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AC CIRCUITS Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 1/34 AC wave forms AC = A lternating C urrent peak peak-to- peak T f = 1/T frequency [Hz] ω = 2πf angular frequency [rad/s] Instantaneous value: i(t) = I m sin(ωt + φ)

Electrotehnica si masini electrice

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Electrotehnica si masini electrice

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Page 1: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 1/34

AC wave forms

AC = Alternating Currentpeak

peak-to-peak T

f = 1/T frequency [Hz]ω = 2πf angular frequency

[rad/s]

Instantaneous value: i(t) = Im sin(ωt + φ)

Page 2: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 2/34

Sinus wave

Average value: Iav = (1/T)∫i(t)dt = 0

Root Mean Squared value: I = √(1/T)∫i2(t)dt = Im/√2 = 0.707 Im

Instantaneous value: i(t) = √2 I sin(ωt + φ)

I = RMS value√2I = peak value (amplitude)f = frequencyT = 1/f = periodω = 2πf = angular frequencyφ = phase

325 V

230 V

0t

[ms]2010

u(t) = 325 sin(314t)

Page 3: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 3/34

Phase difference

ωt

φ=π/6

2π+φ

The two waveforms are π/6 (30o) “out-of phase”.The current lags the voltage by angle φ.

The voltage leads the current by angle φ.

The current cross the horizontal reference axis reaching its peak and zero values after the voltage waveform.

u(t) = Um sin(ωt)

i(t) = Im sin(ωt - φ)

φ = phase difference (phase shift)

Page 4: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 4/34

Complex quantities. Phasor diagram

the current phasor lags behind the voltage phasor

U = U·ej0 I =I·e-jπ/6

A Phasor Diagram can be used to represent two or more stationary sinusoidal quantities: voltage, current, or some other alternating quantity of the same frequency.

√2·A sin(ωt + φ) → A = A ejφ (j = √-1)

ejφ = cosφ + j sinφ

A = A (cosφ + j sinφ)

A = A cosφ + j A sinφ

reference axis

φ=- π/6

U

I

real axis

φ

A

Acosφ

Asi

imaginary axis

Page 5: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 5/34

Phasor addition

u1(t) = √2U1 sin(314t + φ1) u2(t) = √2U2 sin(314t + φ2)

u(t) = u1(t) + u2(t)

u (t) = √2U sin(314t + φ)

U = U1 + U2

U2 = U12+U2

2+2U1U2cos(φ1-φ2)

tan φ = (U1sinφ1+ U2sinφ2)/(U1cosφ1+ U2cosφ2)

u1(t) = √2·30 sin(314t + π/3)u2(t) = √2·20 sin(314t)u(t) = u1(t) + u2(t) = √2U sin(314t + φ)U = ?, φ = ?

U = 43.6 V, φ = 23.4 °

Page 6: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 6/34

Impedance, Phase angle, Resistance, Reactance

ImpedanceZ = U/I [Ω]

Phase angle φ = (φu - φi) [rad]

i(t) = √2(U/Z) sin(ωt + φu - φ)

u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)Z = ?φ = ?

u(t)

i(t)

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt + φi)

~Z, φ

Page 7: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 7/34

Impedance, Phase angle, Resistance, Reactance

ResistanceR = Z cosφ [Ω]

Reactance X = Z sinφ [Ω]

φ = (φu - φi) [rad]Z = √(R2 + X2)tan φ = X/R

u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)R = ?X = ?

u(t)

i(t)

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt + φi)

Z, φ

~ R, X

Page 8: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 8/34

Complex Impedance

Z = Uejφu/Iejφi Z = (U/I)ejφu-φi

Z = Zejφ

Z = Z(cos φ + j sin φ)Z = R + j X

Complex ImpedanceZ = U/I 1

j

R

X

Z

φ

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt + φi)

Z, φ

R, X~Z

Page 9: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 9/34

Power in AC circuits

u(t)

i(t)

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt + φi)

~

Instantaneous powerp(t) = u(t)·i(t)p(t) = 2UI sin(ωt + φu)· sin(ωt + φi)

Page 10: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 10/34

True power

P = (1/T)∫p(t)dt

P = UI cosφ [W] φ = φu - φi

Power in AC circuits

U = Z IP = Z I2

cosφP = R I2

Apparent power

S = U I [VA]

U = Z IS = Z I2

Power factor

kp = P/S = cos φ

Reactive power

Q = UI sinφ

[VAr]

U = Z IQ = Z I2

sinφQ = X I2

S = √(P2 + Q2)

S = U I* = Uejφu Ie-jφi = (UI)ej(φu-φi) = Sejφ = P + jQ

Page 11: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 11/34

Power in AC circuits. Applications

u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)P = ?Q = ?S = ?

u(t)

i(t)

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt + φi)

~

Page 12: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 12/34

AC Resistor with a sinusoidal supply

u(t)

i(t)

u(t) = √2U sin(ωt) (φu= 0)

i(t) = u(t)/R

i(t) = √2 (U/R) sin(ωt)

I = U/R → Z = U/I = Rφi = φu → φ = φu - φi = 0

R = Z cos φ = RX= Z sin φ = 0 Z = R + jX = R

Page 13: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 13/34

AC Resistor with a sinusoidal supply

P = RI2 > 0Q = XI2 = 0

S = ZI2 = RI2 = P (kP = 1)

Page 14: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 14/34

AC Inductor with a sinusoidal supply

u(t) = √2U sin(ωt + φu)

i(t) = √2I sin(ωt) ( φi = 0)u + e = Ri = 0 → u = - e

u(t) = dφ/dt = Ldi/dt

u(t) = √2ωLI cos(ωt)

u(t)

i(t)

uL(t)

L = ct u(t) = √2 ωLI sin(ωt + π/2)

U = ωLI → Z = U/I = ωL

φu = π/2 → φ = φu - φi = π/2

R = Z cos φ = 0X= Z sin φ = ωL > 0

Z = R + jX = jωL

Page 15: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 15/34

AC Inductor with a sinusoidal supply

U

I

P = RI2 = 0Q = XI2 = ωLI2 > 0

S = ZI2 = ωLI2 I=U/(2πfL)

XL=2πfL

Page 16: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 16/34

AC Capacitor with a sinusoidal supply

uC(t)

iC(t)

u(t)

u(t) = √2U sin(ωt) (φu = 0)

i(t) = √2I sin(ωt + φi)

i(t) = dq/dt = C du/dt

u(t) = (1/C)∫i(t)dt

u(t) = -√2I(1/ωC) cos(ωt + φi)

u(t) = √2I(1/ωC) sin(ωt + φi - π/2)

U = I/(ωC) Z = U/I = 1/(ωC)

φu = φi -π/2 = 0→ φi = π/2φ = φu - φi = -π/2, (kP = 0)

R = Z cos φ = 0X= Z sin φ = -1/(ωC) < 0

Z = R + jX = -j/(ωC)

Page 17: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 17/34

AC Capacitor with a sinusoidal supply

P = RI2 = 0Q = XI2 = -1/(ωC)I2 < 0

S = ZI2 = I2/(ωC) XC=1/(2πfC)

I=2πfCU

Page 18: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 18/34

Applications

ZC = 1/(2πfC) = 3.18 kΩ IC = 2πfCU = 15.7 mA

U = 50 Vf = 50 Hz 1 μFU = 50 Vf = 5 kHz

ZC = 1/(2πfC) = 31.8 Ω IC = 2πfCU = 1.57 A

U = 50 Vf = 5o Hz

ZL = 2πfL = 691.15 Ω IL = U/2πfL = 72.3 mA

U = 50 Vf = 5 kHz

ZL = 2πfL = 69.115 kΩ IL = U/2πfL = 0.723 mA

Page 19: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 19/34

Parallel RLC circuit

I

U

u(t) = √2U sin(ωt) (φu = 0)

i(t) = √2I sin(ωt + φi)

i(t) = iR(t) + iL(t) + iC(t)

I = IR + IL + IC

IR = U/ZR = U/R

IL = U/ZL = U/jωL = -jU/ωL

IC = U/ZC = U/(-j/ωC) =jωCU

I = U[1/R - j(1/ωL - ωC)]

1/Z = [1/R - j(1/ωL - ωC)]

1/Z = √[(1/R)2 + (1/ωL - ωC)2]

tanφ = (1/ωL - ωC)/(1/R)

Page 20: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 20/34

Parallel RLC circuit

realaxis

imaginaryaxis

UIR

IL

IC

I

φ

u(t) = √2·50 sin(628t)R = 50 ΩL = 20 mHC = 5 μFZ = ?, IR = ?, IL = ?, IC = ? I = ?, φ = ?

ZL = ωL = 628·0.02 = 12.6 ΩZC = 1/ωC = 1/(628·5·10-6)= 318.3 ΩZ = 1/√[(1/R)2 + (1/ωL - ωC)2] = 12.7 ΩIR = U/R = 1 A, IL = U/ZL = 3.9 A, IC = U/ZC = 0.16 A, I = √[IR

2 + (IL – IC)2] = 3.87 A, φ = arctg[(1/ωL - ωC)/(1/R)] = 75.3 °

Page 21: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 21/34

Parallel resonance circuit

I

U

Resonance : φ = 0

tanφ = (1/ωL - ωC)/(1/R)

1/ωL = ωC

ω0 = 2πf0 = 1/√LCf0 = 1/ (2π √LC)

1/Z0 = √(1/R)2

Z0 = R = Zmax

Page 22: Electrotehnica si masini electrice

AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 22/34

Parallel resonance circuit

I0 = Imin = U/R

realaxis

imaginaryaxis

UIR = I

ILIC

φ = 0

Page 23: Electrotehnica si masini electrice

3 PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 23/34

Three phase wave form

u1(t) = √2U sin(ωt)

u2(t) = √2U sin(ωt – 2π/3)

u3(t) = √2U sin(ωt – 4π/3) 120° 120° 120°

u2(t) = √2U sin(ωt + 4π/3)

u3(t) = √2U sin(ωt + 2π/3)

U1 = U ej0

U2 = U ej4π/3 = a2U1 a = ej2π/3

U3 = U ej2π/3 = aU1

U1

U2

U3

2π/32π/3

2π/3

Page 24: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 24/34

Three phase voltage

U1

U2

U3

2π/32π/3

2π/3

U1 – U2 = U1(1 – a2)a2 = ej4π/3 = = cos(4π/3) + j sin(4π/3)(1 - a2) = 3/2 + j√3/2 = = √3(√3/2 + j1/2) == √3[cos(π/6) + j sin(π/6)]

U1 – U2 = √3 U1 ejπ/6

U1 –U2

π/6

Page 25: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 25/34

Three phase connections

Uph = Phase VoltageUl = Line VoltageIph = Phase CurrentIl = Line Current

Il = Iph

Ul ≠ Uph Il ≠ Iph

STAR (Y) connection

Uph

Ul

Il Iph

neutral

line

line

Ul

Il

Iph

DELTA (∆) connection

Page 26: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 26/34

Unbalanced 3-phase load, STAR connection

L1

L2

L3

N

Z1

Z2

Z3

Z0

I1

I2

I3

I0

U1 U2 U3U0

U’1

U’2

U’3

U1, U2 = a2U1, U3 = aU1

U’1= U1 – U0

U’2 = U2 – U0

U’3 = U3 – U0

I1 + I2 + I3 = I0

I1 = U’1/Z1 = (U1 – U0)/Z1

I2 = U’2/Z2 = (U2 – U0)/Z2

I3 = U’3/Z3 = (U3 – U0)/Z3

(U1 – U0)/Z1 + (U2 – U0)/Z2 + + (U3 – U0)/Z3 = U0/Z0

U0 = (U1/Z1 + U2/Z2 + U3/Z3)/(1/Z1 +

1/Z2+ 1/Z3+ 1/Z0)

Page 27: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 27/34

Unbalanced 3-phase load, STAR connection

L1

L2

L3

N

Z1

Z2

Z3

Z0

I1

I2

I3

I0

U1 U2 U3U0

U’1

U’2

U’3

U’1 = U1 U2’ = U2 U’3 = U3

U0 = 0

U1

U2

U3

U’1

U’3

U’2

U0

N

N’

N’

Page 28: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 28/34

Balanced 3-phase load, STAR connection

L1

L2

L3

N

Z

Z

Z

Z0

I1

I2

I3

I0

U1 U2 U3U0

U’1

U’2

U’3

N’

Z1 = Z2 = Z3 = Zejφ

U0 = [(U1 + U2 + U3)/Z]/(3/Z + 1/Z0) = 0

I1 = U1/Z = U1/(Zejφ) = (U1/Z)e-jφ

I2 = U2/Z = a2U1/(Zejφ) = a2I1

I3 = U3/Z = aU1/(Zejφ) = aI1

U1, U2 = a2U1, U3 = aU1

I1, I2 = a2I1 I3 = aI1

U1

U2

U3

I1

I2

I3

U12

U23

U31

U12

U23

U31

Ul =√3 Uph

Page 29: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 29/34

Balanced 3-phase load, DELTA connection

Z12 = Z23 = Z31 = Zejφ

U12, U23 = a2U12, U31 = aU12

I12, I23 = a2I12 I31 = aI12

U12

U23

U31

I12I23

I31

I2

I3 I1

Il =√3 Iph

L1

L2

L3

U12

U23

U31

Z

Z

Z

I12

I23

I31

I1

I2

I3

I12 = U12/Z = U12/(Zejφ) = (U12/Z)e-jφ

I23 = U23/Z = a2U12/(Zejφ) = a2I12

I31 = U31/Z = aU12/(Zejφ) = aI12

Page 30: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 30/34

Power in 3 - phase circuits

L1

L2

L3

N

Z

Z

Z

Z0

I1f

I2f

I3f

I0

U1 U2 U3U0 = 0

U1

U2

U3

U1 = Uf, U2 = a2Uf, U3 = aUf

I1f = U1/Z = Uf/(Zejφ) =

= (Uf/Z)e-jφ = If e-jφ

I*1f = If ejφ

I2f = U2/Z = a2Uf/(Zejφ) = a2If e-jφ

I*2f = aIf ejφ

I3f = U3/Z = aUf/(Zejφ) = aIf e-jφ

I*3f = a2If ejφ

Z1 = Z2 = Z3 = Zejφ S = U1f I*1f + U2f I*2f + U3f I*3f

Page 31: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 31/34

Power in 3 - phase circuits

L1

L2

L3

N

Z

Z

Z

Z0

I1f

I2f

I3f

I0

U1f U2f U3fU0 = 0

U1

U2

U3

S = Uf If ejφ + a2Uf aIf ejφ +

+ aUf a2If ejφ = 3UfIf ejφ

(a3 = e2π = 1)

S = P + jQ

ejφ = cosφ + j sinφ

S = 3UfIf cosφ + j 3UfIf sinφ

P = 3UfIf cosφ

Q = 3UfIf sinφ

P = √3UlIl cosφ

Q = √3UlIl sinφ

“STAR” Uf = Ul/√3, If = Il

“DELTA”Uf = Ul, If = Il/√3

Page 32: Electrotehnica si masini electrice

THREE PHASE AC CIRCUITS

Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 32/34

Power in 3 - phase circuits. Application

L1

L2

L3

N

R

R

R

I1

I2

I3

U1 U2 U3

U1

U2

U3

Z1 = Z2 = Z3 = R = 100 Ω

PY = √3UlIl cosφ = = √3·400·(230/100)·cos0 = = 1600 W

L1

L2

L3

U12

U23

U31

R

R

R

I12

I23

I31

I1

I2

I3

Z1 = Z2 = Z3 = R = 100 Ω

P∆ = √3UlIl cosφ = = √3·400·(√3·400/100)·cos0 = = 4800 W

U1 = U2 = U3 = 230 V

U12 = U23 = U31 = 400 V