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Physics 212
Lecture 21
Physics 212 Lecture 21, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Earth, Wind & Fire
War
Tower of Power
Average White Band
Chicago
Why?
Topic for today = Power
Your Comments
“Oh my god that was horrible please don’t
make me memorize all that.”
“none of this makes sense please be thurough
in lecture”
“What significance does Q2=L/R2C have in
electrical components?”
“This doesn’t seem conceptually hard but
there are tons of new equations and
relations between variables.”
“Could you better explain the last
lecture and how it leads into this? It
looks like its just something you have to
stare at for a while but once it clicks
then its really easy. I’m having trouble
with visualizing the phase diagram and
this whole pre lecture was gibberish to
me. Not one thing sunk in.”
“i dont see how the transformer
relates..”
05
We’ll do our best to restate the
Big Ideas and give examples
Shape of resonance curve (I vs. w)
It’s all still based on the phasor diagram!
ABSOLUTELY
We will connect the two today..
Short review + demo
Physics 212 Lecture 21, Slide 3
What Did We Learn in
Office Hours This Week?
PHASORS ARE THE KEY !!
FORMULAS ARE NOT !
Imax XC
emax
C
L
Imax XL
R
START WITH PHASOR
DIAGRAM
Imax R
DEVELOP FORMULAS FROM
THE DIAGRAM !!
emax = Imax Z
Imax(XL-XC)
V = Projection along Vertical
Imax R
Imax XL
w
XL = wL
XC = 1/wC
Imax XC
Imax R
XL-XC
f
R
Physics 212 Lecture 21, Slide 4
General RLC Example
07
• “Ohms” Law for each element
– NOTE: Good for PEAK values only)
–
–
–
–
C
Vgen
= Imax Z Z  R 2  X  X 2
 L C
VResistor = Imax R
XL  wL
Vinductor = Imax XL
VCapacitor = Imax XC
1
• Typical Problem
XC 
L
R
wC
A generator with peak voltage 15 volts and angular
frequency 25 rad/sec is connected in series with an 8
Henry inductor, a 0.4 mF capacitor and a 50 ohm
resistor. What is the peak current through the circuit?
X L  w L  200 
1
XC 
 100 
wC
XL
Z  R2   X L  X C   112 
2
I max 
Vgen
Z
R
 0.13 A
XC
Physics 212 Lecture 21, Slide 5
General RLC Example
• “Ohms” Law for each element
2
2
Z

R

X

X
– Vgen
= IZ
 L C
– VResistor = I R
– Vinductor = I XL X L  w L
– VCapacitor = I XC X C  1
wC
12
C
L
R
• Typical Problem
A generator with peak voltage 15 volts and angular frequency 25 rad/sec is
connected in series with an 8 Henry inductor, a 0.004 Farad capacitor and a 50
ohm resistor. What is the peak current through the circuit?
X L  w L  200 
Which element has the largest peak voltage across XL
it?
A) Generator
B) Inductor
C) Resistor
D) Capacitor
XC 
1
 100 
wC
Z  R2   X L  X C   112 
2
E) All the same.
Vmax  I max X
R
XC
I max 
Vgen
Z
 0.13 A
Physics 212 Lecture 21, Slide 7
General RLC Example
14
• “Ohms” Law for each element
2
2
Z

R

X

X
– Vgen
= IZ
 L C
– VResistor = I R
– Vinductor = I XL X L  w L
– VCapacitor = I XC X C  1
wC
C
L
R
• Typical Problem
A generator with peak voltage 15 volts and angular frequency 25 rad/sec is
connected in series with an 8 Henry inductor, a 0.4 mF capacitor and a 50 ohm
resistor. What is the peak current through the circuit?
What happens to the impedance if we decrease the
angular frequency to 20 rad/sec?
XL
XL
Z25
Z20
A) Z increases
R
B) Z remains the same
C) Z decreases
(XL-XC): (200-100)  (160-125)
XC
R
XC
Physics 212 Lecture 21, Slide 8
Resonance
Physics 212 Lecture 21, Slide 9
Resonance
Frequency at which voltage across inductor and capacitor cancel
Resonance in AC Circuits
XL increases with w
X L  wL
Z = R at resonance
XC increases with 1/w
1
XC 
wC
Z  R  (X L  XC )
2
Impedance
R is independent of w
Z
2
is minimum at resonance
Resonance: XL = XC
XL
w0
1
w0 
LC
XC
R
frequency
Physics 212 Lecture 21, Slide
10 10
Off Resonance
Z
In general
Q  2
U max
U
Umax = max energy stored
U = energy dissipated
in one cycle at resonance
Physics 212 Lecture 21, Slide 11
Off Resonance
Physics 212 Lecture 21, Slide 12
Checkpoint 1a
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the resistor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
“Circuit II has twice the impedance and therefore half the peak current and
resistor voltage.”
“They would be the same because the circuit has the same impedance.”
“V must be less across the resistor to account for increased voltage in the
capacitor and inductor”
Physics 212 Lecture 21, Slide 13
Checkpoint 1a
Imax XL
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the resistor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
Resonance: XL = XC
Z = R
Same since R doesn't
change
Imax XL
Imax R
Imax R
Imax XC
Case 1
Case 2
Imax XC
Physics 212 Lecture 21, Slide 14
Checkpoint 1b
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the inductor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
“The peak voltage across the inductor is Imax/(wL) which is half as big in the
second case.”
“V is dependent on change in current, which is dependent on E, which has
the same formula in both cases”
“voltage is I times wL i think and inductor 2 has twice the inductance while
the w is the same in both cases”
Physics 212 Lecture 21, Slide 15
Checkpoint 1b
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the inductor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
Imax XL
Imax XL
Imax R
Imax R
Imax XC
Case 1
Voltage in second circuit will be twice that of
the first because of the 2L compared to L
Case 2
Imax XC
Physics 212 Lecture 21, Slide 16
Checkpoint 1c
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the capacitor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
“The current will decrease by a factor greater than 2 which decreases peak
voltage.”
“they must equal each other because the circuit is resonant”
“Xc would be twice as higher as C is halved in the second case. As such, the
peak voltage across the capacitor would be higher in the second case.”
Physics 212 Lecture 21, Slide 17
Checkpoint 1c
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the capacitor in the two circuits
A. VI > VII
B. VI = VII
C. VI < VII
Imax XL
Imax XL
Imax R
Imax R
Imax XC
Case 1
The peak voltage will be greater in circuit
2 because the value of XC doubles.
Case 2
Imax XC
Physics 212 Lecture 21, Slide 18
Checkpoint 1d
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
At the resonant frequency, which of the following is true?
A. Current leads voltage across the generator
B. Current lags voltage across the generator
C. Current is in phase with voltage across the generator
“the inductor has more voltage than the capacitor ”
“The current is in phase with the resistor which means it lags the voltage of
the generator.”
“At the resonant frequency, XL=XC, so the circuit behaves as if the resistor is
the only element present.”
Physics 212 Lecture 21, Slide
19
Checkpoint 1d
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
At the resonant frequency, which of the following is true?
A. Current leads voltage across the generator
Imax XL
B. Current lags voltage across the generator
C. Current is in phase with voltage across the generator
Imax R
Imax XL
Imax R
Imax XC
The voltage across the inductor and the
capacitor are equal when at resonant
frequency, so there is no lag or lead.
Case 1
Case 2
Imax XC
Physics 212 Lecture 21, Slide 20
Power
• P = IV instantaneous always true
– Difficult for Generator, Inductor and Capacitor because of phase
– Resistor I,V are ALWAYS in phase!
C
P = IV
= I2 R
L
R
• Average Power
Inductor and Capacitor = 0 ( < sinwtcoswt > = 0 )
Resistor
<I2R> = <I2 > R = ½ I2peak R
RMS = Root Mean Square
Ipeak = Irms sqrt(2)
< I2 R > = Irms2 R
Physics 212 Lecture 21, Slide 21
Power Line Calculation
• If you want to deliver 1500 Watts at 100 Volts over
transmission lines w/ resistance of 5 Ohms. How much
power is lost in the lines?
– Current Delivered: I = P/V = 15 Amps
– Loss = IV (on line) = I2 R = 15*15 * 5 = 1,125 Watts!
• If you deliver 1500 Watts at 10,000 Volts over the same
transmission lines. How much power is lost?
– Current Delivered: I = P/V = .15 Amps
– Loss = IV (on line) = I2R = 0.125 Watts
DEMO
Physics 212 Lecture 21, Slide 22
Transformers
(Robots in Disguise, as many of you said)
• Application of Faraday’s Law
– Changing EMF in Primary creates changing flux
– Changing flux, creates EMF in secondary
V p Vs

N p Ns
• Efficient method to change voltage for AC.
– Power Transmission Loss = I2R
– Power electronics
Physics 212 Lecture 21, Slide 23
Follow Up from Last Lecture
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V (= 80 sqrt(2))
The current leads generator voltage by 45o (cos=sin=1/sqrt(2))
L and R are unknown.
C
V ~
L
R
What does the phasor diagram look like at t = 0?
(assume V = Vmaxsinwt)
(A)
X
(B)
(C)
X
(D)
X
V = Vmax sinwt  V is horizontal at t = 0 (V = 0)
V  VL  VC  VR
VL < VC if current leads generator voltage
Physics 212 Lecture 21, Slide 24
Follow Up: Resonance
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V (= 80 sqrt(2))
The current leads generator voltage by 45o (cos=sin=1/sqrt(2))
L and R are unknown.
C
V ~
L
R
How should we change w to bring circuit to resonance?
(A) decrease w
Original w
f
(B) increase w
At resonance
(w0)
(C) Not enough info
At resonance
XL = X C
XL increases
XC decreases
w increases
Physics 212 Lecture 21, Slide 25
Follow Up: Resonance
C
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
V ~
Imax = 2 mA
X C  40 2 k
VCmax = 113 V (= 80 sqrt(2))
The current leads generator voltage by 45o (cos=sin=1/sqrt(2))
L and R are unknown.
f  2
If w is
increased by
a factor of f:
At resonance
XL = X C
(B) f  2 2
(C)
8
f 
3
XL increases by factor of f
XC decreases by factor of f
15 f 
40
f
40
f 
15
2
R
R 25 2 k
X L  15 2 k
By what factor should we increase w to bring circuit to
resonance? i.e. if w0 = fw, what is f?
(A)
L
(D) f 
8
5
X L  f 15 2
X C  (1/ f )  40 2
f 
8
3
Physics 212 Lecture 21, Slide 26
Follow Up: Resonance
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
V ~
Imax = 2 mA
X C  40 2 k
VCmax = 113 V (= 80 sqrt(2))
The current leads generator voltage by 45o (cos=sin=1/sqrt(2))
L and R are unknown.
What is the maximum current at resonance ( Imax(w0) )
C
L
R
R 25 2 k
X L  15 2 k
w0 
(A) I max (w0 )  2 mA
At resonance
XL = X C
(B) I max (w0 )  2 2 mA
ZR
I max (w0 ) 
8
w
3
(C) I max (w0 )  8 / 3 mA
Vmax
100

 2 2 mA
R
25 2
Physics 212 Lecture 21, Slide 27