Chapter 22 – Alternating Current Part 1 2/18/2019 PHY-2054 J. B. Bindell Chapter 22 – Alternating Current Part 1

Upcoming items for your consideration No problem session on Monday (Not much to do!) Watch for a Mastering Physics Assignment Quiz next Friday on AC … will it ever end? Today we start AC Circuits We will review the exam when they are returned.

3 (similar or exactly from past.) Exam STructure Question Number Section 003 Section 004 1 Multiple Choice (5) 2 (seen before) 2-Wire Problem – Different currents in each class and different from the quiz 3 (similar or exactly from past.) Inverted coil problem. Coil problem from class. 4 (New) Charged particle orbit. Electron & proton in same orbit.

We will retain this schedule. PHY2054 Problem Solving/Office Hours Schedule Room MAP-318 Monday Tuesday Wednesday Thursday Friday Bindell 8:30-9:15AM   11:00-12:00PM 10:30 - 11:15 AM* 10:30-11:30AM Dubey 12:00-1:00PM 1:30-2:45PM These sessions will be used both for office hours and problem solving. Students from any section of 2054 are invited to stop by for assistance in course materials (problems, etc.) Note: There will be times when the room may not be available. In that case we will use our individual offices. * In Office Dr. Dubey's hours are for problem solving only. If something else is going on in the room, come to my office! If I am not there …. come to my office.

OK … How was the test? Easy OK Difficult Impossible

How did you do? great less than stellar ok poor bombed

This time I did better than last time about the same as last time worse than last time

ac generator

“Output” from the previous diagram

Nuclear

DC /AC

home generators FUEL

what works on ac? EVERYTHING!

AC -All is “in-phase”

But not always! (capacitor)

Let’s talk about phase y=f(x)=x2

y=f(x-2)=(x-2)2 y x2 (x-2)2 2 x

the “rule” f(x-b) shift a distance b in the POSITIVE direction f(x+b) shift a distance n in the NEGATIVE direction. The signs switch!

The Sine

Let’s talk about PHASE f(t)=A sin(wt) A=Amplitude (=1 here) f(t)=A sin(wt-[p/2]) A=Amplitude (=1 here)

For the future

AC Applied voltages This graph corresponds to an applied voltage of V cos(wt). Because the current and the voltage are together (in-phase) this must apply to a Resistor for which Ohmmmm said that I~V.

phasor

oops – the ac phaser

the resistor

Phasor diagram Pretty Simple, Huh??

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

From the last chapter: HUH??* * unless you have taken calculus.

check it out---

so- cancel When Dt gets very small, cos (wDt) goes to 1. ??

this leaves The resistor voltage looked like a cosine so we would like the inductor voltage to look as similar to this as possible. So let’s look at the following graph again (~10 slides back): f(t)=A sin(wt) A=Amplitude (=1 here) f(t)=A sin(wt-[p/2]) A=Amplitude (=1 here)

result

(wL) looks like a resistance Resistor inductor (wL) looks like a resistance XL=wL Reactance - OHMS comparing

slightly confusing point We will use the CURRENT as the basis for calculations and express voltages with respect to the current. What that means?

back to the phasor thing

What about the capacitor?? Without repeating what we did, the question is what function will have a Df/Dt = cosine? Obviously, the sine! So, using the same process that we used for the inductor,

capacitor phasor diagram

NOTICE THAT The voltage lags the current by 90 deg I and V are represented on the same graph but are different quantities.

SUMMARY