AC Circuits with RLC ComponentsPHY2049: Chapter 31 18 AC Circuits with RLC Components ÎEnormous impact of AC circuits Power delivery Radio transmitters and receivers Tuners Filters

PHY2049: Chapter 31 18

AC Circuits with RLC ComponentsEnormous impact of AC circuits

Power deliveryRadio transmitters and receiversTunersFiltersTransformers

Basic componentsRLCDriving emf

Now we will study the basic principles

PHY2049: Chapter 31 19

AC Circuits and Forced OscillationsRLC + “driving” EMF with angular frequency ωd

General solution for current is sum of two terms

sinm dtε ε ω=

sinm ddi qL Ri tdt C

ε ω+ + =

“Transient”: Fallsexponentially & disappears

“Steady state”:Constant amplitude

Ignore

/ 2 costR Li e tω− ′∼

PHY2049: Chapter 31 20

Steady State SolutionAssume steady state solution of form

Im is current amplitudeφ is phase by which current “lags” the driving EMFMust determine Im and φ

Plug in solution: differentiate & integrate sin(ωt-φ)

( )sinm di I tω φ= −

( ) ( ) ( )cos sin cos sinmm d d m d d m d

d

II L I R t t tC

ω ω τ φ ω φ ω φ ε ωω

− + − − − =

sinmdi qL Ri tdt C

ε ω+ + =

( )sinm di I tω φ= −

( )cosd m ddi I tdt

ω ω φ= −

( )cosmd

d

Iq tω φω

= − −

Substitute

PHY2049: Chapter 31 21

Steady State Solution for AC Current (2)

Expand sin & cos expressions

Collect sinωdt & cosωdt terms separately

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

( )( )

1/ cos sin 0

1/ sin cosd d

m d d m m

L C R

I L C I R

ω ω φ φ

ω ω φ φ ε

− − =

− + =

( )( )

sin sin cos cos sin

cos cos cos sin sind d d

d d d

t t t

t t t

ω φ ω φ ω φ

ω φ ω φ ω φ

− = −

− = +High school trig!

cosωdt terms

sinωdt terms

( ) ( ) ( )cos sin cos sinmm d d m d d m d

d

II L I R t t tC

ω ω τ φ ω φ ω φ ε ωω

− + − − − =

PHY2049: Chapter 31 22

Solve for φ and Im

R, XL, XC and Z have dimensions of resistance

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

Steady State Solution for AC Current (3)

1/tan d d L CL C X XR R

ω ωφ − −= ≡ m

mIZε

=

( )22L CZ R X X= + −

L dX Lω=

1/C dX Cω=

Inductive “reactance”

Capacitive “reactance”

Total “impedance”

PHY2049: Chapter 31 23

Graphical Representation: R, XL, XC, Z

tan L CX XR

φ−

=

L CX X−φ

( )22L CZ R X X= + −

PHY2049: Chapter 31 24

Understanding AC Circuits Using PhasorsPhasor

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)

εmIm

ωdt − φ

( )sinm di I tω φ= −

sinm dtε ε ω= i

ε

Current “lags” voltage by φ

PHY2049: Chapter 31 25

AC Source and Resistor OnlyVoltage is

Relation of current and voltage

Current is in phase with voltage (φ = 0)

IR

i

ε R~sin /R d R Ri I t I V Rω= =

VR

sinR R dv iR V tω= =

ωdt

PHY2049: Chapter 31 26

AC Source and Capacitor OnlyVoltage is

Differentiate to find current

Rewrite using phase

Relation of current and voltage

“Capacitive reactance”:Current “leads” voltage by 90°

sinC dq CV tω=

VCIC

i

ε C~/ cosd C di dq dt CV tω ω= =

/ sinC C dv q C V tω= =

( )sin 90d C di CV tω ω= + °

( )sin 90C di I tω= + ° ωdt + 90

ωdt1/C dX Cω=

PHY2049: Chapter 31 27

AC Source and Inductor OnlyVoltage is

Integrate di/dt to find current:

Rewrite using phase

Relation of current and voltage

“Inductive reactance”:Current “lags” voltage by 90°

( )/ / sinL ddi dt V L tω=

VL

IL

i

ε L~( )/ cosL d di V L tω ω= −

/ sinL L dv Ldi dt V tω= =

( ) ( )/ sin 90L d di V L tω ω= − °

ωdt − 90

( )sin 90L di I tω= − °ωdt

L dX Lω=

PHY2049: Chapter 31 28

What is Reactance?Think of it as a frequency-dependent resistance

Shrinks with increasing ωd

1C

d

XCω

=

L dX Lω=

( " " )RX R=

Grows with increasing ωd

Independent of ωd

PHY2049: Chapter 31 29

QuizThree 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(4) Can’t tell without more info

fd

Im

a

c

b

L dX Lω= For inductor, higher frequency gives higherreactance, therefore lower current

PHY2049: Chapter 31 30

AC Source and RLC CircuitVoltage is

R, L, C all present

Relation of current and voltage

Current “lags” voltage by φImpedance: Due to R, XC and XL

Calculate Im and φSee next slide

VR

VC

sinm dtε ε ω=

VL

εm

Im

ωdt − φ

( )sinm di I tω φ= −

I

PHY2049: Chapter 31 31

AC Source and RLC Circuit (2)Solution using trig (or geometry)

Magnitude = Im, lags emf by phase φ)

( ) ( )2 22 2

/1/tan

1/

m m

L C d d

L C d d

I ZX X L C

R R

Z R X X R L C

εω ωφ

ω ω

=− −

= =

= + − = + −

( )sinm di I tω φ= −

PHY2049: Chapter 31 32

AC Source and RLC Circuit (3)

When XL = XC, then φ = 0Current in phase with emf, “Resonant circuit”:Z = R (minimum impedance, maximum current)

When XL < XC, then φ < 0Current leads emf, “Capacitive circuit”: ωd < ω0

When XL > XC, then φ > 0Current lags emf, “Inductive circuit”: ωd > ω0

tan L CX XR

φ −=

0 1/d LCω ω= =

( )22L CZ R X X= + −

PHY2049: Chapter 31 33

Graphical Representation: VR, VL, VC, εm

tan L C L C

R

X X V VR V

φ− −

= =

L CV V−φ

mε

( )22m R L CV V Vε = + −

PHY2049: Chapter 31 34

RLC Example 1Below 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

εI

t

PHY2049: Chapter 31 35

QuizWhich one of these phasor diagrams corresponds to an RLC circuit dominated by inductance?

(1) Circuit 1(2) Circuit 2(3) Circuit 3

εm εm εm1 32

Inductive: Current lags emf, φ>0

PHY2049: Chapter 31 36

QuizWhich one of these phasor diagrams corresponds to an RLC circuit dominated by capacitance?

(1) Circuit 1(2) Circuit 2(3) Circuit 3

εm εm εm1 32

Capacitive: Current leads emf, φ<0

PHY2049: Chapter 31 37

RLC Example 2R = 200Ω, C = 15μF, L = 230mH, εm = 36v, fd = 60 Hz

ωd = 120π = 377 rad/s

Natural frequency ( )( )60 1/ 0.230 15 10 538rad/sω −= × =

377 0.23 86.7LX = × = Ω

( )61/ 377 15 10 177CX −= × × = Ω

( )22200 86.7 177 219Z = + − = Ω

/ 36 / 219 0.164Am mI Zε= = =

XL < XCCapacitive circuit

1 86.7 177tan 24.3200

φ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

Current leads emf(as expected)

PHY2049: Chapter 31 38

RLC Example 2 (cont)

0.164 86.7 14.2VL m LV I X= = × =

0.164 177 29.0VC m CV I X= = × =

0.164 200 32.8VR mV I R= = × =

( )2232.8 14.2 29.0 36.0 mε+ − = =

L CV V−φ

mε

PHY2049: Chapter 31 39

RLC Example 3Circuit parameters: C = 2.5μF, L = 4mH, εm = 10v

ω0 = (1/LC)1/2 = 104 rad/sPlot Im vs ωd / ω0

R = 5Ω

R = 10Ω

R = 20ΩIm Resonance

0dω ω=

0/dω ω

PHY2049: Chapter 31 40

Power in AC CircuitsInstantaneous power emitted by circuit: P = i2R

More useful to calculate power averaged over a cycleUse <…> to indicate average over a cycle

Define RMS quantities to avoid ½ factors in AC circuits

House currentVrms = 110V ⇒ Vpeak = 156V

( )2 2sinm dP I R tω φ= −

( )2 2 212sinm d mP I R t I Rω φ= − =

rms 2mII = rms 2

mεε = 2ave rmsP I R=

Instantaneous power oscillates

PHY2049: Chapter 31 41

Power in AC Circuits (2)Recall power formula

Rewrite

cosφ is the “power factor”To maximize power delivered to circuit ⇒ make φ close to zeroMost power delivered to load happens near resonanceE.g., too much inductive reactance (XL) can be cancelled by increasing XC (decreasing C)

2ave rmsP I R=

rmsave rms rms rms rms rms cos

ZRP I R I IZ

εε ε φ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

ave rms rms cosP Iε φ=

R

L CX X−Z

φcos R

Zφ =