC Applied voltages

This graph correspondsto an applied voltageof V cos(t).

Because the currentand the voltage aretogether (in-phase) thismust apply to a Resistorfor which Ohm saidthat I~V.

the ac phasor )cos( tIi

The resistor again

)cos( tIRiRv

Phasor diagram- Resistor IRVR

Length =Max CurrentLength=Maximum Voltage

here comes trouble ….

We need the relationship between I (the current through)and vL (the voltage across) the inductor.

From the last chapter:

L

iv L

t

* unless you have taken calculus.

check it out---

initialfinal

initialfinal

tt(t)

thingthing(thing)

.differenceor change means

so-

( cos ) (cos( ) cos( ))

cos( ) cos( ) sin( ) sin( ) cos( )

i I t I t t tv L L LL

t t t

t t t t tv LIL

t

Whent gets very small,cos (t) goes to 1.

cancel

t

ttLIvL

)sin()sin(

:left sat what'look sLet'

??

r

1)sin(

lim 0

this leaves )sin( tvL The resistor voltage looked like a cosine so we would like theinductor voltage to look as similar to this as possible. So let’slook at the following :

f(t)=A sin(t)A=Amplitude (=1 here)

f(t)=A sin(t-[/2])A=Amplitude (=1 here)

)cos()2

sin( tt

)cos()2

sin( tt

)sin( tLIvL

)2

cos(

)2

cos()sin(

tLIv

tt

L

result - inductor

I is the MAXIMUM

current in thecircuit.

comparingResistor

inductor

)cos( tIRiRv )2

cos( tLIvL

RIvR max LIv MaxL

L) looks like a resistance

XL=LReactance - OHMS

For the inductor

LLMaxL VIXLIv

FOR THE RESISTOR

RMaxR VIRv

slightly confusing pointWe will always use the CURRENT as the basis for calculations and express voltages with respect to the current.

What that means?

voltage. theandcurrent

ebetween thshift phase theis where

)tVcos(v

as voltage theand

t)cos( Ii

:as varyingascurrent thedescribe We

the phasor )2

cos( tLIvL

)2

cos( tLIvL

t

t

)sin()2

cos( tt

direction

remember for ac series circuits

In the circuit below, R=30 and L= 30 mH. If the angular frequency of the 60 volt AC source is is 3 K-Hz

WHAT WE WANT TO DO:

(a)calculate the maximum current in the circuit(b)calculate the voltage across the inductor(c)Does Kirchoff’s Law Work?

E=60VR=30

L= 30 mH

=3 KHZ

E=60VR=30

L= 30 mH

=3 KHZ

t

I

R=30XL=L=90

IRVR LL IXV

The instantaneous voltage acrosseach element is the PROJECTIONof the MAXIIMUM voltage ontothe horizontal axis!

t

I

IRVR LL IXV

axVV m

Source voltage leadsthe current by the angle .

22

222222I

Impedance ZIZV

L

L

XRZ

orXIRIZ

Let

)(tan 1

R

L

V

VE=60VR=30

L= 30 mH

=3 KHZ

t

I

IRVR LL IXV

22

222222I

Impedance ZIZV

L

L

XRZ

orXIRIZ

Let

)(tan 1

R

L

V

V

E=60VR=30

L= 30 mH

=3 KHZ

01

22

7130

90tan

9.94)90()30(

90

30

Z

LX

R

L

The drawing is obviouslyNOT to scale.

AZ

VI 632.0

9.94

60

rad

Z

LX

R

L

25.17130

90tan

9.94)90()30(

90

30

01

22

AZ

VI 632.0

9.94

60

source. AC thelikelook should

that voltages two theseof sum at thelook sLet'

)cos(9.18)cos(3063.0

)sin(567)sin(03.03000063.0)sin(

ttxxv

ttxxtLIv

R

L

source. AC thelikelook should

that voltages two theseof sum at thelook sLet'

)cos(9.18)cos(3063.0

)sin(7.56)sin(03.0300063.0)sin(

ttxxv

ttxxtLIv

R

L

wt 56.7Sin(wt) 18.9 cos(wt) SUM0 0 18.9 18.9

0.211.2645510

618.523258327.258707

0.422.0800200

117.40805279 -4.67197

0.632.0152282

415.59884312 -16.4164

0.840.6740903

513.16775681 -27.5063

147.7114048

410.21171358 -37.4997

1.252.8466161

7 6.84856156 -45.9981

1.455.8749996

93.212379001 -52.6626

It Does!

2

For an RLC circuit we so the same thing. We will do this on Friday.

Enough!