AC Circuits with RLC Components

ÎEnormous impact of AC circuits ‹ Power delivery ‹ Radio transmitters and receivers ‹ Tuners ‹ Filters ‹ Transformers

ÎBasic components ‹ R ‹ L ‹ C ‹ Driving emf

ÎNow we will study the basic principles

PHY2049: Chapter 31 18 AC Circuits and Forced Oscillations

ÎRLC + “driving” EMF with angular frequency ωd ε = εωmdsin t di q LRi++=ε sinω t dt C md

ÎGeneral solution for current is sum of two terms

“Transient”: Falls “Steady state”: exponentially & disappears Constant amplitude

Ignore

ie∼ −tR/2 L cosω′ t

PHY2049: Chapter 31 19 Steady State Solution

ÎAssume steady state solution of form iI= mdsin(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=−mdsin()ω tφ di Substitute di q =−ωdmItcos()ωφ d LRi++=ε sinω t dt dt C m ω Im qt=−cos()ωd −φ d

Im Imdω LIRttcos()ωτ d ω φ −+ φ ε ω m sin d −− () cos() d −= m sin d t dC PHY2049: Chapter 31 20

ω Steady State Solution for AC Current (2)

Im Imdω LIRttcos()ωτ d ω φ −+ φ ε ω m sin d −− () cos() d −= m sin d t dC

ÎExpand sin & cos expressions

sin()ω tt−=φωφω sin cos − cos t sinφ dd dω High school trig! () cosωddtt−=φωφω cos cos + sin d t sinφ

ÎCollect sinωdt&cosωdtterms separately

()ωddLC−−=1/ωφφ cos R sin 0 cosωdtterms () ILmdω −+=1/ωφ d C sin IR m φε cos m sinωdtterms

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

PHY2049: Chapter 31 21 Steady State Solution for AC Current (3)

ÎSolve for φ and Im φ ω LCXX−−1/ω ε tan =≡ddLCI = m R R m Z

ÎR, XL, XC and Z have dimensions of resistance

XLLd= ω Inductive “reactance”

XCCd= 1/ω Capacitive “reactance”

2 2 Total “impedance” ZRXX=+−()LC

ÎLet’s try to understand this solution using “phasors”

PHY2049: Chapter 31 22 Graphical Representation: R, XL, XC, Z

X LC− X φ

2 2 ZRXX=+−()LC XX− tanφ = LC R

PHY2049: Chapter 31 23 Understanding AC Circuits Using Phasors

ÎPhasor ‹ Voltage or current represented by “phasor”

‹ Phasor rotates counterclockwise with angular velocity = ωd ‹ Length of phasor is amplitude of voltage (V) or current (I) ‹ y component is instantaneous value of voltage (v) or current (i)

εm ε Im ε = εωsin t md i ωdt − φ

iI=−mdsin()ω tφ Current “lags” voltage by φ

PHY2049: Chapter 31 24 AC Source and Resistor Only i ÎVoltage is viRVRRd==sinω t ε ~ R ÎRelation of current and voltage iI==RdRRsinω tI VR /

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

IR VR ωdt

PHY2049: Chapter 31 25 AC Source and Capacitor Only i ÎVoltage is vqCVCCd==/sinω t ÎDifferentiate to find current qCV= Cdsinω t ε ~ C idqdt==/cosωdC CVω d t ÎRewrite using phase

iCVt=+°ωdCsin(ω d 90 ) ÎRelation of current and voltage VC IC ωdt + 90 iI=+°Cdsin()ω t 90 ωdt ΓCapacitive reactance”: XCCd=1/ω ‹ Current “leads” voltage by 90°

PHY2049: Chapter 31 26 AC Source and Inductor Only i ÎVoltage is vLLd== Ldi/sin dt Vω t ÎIntegrate di/dt to find current:

di//sin dt= () VLd Lω t ε ~ L iV=−()Ld/cosω Lω d t ÎRewrite using phase iV=−°()Ld/sin90ω L(ω d t ) ÎRelation of current and voltage VL iI=−°Ldsin()ω t 90 ωdt ΓInductive reactance”: XLLd= ω ‹ Current “lags” voltage by 90° ω t − 90 d IL

PHY2049: Chapter 31 27 What is Reactance?

Think of it as a frequency-dependent resistance

1 X C = Shrinks with increasing ωd ωdC

Grows with increasing ω X L = ωd L d

("X R "= R ) Independent of ωd

PHY2049: Chapter 31 28 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 vs frequency. Which curve corresponds to an inductive circuit? ‹ (1) a ‹ (2) b ‹ (3) c a ‹ (4) Can’t tell without more info b Im c

fd For inductor, higher frequency gives higher XLLd= ω reactance, therefore lower current

PHY2049: Chapter 31 29 AC Source and RLC Circuit

Im ÎVoltage is ε = εωmdsin t ‹ R, L, C all present

ÎRelation of current and voltage

iI=−mdsin()ω tφ

‹ Current “lags” voltage by φ I ‹ Impedance: Due to R, X and X εm C L VR VL ωdt − φ

ÎCalculate Im and φ ‹ See next slide

VC

PHY2049: Chapter 31 30 AC Source and RLC Circuit (2)

ÎSolution using trig (or geometry)

‹ Magnitude = Im, lags emf by phase φ)

iI=−mdsin()ω tφ

IZmm=φε / XX−−ωω L1/ C tan ==LC d d RR 2222 Z =+−RXX()LC =+ R() d L −1/ d C

PHY2049: Chapter 31ωω 31 AC Source and RLC Circuit (3)

XX− 2 tanφ = LC ZRXX=+−2 () R LC

ÎWhen XL = XC, then φ = 0 ‹ Current in phase with emf, “Resonant circuit”: ωd ==ω0 1/ LC ‹ Z = R (minimum impedance, maximum current)

ÎWhen XL < XC, then φ < 0

‹ Current leads emf, “Capacitive circuit”: ωd < ω0

ÎWhen XL > XC, then φ > 0

‹ Current lags emf, “Inductive circuit”: ωd > ω0

PHY2049: Chapter 31 32 Graphical Representation: VR, VL, VC, εm

εm VVL − C φ

2 2 εmRLC=+−VVV()

XX− VV− tanφ ==L CLC RVR

PHY2049: Chapter 31 33 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)

‹ ωd < ω0

I ε t

PHY2049: Chapter 31 34 Quiz

ÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by inductance? ‹ (1) Circuit 1 ‹ (2) Circuit 2 ‹ (3) Circuit 3 Inductive: Current lags emf, φ>0

εm εm εm 1 2 3

PHY2049: Chapter 31 35 Quiz

ÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by capacitance? ‹ (1) Circuit 1 Capacitive: Current leads emf, φ<0 ‹ (2) Circuit 2 ‹ (3) Circuit 3

εm εm εm 1 2 3

PHY2049: Chapter 31 36 RLC Example 2

ÎR = 200Ω, C = 15μF, L = 230mH, εm = 36v, fd = 60 Hz

‹ ωd = 120π = 377 rad/s

−6 ‹ Natural frequency ω0 =×=1/() 0.230( 15 10) 538rad/s

‹ X L =×377 0.23 = 86.7 Ω XL < XC Capacitive circuit −6 ‹ XC =××=Ω1/() 377 15 10 177

2 2 ‹ Z =+−=Ω200() 86.7 177 219

‹ IZmm==ε / 36/ 219 = 0.164A

−1 ⎛⎞86.7− 177 Current leads emf ‹ φ ==−°tan⎜⎟ 24.3 (as expected) ⎝⎠200 PHY2049: Chapter 31 37 RLC Example 2 (cont)

Î VIXLmL==×=0.164 86.7 14.2V

Î VIXCmC==×=0.164 177 29.0V εm VVL − C φ Î VIRRm==0.164 ×= 200 32.8V

2 Î 2 32.8+−() 14.2 29.0 == 36.0 εm

PHY2049: Chapter 31 38 RLC Example 3

ÎCircuit parameters: C = 2.5μF, L = 4mH, εm = 10v 1/2 4 ‹ ω0 = (1/LC) = 10 rad/s

‹ Plot Im vs ωd / ω0

R = 5Ω R = 10Ω R = 20Ω Im Resonance ωd = ω0

ωd /ω0 PHY2049: Chapter 31 39 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

PHY2049: Chapter 31 40 Power in AC Circuits (2)

2 ÎRecall power formula Pave= IR rms

⎛⎞εrms R ÎRewrite PIRIIave===⎜⎟ rmsε rms rmsε rms rms cosφ ⎝⎠Z Z R Z P = εφI cos cosφ = XX− ave rms rms Z L C φ R Îcosφ is the “power factor” ‹ To maximize power delivered to circuit ⇒ make φ close to zero ‹ Most power delivered to load happens near resonance

‹ E.g., too much inductive reactance (XL) can be cancelled by increasing XC (decreasing C)

PHY2049: Chapter 31 41