Physics-272 Lecture 20. AC Power Resonant Circuits Phasors (2-dim vectors, amplitude and phase)

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1 Physics-7 ecture 0 A Power esonant ircuits Phasors (-dim vectors, amplitude and phase)

2 What is reactance? You can think of it as a frequency-dependent resistance. 1 ω For high ω, χ ~0 - apacitor looks like a wire ( short ) For low ω, χ - apacitor looks like a break ω For low ω, χ ~0 - nductor looks like a wire ( short ) For high ω, χ - nductor looks like a break (inductors resist change in current) ( " " )

3 An circuit is driven by an A generator as shown in the figure. For what driving frequency ω of the generator, will the current through the resistor be largest a) ω large b) ω small c) independent of driving freq. The current amplitude is inversely proportional to the frequency of the generator. ( ω)

4 Alternating urrents: circuit Figure (b) has > and (c) has <. Using Phasors, we can construct the phasor diagram for an ircuit. This is similar to -D vector addition. We add the phasors of the resistor, the inductor, and the capacitor. The inductor phasor is +90 and the capacitor phasor is -90 relative to the resistor phasor. Adding the three phasors vectorially, yields the voltage sum of the resistor, inductor, and capacitor, which must be the same as the voltage of the A source. Kirchoff s voltage law holds for A circuits. Also and are in phase.

5 Phasors Problem: Given drive ε m sin(ωt), find,,,,, ε Strategy: We will use Kirchhoff s voltage law that the (phasor) sum of the voltages,, and must equal drive.

6 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,,,, ε. Next draw the phasor for. Since the inductor voltage always leads draw 90 further counterclockwise. The length of the phasor is ω

7 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,,,, ε 1. Draw phasor along x-axis (this direction is chosen for convenience). Note that since, this is also the direction of the current phasor i. Because of Kirchhoff s current law, (i.e., the same current flows through each element).,

8 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,,,, ε 3. The capacitor voltage always lags draw 90 further clockwise. The length of the phasor is /ω The lengths of the phasors depend on,,, andω. The relative orientation of the,, and phasors is always the way we have drawn it. Memorize it!

9 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,,,, ε The phasors for,, and are added like vectors to give the drive voltage + + ε m : From this diagram we can now easily calculate quantities of interest, like the net current, the maximum voltage across any of the elements, and the phase between the current the drive voltage (and thus the power). ε m

10 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) - cos cos - cos - cos t t t t t v φ ω φ ω φ ω φ ω ω ω φ 1/ tan oltage (t) across A source is called impedance Also: ) cos( cos φ ω ω + t v t i rms rms MA ike:

11 series circuit; Summary of instantaneous urrent and voltages i ( t) cos( ωt) ( t) cos( ωt) v v v v ad 1 ω ( t) cos( ωt 90) cos( ωt 90) ( t) cos( ωt + 90) ω cos( ωt + 90) ( t) ( ) + ( - ) cos( ωt + φ) tanφ ω 1/ ω

12 Alternating urrents: circuit, Fig Y&F Example v ω10000rad/s 300ohm 60mH 0.5µ

13 A licker problem A series circuit is driven by emfε. Which of the following could be an appropriate phasor diagram? ε m ε m ~ (a) (b) (c) ε m B For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b)the drive voltage lags the current. (c) The drive voltage leads the current.

14 A licker problem A series circuit is driven by emfε. Which of the following could be an appropriate phasor diagram? ~ ε m ε m (a) (b) (c) ε m The phasor diagram for the driven series circuit always has the voltage across the capacitor lagging the current by 90. The vector sum of the and phasors must equal the generator emf phasorε m.

15 B licker problem For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b)the drive voltage lags the current. (c) The drive voltage leads the current. ~, ε m φ First, remember that the current phasor is always in the same orientation as the resistor voltage phasor (since the current and voltage are always in phase). From the diagram, we see that the drive phasorε m is lagging (clockwise). Just as lags by 90, in an A driven circuit, the drive voltage will also lag by some angle less than 90. The precise phase lag φ depends on the values of, and ω.

16 ircuits with phasors where... ω 1 ω The phasor diagram gives us graphical solutions for φ and : ε m φ ε tanφ ε m ε m ( ( ) ) + ( ) ε m + φ + ( )

17 series circuit; Summary of instantaneous urrent and voltages i ( t) cos( ωt) ( t) cos( ωt) v v v ε ( t 1 ω ( t) cos( ωt 90) cos( ωt 90) ( t) cos( ωt + 90) ω cos( ωt + 90) ) v ( t) cos( ωt + φ) ε cos( ωt + φ) ad tanφ m ( ) + ( - )

18 agging & eading The phase φ between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. ε m tan φ ω 1 ω φ φ > φ > 0 current AGS applied voltage < φ < 0 current EADS applied voltage φ 0 current N PHASE WTH applied voltage

19 mpedance, From the phasor diagram we found that the current amplitude was related to the drive voltage amplitude ε m by ε m m is known as the impedance, and is basically the frequency dependent equivalent resistance of the series circuit, given by: mpedance Triangle ε φ ( ) m + m or φ cos( φ) Note that achieves its minimum value () when φ 0. Under this condition the maximum current flows in the circuit.

20 esonance For fixed,, the current will be a maximum at the resonant frequency ω which makes the impedance purely resistive ( ). i.e., ε ε reaches a maximum when: m m + m ( ) This condition is obtained when: 1 ω ω ω 1 Note that this resonant frequency is identical to the natural frequency of the circuit by itself! At this frequency, the current and the driving voltage are in phase: tan φ 0

21 esonance Plot the current versus ω, the frequency of the voltage source: ε m 0 0 ω ω For ω very large, >>, φ 90, 0 For ω very small, >>, φ -90, 0 Example: vary 100 v ω1000 rad/s 00, 500, 000 ohm H 0.5 µ

22 licker: a general A circuit containing a resistor, capacitor, and inductor, driven by an A generator. 1) As the frequency of the circuit is either raised above or lowered below the resonant frequency, the impedance of the circuit. a) always increases b) only increases for lowering the frequency below resonance c) only increases for raising the frequency above resonance ) At the resonant frequency, which of the following is true? a) The current leads the voltage across the generator. b) The current lags the voltage across the generator. c) The current is in phase with the voltage across the generator.

23 mpedance sqrt( + ( - ) ) At resonance, ( - ) 0, and the impedance has its minimum value: As frequency is changed from resonance, either up or down, ( - ) no longer is zero and must therefore increase. hanging the frequency away from the resonant frequency will change both the reductive and capacitive reactance such that - is no longer 0. This, when squared, gives a positive term to the impedance, increasing its value. By definition, at the resonance frequency, max is at its greatest and the phase angle is 0, so the current is in phase with the voltage across the generator.

24 Announcements Finish A circuits (review resonance and discuss power) Move on to electromagnetic (EM) waves Mini-quiz on magnetic induction

25 nduction ooking nduction cooking (another application of magnetic induction) Special induction cookware equires cookware that can sustain magnetic flux Stovetop top does does not become not become hot! hot!!

26 E the E man (a mnemonic for phase relationships in A circuits) Also a heavy metal band from Missouri (Eli the ceman). nspired by A/D?

27 ircuits with phasors where... ω 1 ω The phasor diagram gives us graphical solutions for φ and : ε m φ ε tanφ ε m ε m ( ( ) ) + ( ) ε m + φ + ( )

28 agging & eading The phase φ between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. ε m tan φ ω 1 ω φ φ > φ > 0 current AGS applied voltage < φ < 0 current EADS applied voltage φ 0 current N PHASE WTH applied voltage

29 mpedance, From the phasor diagram we found that the current amplitude was related to the drive voltage amplitude ε m by ε m m is known as the impedance, and is basically the frequency dependent equivalent resistance of the series circuit, given by: mpedance Triangle ε φ ( ) m + m or φ cos( φ) Note that achieves its minimum value () when φ 0. Under this condition the maximum current flows in the circuit.

30 esonance For fixed,, the current will be a maximum at the resonant frequency ω which makes the impedance purely resistive ( ). εm i.e., εm m + ( ) reaches a maximum when: This condition is obtained when: 1 ω ω ω 1 Note that this resonant frequency is identical to the natural frequency of the circuit by itself! At this frequency, the current and the driving voltage are in phase: tan φ 0

31 esonance Plot the current versus ω, the frequency of the voltage ε source: m 0 0 ω ω For ω very large, >>, φ 90, 0 For ω very small, >>, φ -90, 0 Example: vary 100 v ω1000 rad/s 00, 500, 000 ohm H 0.5 µ

32 4) Fill in the blanks. This circuit is being driven its resonance frequency. a) above b) below c) exactly at 5) The generator voltage the current. a) leads b) lags c) is in phase with

33 Power in circuit ( t) ( t) cos( ω t) ; v ( t) ( ωt + φ) i ad cos The instantaneous power delivered to -- is: ( t) i( t) v ( t) cos ( ωt + φ) cos( t) P ad ω We can use trig identities to expand the above to, P ( ) [ cos( ωt) cos( φ) sin( ωt) sin( φ) ] cos( ωt) cos ( ωt) cos( φ) sin( ωt) cos( ωt) sin( φ) P( t) cos ( ωt) cos( φ) sin( ωt) cos( ωt) sin( φ) P t ave P P ave ave cos 1 ( ωt) cos( φ) cos( φ) 1 ( ) cos( φ) cos( φ) P t MS MS cos ( φ)

34 Power in circuit, continued 1 P ave P t ( ) cos( φ) cos( φ) MS General result. is voltage across element, MS MS is current through element, and ϕ is phase angle between them. Example; 100Watt light bulb plugged into 10 house outlet, Pure resistive load (no and no ), ϕ 0. P rms P rms rms ave P ave rms rms rms Ω 0.83A MS Note: 10 house voltage is rms and has peak voltage of Question: What is P AE for an inductor or capacitor? φ90 0

35 f you wanted to increase the power delivered to this circuit, which modification(s) would work? Note: ε fixed. Not a clicker question a) increase b) increase c) increase d) decrease e) decrease f) decrease ) Would using a larger resistor increase the current? a) yes b) no

36 Power ~ cosφ ~ (1/)(/) (/ ) To increase power, want to decrease: : decrease decrease : increase decrease : decrease decrease Since power peaks at the resonant frequency, try to get and to be equal by decreasing and. Power also depends inversely on, so decrease to increase Power.

37 Summary Power P( t) ε cos φ rms power factor rms { 1 ε rms ε m rms 1 m ( ) rms + ( ) tanφ Driven Series ircuit: esonance condition esonant frequency ω 1

Course Updates. Reminders: 1) Assignment #10 due Today. 2) Quiz # 5 Friday (Chap 29, 30) 3) Start AC Circuits

Course Updates. Reminders: 1) Assignment #10 due Today. 2) Quiz # 5 Friday (Chap 29, 30) 3) Start AC Circuits ourse Updates http://www.phys.hawaii.edu/~varner/phys272-spr10/physics272.html eminders: 1) Assignment #10 due Today 2) Quiz # 5 Friday (hap 29, 30) 3) Start A ircuits Alternating urrents (hap 31) In this