physics 212 lecture 21, slide 1 physics 212 lecture 21

Physics 212 Lecture 21, Slide Physics 212 Lecture 21, Slide 11

Physics 212Physics 212Lecture 21Lecture 21

Main Point 1

First, we defined the condition of resonance in a driven series LCR circuit to occur at the frequency w 0 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.


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 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 number of turns in the primary.


Physics 212 Lecture 21, Slide Physics 212 Lecture 21, Slide 55Physics 212 Lecture 21, Slide Physics 212 Lecture 21, Slide 55

Peak AC Problems

• “Ohms” Law for each element – NOTE: Good for PEAK values only)

– Vgen = Imax Z

– VResistor = Imax R

– Vinductor = Imax XL

– VCapacitor = Imax XC

• 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?

LX L

22L CZ R X X

1CX C

07

L

R

C


Resonance

Light-bulb DemoLight-bulb Demo


Resonance

R is independent of

XL increases with fLLX

is minimum at resonance

22 )( CL XXRZ

XC increases with 1/1CCX

Resonance: XL = XC 0

1LC

10

Frequency at which voltage across inductor and capacitor cancel

Resonance in AC Circuits

frequency

Imp

edan

ce

0

Resonance


Off Resonance


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 circuitsA. VI > VII B. VI = VII C. VI < VII


Checkpoint 1b


Compare the peak voltage across the inductor in the two circuitsA. VI > VII B. VI = VII C. VI < VII


Checkpoint 1c


Compare the peak voltage across the capacitor in the two circuitsA. VI > VII B. VI = VII C. VI < VII


Checkpoint 1d


At the resonant frequency, which of the following is true?A. Current leads voltage across the generatorB. Current lags voltage across the generator C. Current is in phase with voltage across the generator


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

• Average PowerInductor/Capacitor = 0Resistor<I2R> = <I2 > R = ½ I2peak R

= I2rms R

Average Power Generator = Average Power Resistor

RMS = Root Mean Square

Ipeak = Irms sqrt(2)

L

R

C


Transformers

• Application of Faraday’s Law– Changing EMF in Primary creates changing flux– Changing flux, creates EMF in secondary

• Efficient method to change voltage for AC.– Power Transmission Loss = I2R– Power electronics

p s

p s

V VN N


Follow Up from Last LectureFollow Up from Last LectureConsider the harmonically driven series LCR circuit shown. Vmax = 100 VImax = 2 mAVCmax = 113 V (= 80 sqrt(2))The current leads generator voltage by 45o

(cos=sin=1/sqrt(2))L and R are unknown.

How should we change to bring circuit to resonance?

CC

RR

LLVV ~~

(A) decrease (B) increase (C) Not enough info


Current Follow UpCurrent Follow UpConsider the harmonically driven series LCR circuit shown. Vmax = 100 VImax = 2 mAVCmax = 113 V (= 80 sqrt(2))The current leads generator voltage by 45o


What is the maximum current at resonance ( Imax(0) )

CC

RR

LLVV ~~

(A) (B) (C)max 0I ( ) 2 mA max 0I ( ) 2 2 mA max 0I ( ) 8 / 3 mA


Phasor Follow Up Phasor Follow Up Consider the harmonically driven series LCR circuit shown. Vmax = 100 VImax = 2 mAVCmax = 113 V (= 80 sqrt(2))The current leads generator voltage by 45o


What does the phasor diagram look like at t = 0? (assume V = Vmaxsint)

CC

RR

LLVV ~~

(A) (B) (C) (D)V L

V R

V C

V

VL

VR

VC

V

VL

VR

VC V

V L

V R

V C

V

physics 212 lecture 21, slide 1 physics 212 lecture 21

Documents