Physics 212 Lecture 21 Slide 1 Main Point

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 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. 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 C – NOTE: Good for PEAK values only) – – Vgen = Imax Z VResistor = Imax R Vinductor = Imax XL VCapacitor = Imax XC 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 m. F 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 w XL increases with f Resonance XC increases with 1/w is minimum at resonance w 0 Resonance: XL = XC Physics 212 Lecture 21, Slide 9 10

Off Resonance Physics 212 Lecture 21, Slide 10

Checkpoint 1 a 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 1 b 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 1 c 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 1 d 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 = I 2 R C R • Average Power Inductor/Capacitor = 0 Resistor <I 2 R> = <I 2 > R = ½ I 2 peak R = I 2 rms R L 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 • Efficient method to change voltage for AC. – Power Transmission Loss = I 2 R – 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 m. A VCmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45 o (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 (B) increase w (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 m. A VCmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45 o (cos=sin=1/sqrt(2)) L and R are unknown. L R What is the maximum current at resonance ( Imax(w 0) ) (A) (B) (C) 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 m. A VCmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45 o (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 = Vmaxsinwt) VL VL (A) VL VR (B) V V VC VL VR VC (C) VR VC V VR (D) V VC Physics 212 Lecture 21, Slide 22

Physics 212 Lecture 21, Slide 23