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Transition to AC. W11-3. What’s Happening??. Exam III on Wednesday – Material has already been published for your reference. No Quiz on Friday Current topic is AC Circuits - Unit 15 Let’s look at the calendar. Approximate Calendar. Some Things Aren’t Steady.
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Transition to AC W11-3
What’s Happening?? • Exam III on Wednesday – Material has already been published for your reference. • No Quiz on Friday • Current topic is AC Circuits - Unit 15 • Let’s look at the calendar.
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!
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) IMPORTANT
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 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.
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 : f(t)=A sin(wt) A=Amplitude (=1 here) f(t)=A sin(wt-[p/2]) A=Amplitude (=1 here)
result - inductor I is the MAXIMUM current in the circuit.
comparing Resistor inductor (wL) looks like a resistance XL=wL Reactance - OHMS
For the inductor FOR THE RESISTOR Let’s put these together.
slightly confusing point We will always use the CURRENT as the basis for calculations and express voltages with respect to the current. What that means?
direction wt wt
remember for ac series circuits The current is the same throughout the series circuit. The Maximum Current “I” is also the same for all series circuit elements.
In the circuit below, R=30 W and L= 30 mH. If the angular frequency • of the 60 volt AC source is is 3 K-Hz • WHAT WE WANT TO DO: • calculate the maximum current in the circuit • calculate the voltage across the inductor • Does Kirchoff’s Law Work? E=60V w=3 KHZ R=30 W L= 30 mH
R=30W XL=wL=90W The instantaneous voltage across each element is the PROJECTION of the MAXIIMUM voltage onto the horizontal axis! E=60V I w=3 KHZ R=30 W L= 30 mH wt
Source voltage leads the current by the angle f. I wt E=60V w=3 KHZ R=30 W L= 30 mH
The drawing is obviously NOT to scale. I wt E=60V w=3 KHZ R=30 W L= 30 mH
Let’s look at the NUMBERS
It Does! 2p