Download Cleaning Up - UCF Physics

Cleaning Up
November 28, 2005
TODAY
Last week’s examination
Finish up AC topics
Begin Quick Review of Semesters Work
Wednesday and Friday … Continue
Review and putting things in perspective
DO a few typical problems
Next Monday … FINAL EXAMINATION !
THE TEST
RLC Circuit
Current & Voltage in Phase in R.
CURRENT IS THE SAME THROUGHOUT
THE CIRCUIT!
V=V0Sin(wt)
I=VR/R
FOR EACH TYPE OF CIRCUIT
ELEMENT, THE VOLTAGE AND
CURRENT BEHAVE SOMEWHAT
DIFFERENTLY.
THE VOLTAGE ACROSS ANY
ELEMENT WILL BE OUT OF
PHASE WITH THE APPLIED
VOLTAGE.
What have we done?
 We looked at each element (C, R, L)
separately and looked at the voltages and
currents.
 The results were-
Look at AC RL Circuit (No C)
V=V0Sin(wt)
Remember for a Resistor and an Inductor
RESISTOR
Current in phase with the voltage.
For our circuit, the voltage across the resistor is NOT
the applied voltage!
It also may be out of phase with the applied voltage.
INDUCTOR
The current in the inductor LAGS the voltage
across the inductor by 900 or p/2.
(The voltage also leads the current .. just to
confuse matters more!)
Current
 The current through the
circuit is the same
throughout the circuit!
 IL=IR=Icircuit=I
 The current will not be in
phase with the applied
voltage so we need to
include a phase shift in our
calculations.
The RESISTOR
Since voltage and current are in phase in
a resistor, VR=I(t)R
Using the same symbols, we can assume
that I=I0Sin(wt-f). f is the phase
difference. We chose it to be negative to
agree with the textbook.
Therefore: VR=I0R0Sin(wt-f)
Phasor Notation
V0
Applied
Voltage
Phasor
VR (and current)
wt
wt-f
Notice that plotting a voltage and a
current on the same axes is not
normally permitted. Watch what is
plotted as we go along and be
consistent.
Some math …. (sorry!)
Look at the inductor :
I L  I 0 Sin (wt  f )
Same as in the resistor!
The voltage across the INDUCTOR is given by
dI
VL   L   I 0wLCos(wt  f )
dt
p
 I 0wLSin (wt  f  )
2
Check Graph
The Graph
- Cos(q) = + Sin(q-900)
So … Graphical identification of f
V0
f
Applied
Voltage
Phasor
wt
VR (and current)
wt-f
Moving Along …
VR
VL
wt-f
If current lags the voltage in the inductor by 900 then the voltage
leads the current by 900. VR is the current.
The Geometry
VT
VR
VL
900-f f
wt-f
wt
Back to the Loop
VT
VR
900-f
VL
f
wt
wt-f
1800 -(900 +(wt-f))=900-(wt-f)
p
V0 sin wt  VR sin( wt   )  VL Sin (wt    )
2
VT
VR
900-f
VL
f
wt
wt-f
 VL 
1  IX L 
1  X L 
  tan    tan 
  tan 

 IR 
 R 
 VR 
1
And …
VT  IZ
VT
VR  IR
VR
900-f
VL
f
wt-f
wt
VL  IX L
Z 2  R 2  X L2
Z  R  (X L  XC )
2
2
 X L  XC 
  tan 

R


1
RESONANCE: XL=XC
2
Resonance Curve
What about power??
We have already shown that :
Paverage  I
2
rms
I I
R
R
2 2
Continuing
VT
VR
900-f
VL
I rms
Pavg
Pavg
Erms

Z
Erms

I rms R
Z
R
 Erms I rms  Erms I rms cos 
Z
f
wt-f
wt