AC Power • Resonant Circuits • Phasors (2-Dim Vectors, Amplitude and Phase) What Is Reactance ? You Can Think of It As a Frequency-Dependent Resistance

Physics-272 Lecture 20

• AC Power • Resonant Circuits • Phasors (2-dim vectors, amplitude and phase) What is reactance ? You can think of it as a frequency-dependent resistance.

1 For high ω, χ ~0 X = C C ωC - Capacitor looks like a wire (“short”) For low ω, χC∞ - Capacitor looks like a break

For low ω, χL~0 - Inductor looks like a wire (“short”) XL = ω L For high ω, χL∞ - Inductor looks like a break (inductors resist change in current)

("XR "= R ) An RL circuit is driven by an AC 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 ω generator. (X L= L) Alternating Currents: LRC circuit

Figure (b) has

XL>X C and (c) has XL

Using Phasors, we can construct the phasor diagram for an LRC Circuit. This is similar to 2-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 AC source. Kirchoff’s voltage law holds for AC circuits.

Also V R and I are in phase. Phasors

R ε ω Problem : Given Vdrive = m sin( t), C L find VR, VL, VC, IR, IL, IC ε ∼

Strategy : We will use Kirchhoff’s voltage law that the (phasor) sum of the voltages VR, VC, and VL must equal Vdrive . Phasors, cont. R ε ω Problem : Given Vdrive = m sin( t), C L find VR, VL, VC, IR, IL, IC ε ∼

2. Next draw the phasor for VL. Since the VL = I XL inductor voltage VL always leads IL V = I R draw VL 90˚ further counterclockwise . The R length of the VL phasor is IL XL = I ωωωL Phasors, cont. R Problem : Given V = ε sin( ωt), drive m C L V V V I I I find R, L, C, R, L, C ε ∼

1. Draw VR phasor along x-axis (this direction is chosen for convenience). Note that since VR = IRR, this is also the direction of the current VR, IRR phasor iR. Because of Kirchhoff’s current ≡ law, IL = IC = IR I (i.e., the same current flows through each element). Phasors, cont. R ε ω Problem : Given Vdrive = m sin( t), C L find VR, VL, VC, IR, IL, IC ε ∼

V = IX 3. The capacitor voltage VC always lags L L IC draw VC 90˚ further clockwise. VR = I R The length of the VC phasor is IC XC = I /ωωωC VC = I XC

The lengths of the phasors depend on R, L, C, and ω.

The relative orientation of the VR, VL, and VC phasors is always the way we have drawn it. Memorize it! Phasors, cont. R ε ω Problem : Given Vdrive = m sin( t), C L find VR, VL, VC, IR, IL, IC ε ∼

V • The phasors for VR, VL, and VC are L εm added like vectors to give the drive

voltage VR + VL + VC = εm : VR VC •From this diagram we can now easily calculate quantities of interest, like the net current I , the maximum voltage across any of the elements, and the phase between the current the drive voltage (and thus the power). Voltage V(t) across AC source

2 2 VR = IR v()()()t = VR + VL −VC cos ()ωt +φ VL = IX L 2 2 = ()()IR + IX L - IX C cos ()ωt + φ VC = IX C 2 2 = I ()R + () X L - X C cos ()()ωt + φ = IZ cos ωt + φ

2 2 Z = ()()R + X L - X C Z is called “impedance”

VL − VC ωL − /1 ωC tan φ = = VR R Also: MAX V = V = I Z Like: V R = IR

Vrms = I rms Z i = I cos ωt v = V cos( ωt + φ) LRC series circuit; Summary of instantaneous Current and voltages

VR = IR

VL = IX L

VC = IX C

(ti )= I cos (ωt)

vR ()()t = IR cos ωt 1 v ()()t = IX cos ωt − 90 = I cos ()ωt − 90 C C ωC

vL ()()()t = IX L cos ωt + 90 = IωLcos ωt + 90

2 2 vad ()()()t = I X R + X L - X C cos ()ωt + φ

VL − VC ωL − /1 ωC tan φ = = VR R Alternating Currents: LRC circuit, Fig. 31.11 Y&F Example 31.4

V=50v ωωω=10000rad/s R=300ohm L=60mH C=0.5 µµµC Clicker problem 2A A series RC circuit is driven by

emf ε. Which of the following ~ could be an appropriate phasor diagram? V ε L εm m VC VR

VR VR V C εm VC (a)(b) (c)

2B 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. Clicker problem 2A A series RC circuit is driven by

emf ε. Which of the following ~ could be an appropriate phasor diagram?

V ε L εm m VC VR

VR VR V C εm VC (a)(b) (c)

• The phasor diagram for the driven series RLC circuit always has the voltage across the capacitor lagging the current by 90 °. The vector sum of the VC and VR phasors must equal the generator emf phasor εm. Clicker problem For this circuit which of the following ~ is true? 2B (a)The drive voltage is in phase with the current. (b)The drive voltage lags the current. (c)The drive voltage leads the current.

First, remember that the current phasor I is always in the same orientation as the resistor voltage phasor VR (since the current and voltage VR , I are always in phase). From the diagram, we see φφφ that the drive phasor εm is lagging (clockwise) I. Just as V lags I by 90 °, in an AC driven RC ε C m circuit, the drive voltage will also lag I by some angle less than 90°. The precise phase lag φ φ depends on the values of R, C and ωωω. LRC Circuits with phasors…

I X where . . . R L εm X L ≡ ωL C L ⇒⇒⇒ φφφ IR 1 ∼ X C ≡ I X C ωC ε The phasor diagram gives us graphical solutions for φφφ and I: 2 2 2 2 ε m = I (R + (X L − X C ) ) I X L I X C

εm ⇓⇓⇓ φφφ 2 2 X − X ε m = I R + ()X L − X C = IZ IR tan φ = L C R 2 2 Z ≡ R + ()X L − X C LRC series circuit; Summary of instantaneous Current and voltages

VR = IR

VL = IX L

VC = IX C

(ti )= I cos (ωt)

vR ()()t = IR cos ωt 1 v ()()t = IX cos ωt − 90 = I cos ()ωt − 90 C C ωC

vL ()()()t = IX L cos ωt + 90 = IωLcos ωt + 90

ε (t) = vad ()()()t = IZ cos ωt + φ = ε m cos ωt + φ V V L − C X L − X C 2 2 tan φ = = Z = ()()X R + X L - X C VR R Lagging & Leading The phase φφφbetween the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. X L ε X L − X C L ≡ ω I = m tan φ = R 1 Z X ≡ C ω C XL

Z XL XL φφφ Z R φφφ R R Z XC XC XC

XL > XC XL < XC XL = XC φφφ> 0 φφφ< 0 φφφ= 0 current current current LAGS LEADS IN PHASE WITH applied voltage applied voltage applied voltage Impedance, Z • From the phasor diagram we found that the current amplitude

I was related to the drive voltage amplitude εεεm by ε = I Z m m • Z is known as the “impedance”, and is basically the frequency dependent equivalent resistance of the series LRC circuit, given by: Z “ Impedance IZ X X I X − X L − C Triangle” L C | |φ | |φ I R R

εm 2 2 R Z≡= RXX +()L − C or Z = Im cos(φ )

• Note that Z achieves its minimum value ( R) when φφφ = 0. Under this condition the maximum current flows in the circuit. Resonance • For fixed R, C, L the current I will be a maximum at the resonant frequency ω which makes the impedance Z

purely resistive ( Z = R ). i.e., εm εm Im = = Z 2 2 R +()XL −XC reaches a maximum when: XL= X C

This condition is obtained when: 1 1 ωL = ⇒⇒⇒ ω = ωC LC

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! • At this frequency, the current and the driving voltage are in phase: X − X tan φ = L C = 0 R Resonance Plot the current versus ω, the frequency of the voltage source: εεεm R • For ωωωvery large, XL >> XC, φφφ 90 °, I 0 I • For ωωωvery small, XC >> XL, φφφ -90 °, I 0 0 0 222ωωω ωωω

Example: vary R V=100 v ωωω=1000 rad/s R=200, 500, 2000 ohm L=2 H C=0.5 µµµC Clicker: a general AC circuit containing a resistor, capacitor, and inductor, driven by an AC 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

2) 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. 2 2 • Impedance = Z = sqrt ( R + (X L-XC) ) • At resonance, ( XL-XC) = 0, and the impedance has its minimum value: Z = R

• As frequency is changed from resonance, either up or down, ( XL-XC) no longer is zero and Z must therefore increase.

Changing the frequency away from the resonant frequency will change both the

reductive and capacitive reactance such that XL -XC is no longer 0. This, when squared, gives a positive term to the impedance, increasing its value. By

definition, at the resonance frequency, Imax is at its greatest and the phase angle is 0, so the current is in phase with the voltage across the generator. Announcements

Finish AC circuits (review resonance and discuss power)

Move on to electromagnetic (EM) waves

Mini-quiz on magnetic induction Stove top does does top Stove !! hot become not Stovetop does not not does Stovetop ! hot become

Induction cooking Induction of application (another induction) magnetic Induction Cooking Induction Requires cookware that that cookware Requires flux magnetic sustain can Special induction induction Special cookware ELI the ICE man (a mnemonic for phase relationships in AC circuits)

Also a heavy metal band from Missouri (Eli the Iceman). Inspired by “AC/DC” ? LRC Circuits with phasors…

εm ⇓⇓⇓ φφφ 2 2 X − X ε m = I R + ()X L − X C = IZ IR tan φ = L C R 2 2 Z ≡ R + ()X L − X C Lagging & Leading The phase φφφbetween the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. X L ε X L − X C L ≡ ω I = m tan φ = R 1 Z X ≡ C ω C XL

Z XL XL φφφ Z R φφφ R R Z XC XC XC

εm 2 2 R Z≡= RXX +()L − C or Z = Im cos(φ )

This condition is obtained when: 1 1 ωL = ⇒⇒⇒ ω = ωC LC

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! • At this frequency, the current and the driving voltage are in phase: X − X tan φ = L C = 0 R Resonance Plot the current versus ω, the frequency of the voltage εεε source:m R • For ωωωvery large, XL >> XC, φφφ 90 °, I I 0

• For ωωωvery small, XC >> XL, φφφ -90 °, I 0 0 0 222ωωω ωωω

Example: vary R V=100 v ωωω=1000 rad/s R=200, 500, 2000 ohm L=2 H C=0.5 µµµC 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 Power in LRC circuit

(ti )= I(t)cos (ωt); vad (t) = V cos (ωt + φ)

The instantaneous power delivered to L-R-C is:

P(t)= (ti )vad (t)= V cos (ωt + φ)I cos (ωt) We can use trig identities to expand the above to,

P(t)= V[cos (ωt)cos (φ)− sin (ωt)sin (φ)]I cos (ωt) = VI cos 2 ()()()()()ωt cos φ −VI sin ωt cos ωt sin φ 2 Pave = P()t = VI cos ()ωt cos ()()()φ −VI sin ωt cos ωt sin ()φ  1  = VI cos 2 ()ωt cos ()φ = VI  cos ()φ  2  1 V I Pave = P()t = VI cos ()φ = cos ()φ 2 2 2

Pave = VRMS I RMS cos ()φ Power in LRC circuit, continued 1 P = P()t = VI cos ()()φ = V I cos φ ave 2 RMS RMS

General result. V RMS is voltage across element, I RMS is current through element, and ϕ is phase angle between them. Example; 100Watt light bulb plugged into 120V house outlet, Pure resistive load (no L and no C), ϕ = 0. V 2 P = I V = rms rms rms R V 2 120 2 R = rms = = 144 Ω Pave 100

Pave 100 I rms = = = 83.0 A Vrms 120 Note: 120V house voltage is rms and has peak voltage of 120 √√√2 = 170V

Question: What is P AVE for an inductor or capacitor? φ=90 0 Not a clicker question If you wanted to increase the power delivered to this RLC circuit, which modification(s) would work? Note: ε fixed.

a) increase R d) decrease R b) increase C e) decrease C c) increase L f) decrease L

II) Would using a larger resistor increase the current?

a) yes b) no • Power ~ Icos φ ~ (1/ Z)( R/Z) = ( R/Z2)

• To increase power, want Z to decrease: L • : decrease XL ⇒ decrease L C • : increase XC ⇒ decrease C • R: decrease Z ⇒ decrease R

Since power peaks at the resonant frequency, try to get XL and XC to be equal by decreasing L and C. Power also depends inversely on R, so decrease R to increase Power. Summary • Power “power factor” 1

{ 1 ε rms ≡ ε m I ≡ I 2 rms 2 m 〈P( t ) 〉 φ=ε rms I rms cos 2 2 2 X − X =()Irms R Z ≡ R + ()X − X tan φ = L C L C R

1 ω = • Driven Series LRC Circuit: LC • Resonance condition • Resonant frequency