Chapter 21: RLC Circuits

PHY2054: Chapter 21 1 Voltage and Current in RLC Circuits

ÎAC emf source: “driving frequency” f

ε = εωm sin t ω = 2π f

ÎIf circuit contains only R + emf source, current is simple ε ε iI==sin()ω tI =m () current amplitude R mmR

ÎIf L and/or C present, current is not in phase with emf ε iI=−=sin()ωφ t I m mmZ

ÎZ, φ shown later

PHY2054: Chapter 21 2 AC Source and Resistor Only

ÎDriving voltage is ε = εωm sin t i ÎRelation of current and voltage iR= ε / ε ~ R ε iI==sinω tI m mmR

‹ Current is in phase with voltage (φ = 0)

PHY2054: Chapter 21 3 AC Source and Capacitor Only q ÎVoltage is vt==ε sinω CmC ÎDifferentiate to find current i qC= εm sinω t

idqdtCV==/cosω C ω t ε ~ C ÎRewrite using phase (check this!)

iCVt=+°ω C sin()ω 90

ÎRelation of current and voltage

εm iI=+°=mmsin()ω t 90 I (XC =1/ωC) XC

ΓCapacitive reactance”: XCC =1/ω ‹ Current “leads” voltage by 90°

PHY2054: Chapter 21 4 AC Source and Inductor Only

ÎVoltage is vLm== Ldi/sin dtε ω t ÎIntegrate di/dt to find current: i di//sin dt= ()εm Lω t

iLt=−()εm /cosωω ε ~ L ÎRewrite using phase (check this!)

iLt=−°()εm /sin90ωω( )

ÎRelation of current and voltage

εm iI=−°=mmsin()ω t 90 I (XLL = ω ) X L

ΓInductive reactance”: XLL = ω ‹ Current “lags” voltage by 90°

PHY2054: Chapter 21 5 General Solution for RLC Circuit

ÎWe assume steady state solution of form iI= m sin(ω t−φ ) ‹ Im is current amplitude ‹ φ is phase by which current “lags” the driving EMF

‹ Must determine Im and φ ÎPlug in solution: differentiate & integrate sin(ωt-φ)

iI=−m sin()ω tφ di Substitute di q =−ωItm cos()ωφ LRi++=ε sinω t dt dt C m I qt=−m cos()ω −φ ω

I I ωLtcos()ωφ−+ ωφ IRt ωφ sin ε −− () ω m cos() t −= sin t mmωC m PHY2054: Chapter 21 6 General Solution for RLC Circuit (2) I I ωLtcos()ωφ−+ ωφ IRt ωφ sin ε −− () ω m cos() t −= sin t mmωC m

ÎExpand sin & cos expressions

sin()ωtt−=φωφω sin cos − cos t sinφ High school trig! cos()ωtt−=φωφω cos cos + sin t sinφ

ÎCollect sinωt&cosωtterms separately

()ωLC−−=1/ωφ cos R sin φ 0 cosωtterms () ILmmmω −+=1/ωφ C sin IR φε cos sinωtterms

ÎThese equations can be solved for Im and φ (next slide)

PHY2054: Chapter 21 7 General Solution for RLC Circuit (3)

ÎSolve for φ and Im φ ωLC−1/ω XX− ε tan =≡LCI = m R R m Z

ÎR, XL, XC and Z have dimensions of resistance

XLL = ω Inductive “reactance”

XCC =1/ω Capacitive “reactance”

2 2 Total “impedance” ZRXX=+−()LC

ÎThis is where φ, XL, XC and Z come from!

PHY2054: Chapter 21 8 AC Source and RLC Circuits ε I = m Maximum current m Z XX− tanφ = LC Phase angle R

Inductive reactance XLL ==ω ()ωπ2 f ω Capacitive reactance XCC =1/

2 2 Total impedance ZRXX=+−()LC

φ= angle that current “lags” applied voltage

PHY2054: Chapter 21 9 What is Reactance?

Think of it as a frequency-dependent resistance

1 X = Shrinks with increasing ω C ωC

Grows with increasing ω X L = ωL

("X R "= R ) Independent of ω

PHY2054: Chapter 21 10 Pictorial Understanding of Reactance

2 2 ZRXX=+−()LC

XX− tanφ = LC R

R cosφ = Z

PHY2054: Chapter 21 11 Summary of Circuit Elements, Impedance, Phase Angles

XXLC− 2 2 tanφ = ZRXX=+−()LC R

PHY2054: Chapter 21 12 Quiz

ÎThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured as a function of frequency. Which one of the following curves corresponds to an inductive circuit? ‹ (1) a ‹ (2) b a ‹ (3) c b ‹ (4) Can’t tell without more info Imax c

f XLL ==ω ()ωπ2 f ε For inductor, higher frequency gives higher IXmax= max / L reactance, therefore lower current

PHY2054: Chapter 21 13 RLC Example 1

ÎBelow are shown the driving emf and current vs time of an RLC circuit. We can conclude the following ‹ Current “leads” the driving emf (φ<0)

‹ Circuit is capacitive (XC > XL)

I ε t

PHY2054: Chapter 21 14 RLC Example 2

ÎR = 200Ω, C = 15μF, L = 230mH, εmax = 36v, f = 60 Hz

‹ X L =××2π 60 0.23 = 86.7 Ω XC > XL Capacitive circuit −6 ‹ XC =×××=Ω1/() 2π 60 15 10 177

2 ‹ Z =+−=Ω2002 () 86.7 177 219

‹ IZmax===ε max / 36/ 219 0.164A

−1 ⎛⎞86.7− 177 Current leads emf ‹ φ ==−°tan⎜⎟ 24.3 (as expected) ⎝⎠200

it=+°0.164sin(ω 24.3 )

PHY2054: Chapter 21 15 Resonance

ÎConsider impedance vs frequency

2222 Z =+−RXX()LC =+− R()ωω L1/ C

ÎZ is minimum when ωLC=1/ω ωω==0 1/ LC ‹ This is resonance!

ÎAt resonance ‹ Impedance = Z is minimum

‹ Current amplitude = Im is maximum

PHY2054: Chapter 21 16 Imax vs Frequency and Resonance

ÎCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v 1/2 ‹ f0 = 1 / 2π(LC) = 1590 Hz ‹ Plot Imax vs f 2 2 Imax =+−10/RLC()ωω 1/

R = 5Ω R = 10Ω R = 20Ω Imax Resonance ff= 0

f / f0 PHY2054: Chapter 21 17 Power in AC Circuits

ÎInstantaneous power emitted by circuit: P = i2R 22 PIR=−mdsin ()ω tφ Instantaneous power oscillates ÎMore useful to calculate power averaged over a cycle ‹ Use <…> to indicate average over a cycle

221 2 P =−=IRmdsin (ωφ t) 2 IR m

ÎDefine RMS quantities to avoid ½ factors in AC circuits

Im εm 2 Irms = εrms = Pave= IR rms 2 2

ÎHouse current

‹ Vrms = 110V ⇒ Vpeak = 156V

PHY2054: Chapter 21 18 Power in AC Circuits

Î 2 Power formula Pave= IR rms IIrms= max /2

εrms εrms ÎRewrite using I = PIRIave== rmsε rms rms cosφ rms Z Z

Z R XX− Pave= εφ rmsI rms cos cosφ = L C Z φ R

Îcosφ is the “power factor” ‹ To maximize power delivered to circuit ⇒ make φ close to zero ‹ Max power delivered to load happens at resonance

‹ E.g., too much inductive reactance (XL) can be cancelled by increasing XC (e.g., circuits with large motors)

PHY2054: Chapter 21 19 Power Example 1

ÎR = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz

2 2 ‹ Z =+−=Ω200() 80 150 211.9

‹ IZrms==ε rms / 120/ 211.9 = 0.566A

−1 ⎛⎞80− 150 Current leads emf ‹ φ ==−°tan⎜⎟ 19.3 Capacitive circuit ⎝⎠200

‹ cosφ = 0.944

‹ Pave==××=ε rmsI rms cosφ 120 0.566 0.944 64.1W Same 22 ‹ PIRave== rms 0.566 ×= 200 64.1W

PHY2054: Chapter 21 20 Power Example 1 (cont)

ÎR = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz ÎHow much capacitance must be added to maximize the power in the circuit (and thus bring it into resonance)?

‹ Want XC = XL to minimize Z, so must decrease XC

‹ XfCCC =Ω=150 1/ 2π = 17.7μF

‹ XXCLnew==Ω80 C new = 33.2μF

‹ So we must add 15.5μF capacitance to maximize power

PHY2054: Chapter 21 21 Power vs Frequency and Resonance

ÎCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v 1/2 ‹ f0 = 1 / 2π(LC) = 1590 Hz

‹ Plot Pave vs f for different R values

R = 2Ω R = 5Ω R = 10Ω P Resonance ave R = 20Ω ff= 0

f / f0 PHY2054: Chapter 21 22 Resonance Tuner is Based on Resonance Vary C to set resonance frequency to 103.7 (ugh!) Circuit response Q = 500 Other radio stations. Tune for f = 103.7 MHz RLC response is less

PHY2054: Chapter 21 23 Quiz

ÎA generator produces current at a frequency of 60 Hz with peak voltage and current amplitudes of 100V and 10A, respectively. What is the average power produced if they are in phase? ‹ (1) 1000 W ‹ (2) 707 W 1 ‹ (3) 1414 W Pave==2 εε peakII peak rms rms ‹ (4) 500 W ‹ (5) 250 W

PHY2054: Chapter 21 24 Quiz

ÎThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken? ‹ (1) Increase R ‹ (2) Decrease L ‹ (3) Increase L ‹ (4) Increase C

XX− tanφ = LC R

Current lags applied emf (φ > 0), thus circuit is inductive. Either

(1) Reduce XL by decreasing L or (2) Cancel XL by increasing XC (decrease C).

PHY2054: Chapter 21 25 Example: LR Circuit

ÎVariable frequency EMF source with εm=6V connected to a resistor and inductor. R=80Ω and L=40mH.

‹ At what frequency f does VR = VL?

XLRL ==⇒=ω ωπ2000 f = 2000/ 2 = 318Hz

‹ At that frequency, what is phase angle φ?

tanφ ==⇒=°XRL / 1φ 45

‹ What is the current amplitude and RMS value? 22 Imax=+==ε max / 80 80 6/113 0.053A

IIrms== max / 2 0.037A it= 0.053sin(ω −° 45 )

PHY2054: Chapter 21 26 Transformers

ÎPurpose: change alternating (AC) voltage to a bigger (or smaller) value

Input AC voltage Changing flux in in the “primary” “secondary” turns turns produces a flux induces an emf

ΔΦ ΔΦ B VN= B VNss= ppΔt Δt

Ns VVsp= N p

PHY2054: Chapter 21 27 Transformers

ÎNothing comes for free, however! ‹ Increase in voltage comes at the cost of current. ‹ Output power cannot exceed input power! ‹ power in = power out ‹ (Losses usually account for 10-20%)

iVp pss= iV

i VN s ==pp iVNps s

PHY2054: Chapter 21 28 Transformers: Sample Problem

ÎA transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120 V and the output current is 15.0 A. What is the output voltage and input current?

Ns ⎛⎞1240 “Step-up” VVsp==120⎜⎟ = 451V transformer N p ⎝⎠330

V ⎛⎞451 ii==s 15 = 56.4A iVpp= iV ss ps ⎜⎟ Vp ⎝⎠120

PHY2054: Chapter 21 29 Transformers

¾ This is how first experiment by Faraday was done ¾ He only got a deflection of the galvanometer when the switch is opened or closed ¾ Steady current does not make induced emf.

PHY2054: Chapter 21 30 Applications

Microphone

Tape recorder

PHY2054: Chapter 21 31 ConcepTest: Power lines ÎAt large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit (high V, low i) or (high i, low V)?

‹ (1) high V, low i ‹ (2) low V, high i ‹ (3) makes no difference Power loss is i2R

PHY2054: Chapter 21 32 Electric Power Transmission

i2R: 20x smaller current ⇒ 400x smaller power loss

PHY2054: Chapter 21 33