Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit...

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Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor diagram. Behaviour of basic dipoles (resistor, inductor, capacitor) to an alternating current. RLC series circuit. Impedance and phase lag. Resonance. Filters Niagara Falls Nikola Tesla 1856-1943

Transcript of Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit...

Page 1: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

Unit 10: Alternating-current circuits

Introduction. Alternating current features.

Phasor diagram.

Behaviour of basic dipoles (resistor, inductor,

capacitor) to an alternating current.

RLC series circuit. Impedance and phase lag.

Resonance. Filters

Niagara

Falls

Nikola Tesla 1856-1943

Page 2: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

Period T = 2π/ω (s)

Frequency f = 1/T (Hz)

Angular frequency

w = 2πf (rad/s)

Phase wt+ϕ

Initial phase ϕ (degrees or radians) (phase at t=0)

Amplitude=Maximum voltage Um (V)

is the root mean square value. Is that measured by the

measurement devices on A.C.

ωt

T

ϕ

Um

u(t) = Um cos(ωt + ϕ) u(t)

f Europe: 50 Hz

f North America: 60 Hz

Sinusoidal alternating-current features

2

m

rms

UU =

Page 3: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

To simplify the analysis of A.C. circuits, a graphical representation of sinusoidalfunctions called phasor diagram can be used.

A phasor is a vector whose modulus (length) is proportional to the amplitude ofsinusoidal function it represents.

The vector rotates counterclockwise at an angular speed equal to ω. The anglemade up with the horizontal axis is the phase (ωt+φ).

Therefore, depending if we are working with the function sinus or the functioncosinus, this function will be represented by the vertical projection or thehorizontal projection of the rotating vector.

ωt

T

ϕ

Um

u(t) = Um cos(ωt + ϕ)

u(t)

Phasor diagram

U

ωt+φ

Um

ω

+

+

)sin(:

)cos(:Pr

ϕω

ϕω

tUVertical

tUHorizontalojections

m

m

Page 4: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

As the position of phasor is different for any time considered, the graphicalrepresentations are done on time t=0 and then, the initial phase φ is the anglebetween vector and horizontal axis. In this way, the phasor is a unique vector(not changing on time) for a given function:

ωt

T

ϕ

Um

u(t) = Um cos(ωt + ϕ)

u(t)

Phasor diagram

U

φ

Um

Phasor diagram

Page 5: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

ωtϕϕϕϕu =0

u(t) = Um cos(ωt + ϕu)

u(t)

Initial phase. Examples.

ωt

u(t)

ϕϕϕϕu=90º (ππππ/2 rad)

ωt

u(t)

ϕϕϕϕu=-90º (-ππππ/2 rad)ωt

u(t)

ϕϕϕϕu=-45º (-ππππ/4 rad)

UU

U U

Page 6: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

u(t) = Um cos(ωt + ϕu )i(t) = Im cos(ωt+ϕi)

ϕ ωt

iu ϕϕϕ −=

Phase lag between two waves (voltage and intensity)

Phase lag is defined as

ϕϕϕϕi=0 ϕϕϕϕu<0 0<ϕ

Voltage u(t) goes behind intensity i(t)

Intensity i(t) goes ahead voltage u(t)

To be compared, both functionsmust be sin or cos and with

equal angular frequency

Uφu

Phasor diagram

I

Page 7: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

ϕ ωt

iu ϕϕϕ −=

Phase lag between two waves (voltage and intensity)

ϕϕϕϕi=0ϕϕϕϕu>0

0>ϕ

ϕ ωt

ϕϕϕϕi<0ϕϕϕϕu=0

0>ϕ

Uφ =φu

I

Uφ =φi

I

Page 8: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

Behaviour of basic dipoles. Resistor

Resistor

ωt

i

u

u(t) = R i(t) = RIm cosωt = Um cosωt

i(t) = Im cosωt

Ri(t)

u(t)

Um = R Im

ϕ = 0

Tipler, chapter 29.1

uR = iR

U

I

Page 9: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

Behaviour of basic dipoles. Inductor

i(t) = Im cosωt

Li(t)

u(t)

Um = LωIm

ϕ = π/2

Tipler, chapter 29.1

Inductor

ωt

iu

XL = Lω Inductive reactance (Ω)

U

I

=L

di(t)u L

dt

m m m

di(t)u(t) L L I sen t L I cos( t ) U cos( t )

dt

π πω ω ω ω ω= = − = + = +

2 2

Page 10: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

Behaviour of basic dipoles. Capacitor

Ci(t)

u(t)

φ = - π/2

Tipler, chapter 29.1

Capacitor

ωt

i

u

u(t) = Um cosωt

XC = 1/Cω Capacitive Reactance (Ω)

Cuq =

UI

Cd(u )i(t) C

dt=

m m m

dq(t) Cdu(t)i(t) CU sen t CU cos( t ) I cos( t )

dt dt

π πω ω ω ω ω= = = − = + = +

2 2

mm

IU

Cω=

Page 11: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

R

L

C

)cos( umL wtLwIu ϕ+=

)cos( umR wtRIu ϕ+=

)cos( um

C wtCw

Iu ϕ+=

+=

=

2πϕ

mLLm IXU

=

=

mRm IRU

−=

=

2πϕ

mCCm IXU

Behaviour of basic dipoles. Review

Voltage and intensity go on phase

Voltage goes ahead intensity 90º

Voltage goes behind intensity 90º

Page 12: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

L R C

uLuR uC

i(t)= Im cos (wt)

u(t) = uL (t)+ uR (t)+ uC (t)= Um cos (wt+ϕ)

Let’s take a circuit with resistor, inductor and capacitor in series. If asinusoidal intensity i(t)=Imcos(wt) is flowing through such devices,voltage on terminals of circuit will be the addition of voltages on eachdevice:

RLC series circuit. Impedance of dipole

u(t)

Addition of sinusoidal

functions is another

sinusoidal function

Page 13: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

RLC series circuit. Impedance of dipole

Um cos (wt+ϕ) = LwIm cos (wt +π/2)+RIm cos (wt)+(1/Cw)Im cos (wt -π/2)

UL

I URUC

UL-UC

I UR

U

ϕ Um

(Lω-1/Cω) Im

RIm

ϕ Um

ZXRXXRI

U

CwLwRIUI

CwLwRIU CL

m

mmmmmm =+=−+=−+=−+= 222222222

)()1

(())1

(()(

ϕϕ tgR

X

R

XX

R

CwLw

tg CL ==−

=

=

1 Z Is called Impedance of dipole (Ω)

ϕ is phase lag of dipole

Z and ϕ are depending not only on parameters of R, L and C, but also on frequency of applied current.

ϕ is ranging between - and 2

π2

π

U = UL + UR + UC

Page 14: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

R

ZX

ϕϕϕϕ

X<0 (ϕϕϕϕ<0)

R

ZX

ϕϕϕϕ

Impedance triangle.

All the equations of a RLC dipole can be summarized on Impedance Triangleof a dipole for a given frequency:

2222 1XR

CwLwRZ +=−+= )((

R

X

R

XX

R

CwLw

tg CL =−

=

=

1

ϕ

X=XL-XC=Lw-1/Cw

DipoleReactance

X>0 (ϕϕϕϕ>0)

Page 15: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

22 1)((

CwLwR

I

UZ

m

m −+==

RLC series circuit. Resonance

Drawing Z v.s frequency

Z v.s. freq

0

100

200

300

400

500

600

0 500 1000 1500 2000 2500 3000 3500 4000

frequency (Hz)Z

(O

hm

)

Example taking: R = 80 Ω L = 100 mH C = 20 μF

Resonance: f0=707 Hz Z=80 Ω

On resonance, impedance of circuit is minimum, and

amplitude of intensity reaches a maximum (for a given

voltage). Intensity and voltage on terminals of RLC

circuit go then on phase.

There is a frequency where XL=XC and then the impedance gets its minimum value (Z=R).

This frequency is called Frequency of resonance (f0) and can be easily computed:

LCf

LCCL

1

2

11100

0

ωω

ω ===

Page 16: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

RLC series circuit as a Bandpass filter

C

Ru(t)

L

uR(t)

Input

Outp

ut

22

mmmRouput

)C

1L(R

RU

Z

URRIUU

ωω −+

====

22m

R

input

output

)C

1L(R

R

U

U

U

U

ωω −+

==

2

1

U

U

21 f,fm

R =

Bandwith [f1 , f2]

1 LQ

R C=

The tunning circuit of a

radio is a Bandpass filter

Page 17: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

RLC series circuit as a Highpass filter

Input

Outp

ut

1

1

2

L

m f

U

U=Bandpass [f1 , ∞]

1 LQ

R C=

2 21( )

m m

ouput L m

U L UU U L I L

ZR L

C

ωω ω

ωω

= = = =

+ −

2 21( )

ouput L

input m

U U L

U UR L

C

ω

ωω

= =

+ −

Page 18: Unit 10: Alternating-current circuitsjmas.webs.upv.es/ffi/Unit 10/Slides Unit 10 Alternating...Unit 10: Alternating-current circuits Introduction. Alternating current features. Phasor

RLC series circuit as a Lowpass filter

Input

Outp

ut

1

1

2

C

m f

U

U=Bandpass [∞, f1]

1 LQ

R C=

2 2

1 1

1( )

m m

ouput C m

U UU U I

C C ZC R L

C

ω ωω ω

ω

= = = =

+ −

2 2

1

1( )

ouput C

input m

U U

U UC R L

Cω ω

ω

= =

+ −