Download Lecture 5

Physics 2112
Unit 21
Outline:
 Resonance
 Power/Power Factor
 Q factor


What is means
A useful approximation
 Transformers
From last time…..
“Ohms Law” for each element
Vresistor  I max R
Vinductor  I max X L
Vcapacitor  I max X C
C
I max 
Vgen
L
R
Z
Vgen  I max Z
Z  R 2   X L  X C  2 (impedance)
where: X  L
L
1
XC 
C
 of generator
XL
(inductive reactance)
(capacitive reactance)
R
XC
This means…..
I max 
Vgen
R  (X L  XC )
2
2
C
L
R
Current is largest when
X L  XC
or
d
1
LC
Where have we seen this before?
o 
1
LC
(Natural Frequency)
When d = o  resonance
C
L
Resonance
Frequency at which voltage across inductor and capacitor cancel
R is independent of 
Resonance in AC Circuits
XL increases with 
XC increases with 1/
1
XC 
C
Z  R  ( X L  XC )
2
Z = R at resonance
Impedance
X L  L
Z
2
XL
0
is minimum at resonance
Resonance: XL  XC
0 
1
LC
XC
frequency
R
Example 21.1 (Tuning Radio)
A radio antenna is hooked to
signal filter consisting of a
resistor, a variable capacitor
and a 50uH inductor.
If we would like to get maximum
current for the signal from our
favorite country music station,
US 99 (99.5MHz), what should
we set the capacitor to?
C
L
R
amplifier
Example 21.2 (Peak Current)
C
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?
R
L
CheckPoint 1(A)
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
Resonance: XL  XC
B. VI = VII
Z R
C. VI < VII
Same since R doesn't change
Imax XL
Imax R
Imax R
Imax XC
Case 1
Imax XC
Case 2
CheckPoint 1(B)
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 inductor in the
two circuits
A. VI > VII
Voltage in second circuit will
B. VI = VII
be twice that of the first
because of the 2L compared
C. VI < VII
to L.
Imax XL
Imax R
Imax R
Imax XC
Case 1
Imax XC
Case 2
CheckPoint 1(C)
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 capacitor in the
two circuits
A. VI > VII
B. VI = VII
The peak voltage will be
C. VI < VII
greater in circuit 2 because
the value of XC doubles.
Imax XL
Imax R
Imax R
Imax XC
Case 1
Imax XC
Case 2
CheckPoint 1(D)
Imax XL
At the resonant frequency, which of the following
is true?
A. The current leads the voltage across the
generator
B. The current lags the voltage across the
generator
C. The current is in phase with the voltage
across the generator
Imax XL
Imax R
Imax R
Imax XC
Case 1
Imax XC
Case 2
Resonance….what it means…..
A circuit is in resonance when….
“
 frequency of the signal generator is the same
as the natural frequency of the circuit.
 Imax = e/R
 Voltage drop across the resistor is maximum
 XL = XC
 Z has minimum value
 phase angle, f, is zero
 voltage and current are in sync at the signal
generator.
Power
What is the power put into circuit by
signal?
C
At any instant P=VI  varies with
time.
But V and I not in sync at genarator
R
L
Power
PIN = POUT
RMS  Root Mean Square
I rms  I peak / 2
PAVG  I
2
rms
I rms R 
R  (I
2
peak
Vrms
V
R  rms
Z
cos f
PAVG  I rmsV rmscos f
/ 2) R
C
L
R
Changes electrical
energy into thermal
energy
Example 21.3 (Power lost)
How much electrical energy was turned
into thermal energy every minute in
the three situations we had in
example 20.2?
Recall
=60rad/sec
=400rad/sec
 =206rad/sec
f =-72o
f=55.7o
f = 0o
In the circuit to the right
•
L=500mH
•
Vmax = 6V
•
C=47uF
•
R=100W
Example 21.4
A bright electric light bulb and a small
window air conditioner both are
plugged into a wall socket with puts
out Vrms=120V. Both draw Irms= 1amp
of current. The light bulb has a power
factor of 1 and the AC unit has a
power factor of 0.85.
How much electrical energy do they
consume in 1 hour?
Power Line Calculation
If you want to deliver 1,500 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 1,500 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)  I 2R  0.125 Watts
Lower, but not zero!!!
Cost to you
Utility companies charge a surcharge
for industrial customers with large
power factors (e.g. large AC loads).
Customers can correct power factor
with capacitors.
Assumed power factor for residential >
0.95.
Warning!!!!
About to talk about “Quality Factor”, Q
Just to confuse you….we’re also going to
be talking about the charge on a
capacitor, Q
We’ll go through the math quickly….
Hang in there. We’ll show what it means
conceptually and do an example
problem in the end.
Electricity & Magnetism Lecture 21, Slide 18
Recall Damped Harmonic Motion
Q  QMax e
 t
cos( ' t   )
Damping
factor
R

2L
Natural
oscillation
frequency

1
LC
Damped
oscillation
frequency
 '2  o2   2
Unit 19, Slide 19
What does Q mean algebraically….
Energy

Charge2
Energy Lost  (e
 t 2
) e
Define Quality Factor
I
Q
I
Q
2
rms
2
rms
 t /( L / R )
Q  o
L

R
o 
1
LC
2
L
I rms
L( f o 2 )
o 
2
I
R
rms R
(Maximum energy Stored)
= 2 (Energy Lost per Cycle)
Unit 19, Slide 20
Second way to look at Q
I max 

Vgen
R  (X L  XC )
2
2
Vgen
1
R
( x  1)
1 Q
2
x
2
2
2
How fast you fall
away from current
at resonance

x
o
How far you are
from resonance
Bigger Q, fall off faster
Unit 19, Slide 21
Q=0.5
Q=1.0
Imax/(e/R)
Imax/(e/R)
What does Q mean graphically…..
x=/o
Q=5.0
Imax/(e/R)
Imax/(e/R)
x=/o
x=/o
Q=20.0
x=/o
Useful approximation from Q
Often used approximation for Q>2
o
Q
FWHM
“Full Width
at Half
Maximum”
Q=5.0
Example 21.5 (FWHM)
An electronic filter is made up of an inductor, L=0.5H,
a capacitor, C=2.0uF and resistor, R=20W. It has an
input signal with Vmax = 0.1V.
1) What is the resonant frequency for this filter?
2) What is the average output power at this frequency?
3) What is the quality factor Q for this filter?
4) Use this Q to estimate FWHM.
5) What is the average power output at a frequency
of o + FWHM/2?
Electricity & Magnetism Lecture 21, Slide 24
Transformers
Primary (P)
DVP  DIP
 DBP
 DFS
 DIS
 DVS
 DBS
 DFP
Secondary (S)
Difficult to
calculate
step by step
Transformers
FS
FP

loop loop
eP
NP

eS
NS
NS
eS 
eP
NP
Note by
energy PS
conservation
 PP  I S 
NP
IP
NS
Example 21.6 (Coil in your car)
In the old days, the high voltage needed to fire the
spark plugs in your car was created by “the coil”. On
a typical coil, there were 200 turns the primary side
and 400,000 turns on the secondary side.
If the car’s alternator provided 12 volts to the coil, what
was the voltage provided to the spark plugs?
Uniy 21, Slide 27