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Physics 212 Lecture 21 Physics 212 Lecture 21, Slide 1 Music Who is the Artist? A) B) C) D) E) Earth, Wind & Fire War Tower of Power Average White Band Chicago Why? Topic for today = Power Your Comments “Oh my god that was horrible please don’t make me memorize all that.” “none of this makes sense please be thurough in lecture” “What significance does Q2=L/R2C have in electrical components?” “This doesn’t seem conceptually hard but there are tons of new equations and relations between variables.” “Could you better explain the last lecture and how it leads into this? It looks like its just something you have to stare at for a while but once it clicks then its really easy. I’m having trouble with visualizing the phase diagram and this whole pre lecture was gibberish to me. Not one thing sunk in.” “i dont see how the transformer relates..” 05 We’ll do our best to restate the Big Ideas and give examples Shape of resonance curve (I vs. w) It’s all still based on the phasor diagram! ABSOLUTELY We will connect the two today.. Short review + demo Physics 212 Lecture 21, Slide 3 What Did We Learn in Office Hours This Week? PHASORS ARE THE KEY !! FORMULAS ARE NOT ! Imax XC emax C L Imax XL R START WITH PHASOR DIAGRAM Imax R DEVELOP FORMULAS FROM THE DIAGRAM !! emax = Imax Z Imax(XL-XC) V = Projection along Vertical Imax R Imax XL w XL = wL XC = 1/wC Imax XC Imax R XL-XC f R Physics 212 Lecture 21, Slide 4 General RLC Example 07 • “Ohms” Law for each element – NOTE: Good for PEAK values only) – – – – C Vgen = Imax Z Z R 2 X X 2 L C VResistor = Imax R XL wL Vinductor = Imax XL VCapacitor = Imax XC 1 • Typical Problem XC L R wC 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 R2 X L X C 112 2 I max Vgen Z R 0.13 A XC Physics 212 Lecture 21, Slide 5 General RLC Example • “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 12 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 R2 X L X C 112 2 E) All the same. Vmax I max X R XC I max Vgen Z 0.13 A Physics 212 Lecture 21, Slide 7 General RLC Example 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 8 Resonance Physics 212 Lecture 21, Slide 9 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 10 10 Off Resonance 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 11 Off Resonance Physics 212 Lecture 21, Slide 12 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 “Circuit II has twice the impedance and therefore half the peak current and resistor voltage.” “They would be the same because the circuit has the same impedance.” “V must be less across the resistor to account for increased voltage in the capacitor and inductor” Physics 212 Lecture 21, Slide 13 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 14 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 “The peak voltage across the inductor is Imax/(wL) which is half as big in the second case.” “V is dependent on change in current, which is dependent on E, which has the same formula in both cases” “voltage is I times wL i think and inductor 2 has twice the inductance while the w is the same in both cases” Physics 212 Lecture 21, Slide 15 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 16 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 “The current will decrease by a factor greater than 2 which decreases peak voltage.” “they must equal each other because the circuit is resonant” “Xc would be twice as higher as C is halved in the second case. As such, the peak voltage across the capacitor would be higher in the second case.” Physics 212 Lecture 21, Slide 17 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 18 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 “the inductor has more voltage than the capacitor ” “The current is in phase with the resistor which means it lags the voltage of the generator.” “At the resonant frequency, XL=XC, so the circuit behaves as if the resistor is the only element present.” Physics 212 Lecture 21, Slide 19 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 20 Power • P = IV instantaneous always true – Difficult for Generator, Inductor and Capacitor because of phase – Resistor I,V are ALWAYS in phase! C P = IV = I2 R L R • Average Power Inductor and Capacitor = 0 ( < sinwtcoswt > = 0 ) Resistor <I2R> = <I2 > R = ½ I2peak R RMS = Root Mean Square Ipeak = Irms sqrt(2) < I2 R > = Irms2 R Physics 212 Lecture 21, Slide 21 Power Line Calculation • 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 = IV (on 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 = .15 Amps – Loss = IV (on line) = I2R = 0.125 Watts DEMO Physics 212 Lecture 21, Slide 22 Transformers (Robots in Disguise, as many of you said) • 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 23 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 What does the phasor diagram look like at t = 0? (assume V = Vmaxsinwt) (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 24 Follow Up: Resonance 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 w to bring circuit to resonance? (A) decrease w Original w f (B) increase w At resonance (w0) (C) Not enough info At resonance XL = X C XL increases XC decreases w increases Physics 212 Lecture 21, Slide 25 Follow Up: Resonance C Consider the harmonically driven series LCR circuit shown. Vmax = 100 V V ~ Imax = 2 mA X C 40 2 k VCmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45o (cos=sin=1/sqrt(2)) L and R are unknown. f 2 If w is increased by a factor of f: At resonance XL = X C (B) f 2 2 (C) 8 f 3 XL increases by factor of f XC decreases by factor of f 15 f 40 f 40 f 15 2 R R 25 2 k X L 15 2 k By what factor should we increase w to bring circuit to resonance? i.e. if w0 = fw, what is f? (A) L (D) f 8 5 X L f 15 2 X C (1/ f ) 40 2 f 8 3 Physics 212 Lecture 21, Slide 26 Follow Up: Resonance Consider the harmonically driven series LCR circuit shown. Vmax = 100 V V ~ Imax = 2 mA X C 40 2 k VCmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45o (cos=sin=1/sqrt(2)) L and R are unknown. What is the maximum current at resonance ( Imax(w0) ) C L R R 25 2 k X L 15 2 k w0 (A) I max (w0 ) 2 mA At resonance XL = X C (B) I max (w0 ) 2 2 mA ZR I max (w0 ) 8 w 3 (C) I max (w0 ) 8 / 3 mA Vmax 100 2 2 mA R 25 2 Physics 212 Lecture 21, Slide 27