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Physics 212
Lecture 21
Resonance and power in AC circuits
Physics 212 Lecture 21, Slide 1
Imax XC
Imax XL
emax
w
XL = wL
XC = 1/wC
C
L
Imax XL
R
Imax R
Imax R
Imax XC
emax = Imax Z
Imax(XL-XC)
Imax R
XL-XC
f
R
Physics 212 Lecture 21, Slide 2
Peak AC Problems
07
• “Ohms” Law for each element
– PEAK values
–
–
–
–
C
Vgen
= Imax Z
VResistor = Imax R
Vinductor = Imax XL
VCapacitor = Imax XC
L
R
• 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?
X L  w L  200 
1
XC 
 100 
wC
XL
Z  R 2   X L  X C   112 
2
I max 
V gen
Z
R
 0.13 A
XC
Physics 212 Lecture 21, Slide 3
Peak AC Problems
12
• “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 0.004 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  R 2   X L  X C   112 
2
E) All the same.
Vmax  I max X
R
XC
I max 
V gen
Z
 0.13 A
Physics 212 Lecture 21, Slide 5
Peak AC Problems
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 6
Resonance
Light-bulb Demo
Physics 212 Lecture 21, Slide 7
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
8
10
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 9
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 10
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 11
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 12
Quality factor Q
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 13
Larger Q, sharper resonance
Physics 212 Lecture 21, Slide 14
Series LCR circuit
1. The current leads generator voltage by 45o.
C
V ~
2. Assume V = Vmaxsinwt
L
R
What does the phasor diagram look like at t = 0?
(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 15
Series LCR circuit
C
The current leads generator voltage by 45o
V ~
R
How should we change w to bring circuit to resonance?
(A) decrease w
Original w
f
(B) increase w
At resonance
(w0)
L
(C) Not enough info
At resonance
XL = X C
XL increases
XC decreases
Increase w
Physics 212 Lecture 21, Slide 16
Series LCR circuit
1. The current leads generator voltage by 45o.
2. Suppose XC = 2XL at frequency w.
C
V ~
L
R
By what factor should we increase w
to bring circuit to resonance w0 ?
w0
(A)
 2
w
XC  2X L
w0
(B)
2 2
w
w0
(C)
2
w
w0
(D)
4
w
1
1
1 2
2
 2w L  w 
 w0
wC
2 LC 2
Physics 212 Lecture 21, Slide 17
Series LCR circuit
C
Consider the harmonically driven series LCR circuit shown.
At some frequency w, XC = 2XL = R
Vmax = 100 V
R = 50 k
V ~
L
R
If we change w, what is the maximum current that can flow?
(A) I max  2 mA
At resonance
XL = X C
(B) I max  2 mA
Z R
(C) I max  2 2 mA
Vmax
100
I max (w0 ) 

 2 mA
3
R
50 x 10
Physics 212 Lecture 21, Slide 18
Instantaneous power
C
P(t) = I(t)V(t)
L
R
P(t) varies in time. We want the average power.
Average, over one cycle, of any periodic function :
P t  
T
1
P t  d t

T 0
T
Circuit element:
1
P  t    V  t  I  t  dt
T 0
For harmonically-varying functions like sin wt or cos wt , T = 2π/ω .
Physics 212 Lecture 21, Slide 19
For L or C, voltage and
current are /2 out of phase:
I  sin wt, V  cost wt
PL ,C  t 
T
1
sin w t  cos w t  dt  0

T 0
Average power for R is not zero since V and I are in phase.
T
2
1
1 2
2
PR  t   I R    I peak sin w t   R dt  I peak R
T 0
2
e peak
1
1
 I peak
R  I peak e peak cos f
2
Z
2
Z
f
XL-XC
R
Define root mean square (RMS)
1 2
2
I R  I peak R  I RMS
R
2
1
I RMS 
I peak  .707 I peak
2
2
Average power dissipated in R for LCR circuit
1
Power  I peak e peak cos f
2
Z
f
R
Maximum power delivered when f = 0 … at resonance !
Imax w0L
At resonance,
VC VL w 0 LI max w 0 L



Q
VR VR
RI max
R
Imax R = emax
Imax / w0C
w0L = 1/w0C
f=0
XL-XC
Power Transmission
• 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 from power 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 = 0.15 Amps
– Loss from power line = I2R = 0.15*0.15*5 = 0.113 Watts
DEMO
Physics 212 Lecture 21, Slide 22
How do we transform 10,000 V into 120 V for your toaster ?
Transformer
Time-varying voltage in primary causes
time-varying B in iron-core.
Flux per turn = B (Area) = B
Iron core maintains same B throughout.
Primary coil captures flux P = NP B
Secondary coil captures flux S = NS B
7200 V / 240 V
Faraday’s Law:
VP = d P /dt = NP B
VS = d S /dt = NS B
Vp N p

Vs N s
Physics 212 Lecture 21, Slide 23