Unit 9: Alternating-current circuits - UPV 9/Slides Unit 9 Alternating... · Unit 9:...

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Unit 9: Alternating-current circuits Introduction. Alternating current features. Behaviour of basic dipoles (resistor, inductor, capacitor) to an alternating current. RLC series circuit. Impedance and phase lag. Power on A.C. Resonance. Niagara Falls Nikola Tesla 1856-1943

Transcript of Unit 9: Alternating-current circuits - UPV 9/Slides Unit 9 Alternating... · Unit 9:...

Page 1: Unit 9: Alternating-current circuits - UPV 9/Slides Unit 9 Alternating... · Unit 9: Alternating-current circuits Introduction. Alternating current features. Behaviour of basic dipoles

Unit 9: Alternating-current circuits

Introduction. Alternating current features.

Behaviour of basic dipoles (resistor, inductor,

capacitor) to an alternating current.

RLC series circuit. Impedance and phase lag.

Power on A.C.

Resonance.

Niagara

Falls

Nikola Tesla 1856-1943

Page 2: Unit 9: Alternating-current circuits - UPV 9/Slides Unit 9 Alternating... · Unit 9: Alternating-current circuits Introduction. Alternating current features. Behaviour of basic dipoles

Coil turning inside a magnetic field B.

NBSwsenwtdt

d=−=

φε

N S

S

B

ω ω ω ω

ωωωω·t

wtNBS cos== SBNrr

φ

%ωωωω

Um

Symbol:

Alternating-current generation

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N1 N2

V1 ~ V2 ~

The transformer The current on primary produces a current and then a flux varying on time.

This flux is completely driven by the ferromagnetic material, producing an inducedelectromotive force on terminals of secondary. The flux on each loop φu is the same forprimary and secondary.

It works due to electromagnetic induction and then it doesn’t work on D.C.

Primary

winding

Secondary

windingdt

dNV u

111

Φ==ε

dt

dNV u

222

Φ==ε

1

2

1

2

2

2

1

1 V

N

N

VN

V

N

V==

Voltage ratio or ratio

of transformation

equals the turns ratio

On an ideal transformer the power on primary equals the power on secondary

On an real transformer there are three type of losses:

On windings: Joule heating

On core: Magnetic Histeresys

and Eddy currents

In order to minimize Eddy currents the core is built with

sheets isolated between them

B

r

i

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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)

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

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ωt

ϕϕϕϕu =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)

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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 repreented 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

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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

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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 functions

must be sin or cosand with equal

angular frequency

Uφu

Phasor diagram

I

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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

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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

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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

)2

tcos(U)2

tcos(ILtsenILdt

)t(diL)t(u mmm

πωπωωωω +=+=−==

XL = Lω Inductance (Ω)

dt

)t(diLuL =

U

I

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Behaviour of basic dipoles. Capacitor

Ci(t)

u(t)

φ = - π/2

Tipler, chapter 29.1

Capacitor

ωt

i

u

u(t) = Um cosωt

)2

tcos(I)2

tcos(CU)t(senCUdt

)t(Cdu

dt

)t(dq)t(i mmm

πω

πωωωω +=+=−===

ω=

C

IU m

mXC = 1/Cω Capacitance (Ω)

Cuq =

U

I

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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

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Sinusoidal functions are interesting because additionof (S.F.) is another S.F.

On the other hand, Fourier’s law says that anyperiodic function can be split in S.F. of differentfrequencies. This is the main reason why behaviourof S.F. is studied.

Knowing response of basic dipoles to a sinusoidalcurrent, we know response to any (no sinusoidal)current.

Sinusoidal functions (S.F.) and Fourier’s law.

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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 inseries. If a sinusoidal intensity i(t)=Imcos(wt) is flowingthrough such devices, voltage on terminals of circuit will bethe addition of voltages on each device:

RLC series circuit. Impedance of dipole

u(t)

Addition of sinusoidal

functions is another

sinusoidal function

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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

CwLwRIU CL

m

mmm =+=−+=−+= 222222 1

)()((

ϕϕ tgR

X

R

XX

R

CwLw

tg CL ==−

=

=

1Z 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.

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R

XL=Lw

ZX

ϕϕϕϕ

X<0 (ϕϕϕϕ<0)

R

ZX

ϕϕϕϕ

Impedance triangle.

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

2222 1XR

CwLwRZ +=−+= )((

R

X

R

XX

R

CwLw

tg CL =−

=

=

1

ϕ

XC=1/Cw X=XL-XC=Lw-1/Cw

InductiveReactance

CapacitiveReactance

Dipole Reactance

X>0 (ϕϕϕϕ>0)

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Power on A.C.

Power consumed on a device on A.C. can be computed multiplying voltage and

intensity on each time. This is the Instantaneous power:

)tsin(I)t(i m ω=

)tsin(u)t(u m ϕω +==⋅+=⋅= )tsin(I)tsin(U)t(i)t(u mm ωϕω

t2sinsin2

IUtsincosIUtsin]sintcoscost[sinIU mm2

mmmm ωϕωϕωϕωϕω⋅

+⋅=⋅+⋅⋅=

Reactive powerActive or

Real power

p(t)

)(tp

Instantaneous power = Active power + Reactive power

Frequency of reactive power doubles that of active power

tIU mm ωϕ 2sincos⋅ tIU mm ωϕ 2

2sinsin

⋅= +

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Power on A.C.

> 0 and < 0 zero on a cycleAlways > 0

Active power + Reactive power = Instantaneous power

=+

t2sinsin2

IU mm ωϕ⋅

)t(p+ =

Example taking: Im = 1 A Um = 1 V ω = 1 rad/s ϕ = 0,6 rad

Average value on a cycle:

ϕωϕ cos2

IUtdtsincosIU

T

1dt)t(p

T

1PP mm

T

0

2

mm

T

0

a

av

⋅=⋅===

Power on A.C.

-0,4

-0,2

0

0,2

0,4

0,6

0,8

1

0 2 4 6 8 10 12

wt (rad)

Po

wer

(w)

Active pow er

Power on A.C.

-0,4

-0,2

0

0,2

0,4

0,6

0,8

1

0 2 4 6 8 10 12

wt (rad)P

ow

er

(w)

Reactive pow er

Power on A.C.

-0,4

-0,2

0

0,2

0,4

0,6

0,8

1

0 2 4 6 8 10 12

wt (rad)

Po

wer

(w)

Inst. pow er

cos ϕ ≡ Power factor

tsincosIU mm ωϕ 2⋅

=

TT

dtwt0

2

2)(sin

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DipoleActive

power Pa(t)Average

Paav

Reactive power Pr(t)

Average Prav

P(t) Pav

Rϕ=0

Cos ϕ=1

Sin ϕ=00 0

Lϕ=90º

Cos ϕ=0

Sin ϕ=10 0 0 0

Cϕ=-90º

Cos ϕ=0

Sin ϕ=-10 0 0 0

Rms magnitudes. Power for basic dipoles

Consumed power on basic dipoles (R, L and C):

)(2sinsin2

sincos 2tpt

IUtIU mm

mm =⋅

+⋅ ωϕωϕ

An A.C. current consumes the same power than a D.C. current having the rms magnitudes

ϕϕ cosIUcos2

IUpowerAverage rmsrms

mm =⋅

=2

mrms

UU =

2

mrms

II =

tsinIU2

mm ωR

Um

2

2

tsinIU2

mm ωR

Um

2

2

t2sinL2

U2

m ωω

t2sinL2

U2

m ωω

t2sinC2

U2

m ωω− t2sinC2

U2

m ωω−

Tipler, chapter 29.1

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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. cos ϕ=1 and consumed

power on RLC dipole is maximum.

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 ===