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Physics 212
Lecture 21
Physics 212 Lecture 21, Slide 1
Main Point 1
First, we defined the condition of resonance in a driven series LCR circuit to occur
at the frequency w0 that produces the largest value for the peak current in the
circuit. At this frequency, which is also the natural oscillation frequency of the
corresponding LC circuit, the inductive reactance is equal to the capacitive
reactance so that the total impedance of the circuit is just resistive.
Physics 212 Lecture 21, Slide 2
Main Point 2
Second, we determined that the average power delivered to the circuit by the
generator was equal to the product of the rms values of the emf and the current
times the cosine of the phase angle phi. The sharpness of the peak in the
frequency dependence of the average power delivered was determined by the Q
factor, which was defined in terms of the ratio of the energy stored to the energy
Physics 212 Lecture 21, Slide 3
dissipated at resonance.
Main Point 3
Third, we examined the properties of an ideal transformer and determined that
the ratio of the induced emf in the secondary coil to that of the generator was
just equal to the ratio of the number of turns in secondary to the number of turns
in the primary. We also determined that when a resistive load is connected to the
secondary coil, the ratio of the induced current in the primary to that in the
secondary is also equal to the ratio of the number of turns in secondary to the
Physics 212 Lecture 21, Slide 4
number of turns in the primary.
Peak AC Problems
07
• “Ohms” Law for each element
– NOTE: Good for PEAK values only)
–
–
–
–
Vgen
= Imax Z Z  R2  X  X 2
 L C
VResistor = Imax R
XL  L
Vinductor = Imax XL
VCapacitor = Imax XC
1
• Typical Problem
XC 
C
L
R
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?
Physics 212 Lecture 21, Slide 5
Physics 212 Lecture 21, Slide 6
Physics 212 Lecture 21, Slide 7
Resonance
Light-bulb Demo
Physics 212 Lecture 21, Slide 8
Resonance
Frequency at which voltage across inductor and capacitor cancel
R is independent of 
XC increases with 1/
1
XC 
C
Resonance
Impedance
XL increases with f
X L  L
Resonance in AC Circuits
Z  R 2  ( X L  X C )2
is minimum at resonance
Resonance: XL = XC
0
1
0 
LC
frequency
Physics 212 Lecture 21, Slide
9
10
Off Resonance
Physics 212 Lecture 21, Slide 10
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
Physics 212 Lecture 21, Slide 11
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
Physics 212 Lecture 21, Slide 12
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
Physics 212 Lecture 21, Slide 13
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
Physics 212 Lecture 21, Slide 14
Physics 212 Lecture 21, Slide 15
Power
• P = IV instantaneous always true
– Difficult for Generator, Inductor and Capacitor because of phase
– Resistor I,V are ALWAYS in phase!
P = IV
= I2 R
C
L
R
• Average Power
Inductor/Capacitor = 0
Resistor
<I2R> = <I2 > R = ½ I2peak R
= I2rms R
RMS = Root Mean Square
Ipeak = Irms sqrt(2)
Average Power Generator = Average Power Resistor
Physics 212 Lecture 21, Slide 16
Transformers
• 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 17
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
How should we change  to bring circuit to resonance?
(A) decrease 
(B) increase 
(C) Not enough info
Physics 212 Lecture 21, Slide 18
Physics 212 Lecture 21, Slide 19
Current Follow Up
C
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
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.
L
R
What is the maximum current at resonance ( Imax(0) )
(A) Imax (0 )  2 mA
(B) Imax (0 )  2 2 mA
(C) Imax (0 )  8/ 3 mA
Physics 212 Lecture 21, Slide 20
Physics 212 Lecture 21, Slide 21
Phasor Follow Up
C
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.
V ~
L
R
What does the phasor diagram look like at t = 0?
(assume V = Vmaxsint)
(A)
(B)
(C)
(D)
Physics 212 Lecture 21, Slide 22
Physics 212 Lecture 21, Slide 23