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Alternating currentsingle-phase

circuits

Unit 2. AC SINGLE-PHASE CIRCUITS

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Unit 2. AC SINGLE-PHASE CIRCUITS

Basics of AC circuits Importance of AC RMS and mean values Phasorial magnitudes Ohm’s law for AC circuits AC loads Instantaneous power Power triangle AC power: P, Q and S Power factor and cos Power factor improvement Electrical measurements Exercises

CONTENTS:

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Direct Current

A current flowing in a constant direction

Voltage with constant polarity.

Unit 2. AC SINGLE-PHASE CIRCUITSBASICS OF AC CIRCUITS

http://www.pbs.org/wgbh/amex/edison/sfeature/acdc.html

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Unit 2. AC SINGLE-PHASE CIRCUITSBASICS OF AC CIRCUITS

Alternating Current Current flowing with alternating polarity, reversing positive and negative over time. Voltages with alternating polarity.

http://www-tc.pbs.org/wgbh/amex/edison/sfeature/images/acdc_inside_generator.gif

tsin)tcos(

dtd

dtde(t)

0

00 E

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■ Electric power is generated, transmitted, distributed and consumed in AC

■ ≈ 90% of the total electric power is consumed as AC

■ AC amplitude can be changed (stepped up or stepped down) easily bysimple and cost effective electrical machines called transformers.

■ Electric power can be transmitted efficiently and economically as highvoltage AC, to minimize power losses

■ Three-phase AC motors/generators perform better (higher efficiency,lower maintenance, etc.) than DC motors/generators

Unit 2. AC SINGLE-PHASE CIRCUITSWHY AC? Importance of AC

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Peak, peak-to-peak and average valuesSAME VALUES for sinusoidal, rectangular and triangular waveforms

Average value = 0

Root Mean Square Value (RMS)For sinusoidal waveform RMS = Amplitude/sqrt(2)

T: period of signal (s)

Unit 2. AC SINGLE-PHASE CIRCUITSMEASUREMENTS OF AC MAGNITUDES

2V(t)dtv

T1V 0T

0

2RMS 2

I(t)dtiT1I 0T

0

2RMS

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Unit 2. AC SINGLE-PHASE CIRCUITSMATHEMATICAL FORMULATION OF AC MAGNITUDES

v(t) = Vocos(t + V)Vo

= 2··f rad/s

ºRMS

VVV

Phasorial expression (phasor = phase vector)

Modulus: RMS value of the AC magnitude

Phase: initial phase angle of the AC magnitude

Frequency: does not appear in the phasor

A phase vector ("phasor") is a representation of a sine wave whose amplitude (A), phase (), and frequency (ω) are time-invariant.

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Unit 2. AC SINGLE-PHASE CIRCUITSOHM’S LAW FOR AC CIRCUITS

Ohm’s law for DC circuits: V = I.R

Ohm’s law for AC circuits:

being

Z: impedance R: resistance X: reactance

All quantities are expressed in complex, not scala,r form

= = V - I

Z IV ZjXRZ

v(t) = Vocos(t + V)i(t) = Iocos(t + I)

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Unit 2. AC SINGLE-PHASE CIRCUITSPHASE IN AC CIRCUITS Out-of-phase waveforms

Phase shift of 90 degrees:

A leads B

B lags A

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Unit 2. AC SINGLE-PHASE CIRCUITSAC PURE RESISTIVE CIRCUITS Voltage and current are “in phase”

Instantaneous AC power is always positive.

= 0º phase shift between voltage and current cos = 1

vR(t) = R·i(t)

v(t) = Vocos(t + )

i(t) = (V0/R)cos(t + )

i(t) = I0cos(t + )

Z = R + j0V = I

Z = R + j0

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Unit 2. AC SINGLE-PHASE CIRCUITSAC PURE INDUCTIVE CIRCUITS

Inductor current lags inductor voltage by 90º.

Instantaneous AC power may be positive or negative

= 90º phase shift between voltage and current cos = 0

vL(t) = L·di(t)/dt

v(t) = Vocos(t + )

i(t) = I0cos(t + )

)º90tcos(V)tsin(Vv(t)dtL1i(t) 00 VV LL

V = I+90º

Z = 0 + jL

V = I+90º

Z = 0 + jL 12

Unit 2. AC SINGLE-PHASE CIRCUITSAC PURE CAPACITIVE CIRCUITS

Capacitor voltage lags capacitor current by 90º.

Instantaneous AC power may be positive or negative

= -90º phase shift between voltage and current cos = 0

iC(t) = C·dv(t)/dt

v(t) = Vocos(t + )

i(t) = I0cos(t + )

)º90tcos(CV)tsin(CVdt

dv(t)Ci(t) 00 VV

V = I-90ºZ = 0 - j/(C)V = I-90º

Z = 0 - j/(C)

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Unit 2. AC SINGLE-PHASE CIRCUITSAC LOADS SUMMARY PURE RESISTANCE

V and I “in phase” = 0ºP consumption but Q = 0cos = 1 Z = R + j0

PURE INDUCTORV leads I by 90º, =90ºP =0 and Q > 0cos = 0 Z = 0 + jL

PURE CAPACITORV lags I by 90º, =-90ºP =0 and Q < 0cos = 0Z = 0 - j/(C)

V

I

V

I

VI

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Unit 2. AC SINGLE-PHASE CIRCUITSPOWER IN THE TIME DOMAIN

Instantaneous power:

p(t) = v(t)·i(t) = V0cos(wt + V)·I0cos(wt + )

cosA·cosB = 1/2·[cos(A+B) + cos(A - B)]

p(t) = ½V0I0cos(V - I) + ½V0I0cos(2wt + V + I) watt

The average value is the active or true power:

P = ½·V0·I0·cos(V - ) = VRMS·IRMS·cos watts

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Unit 2. AC THREE-PHASE CIRCUITSINSTANTANEOUS POWER

v(t)

i(t)

p(t)Average power/Active power/True power

p(t) = ½·Vo·Io·cos(V - I) + ½·Vo·Io·cos(2wt + V + I) 16

Unit 2. AC SINGLE-PHASE CIRCUITSAC POWER

Active/true power: P = V·I·cos P = I2·R (Watt)

Reactive power: Q = V·I·sin Q = I2·X (VAr)

Apparent power S = V·I =sqrt(P2+Q2) S = I2·Z (VA)

Complex power (VA)jQPZ·II VS 2*

V, I: RMS values

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Unit 2. AC SINGLE-PHASE CIRCUITSTHE POWER TRIANGLE

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Unit 2. AC SINGLE-PHASE CIRCUITSAC POWER: PARALLEL-CONNECTED NETWORKS

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Unit 2. AC SINGLE-PHASE CIRCUITSQUESTION

Is it safe to close the breaker between these two alternators if their outputfrequencies are different? Explain why or why not.

Solution: When the frequencies of two or more AC voltage sources aredifferent, the phase shift(s) between them are constantly changing.

http://powerelectrical.blogspot.com/2007/04/short-questions-and-solved-problems-in.html 20

Unit 2. AC SINGLE-PHASE CIRCUITSQUESTION

Given the output voltages of the two alternators, it is not safe to close thebreaker. Explain why.

Solution: The greatest problem with closing the breaker is the 37º phaseshift between the two alternators’ output voltages.http://powerelectrical.blogspot.com/2007/04/short-questions-and-solved-problems-in.html

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Unit 2. AC SINGLE-PHASE CIRCUITSQUESTION

Are the voltmeters readings possible? If so, how would you represent thetree voltages in this circuit in rectangular and polar forms?.

http://powerelectrical.blogspot.com/2007/04/short-questions-and-solved-problems-in.html

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Unit 2. AC SINGLE-PHASE CIRCUITSPOWER FACTOR AND COS

Power factor definition: PF = P/S

cos definition: cos = cos(V – I)

When dealing with single-frequency single-phase circuits:

PF = P/S = (V·I·cos)/(V·I) = cos

> o lagging

< 0 leading

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE

Example. A circuit has equivalent impedance Z = 3 + j·4 andan applied voltage v(t) = 42.5·cos(1000t + 30º) (volt). Givecomplete power information.

Hence, P = 108.4 W, Q = 144.5 VAr (i), S = 180.6 VA, and PF = cos 53.13º = 0.6 (i).

V 2

42.5V30º

RMS

A6.015

242.5/ 4·3242.5/

ZVI 23.13º-

º13.53

º30º30

RMSRMS

j

VA180.616.01·242.5/*I·VS 53.13º23.13ºº30

RMSRMS

r)j·144.5(VA108.4(W)180.61S 53.13º

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Unit 2. AC SINGLE-PHASE CIRCUITSVOLTAGE AND FREQUENCY WORLDWIDE

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Unit 2. AC SINGLE-PHASE CIRCUITSQUESTIONS REGARDING PF IMPROVEMENT Transformers, distribution systems, and utility company alternators

are all rated in kVA or MVA.

Consequently, an improvement in the power factor, with itscorresponding reduction in kVA, releases some of this generationand transmission capability so that it can be used to serve othercustomers.

This is why it is more costly for an industrial customer tooperate with a lower power factor.

Example. A load of P = 23 kW with PF = 0.5 (i) is fed by a 230 Vsource. A capacitor is added in parallel so that the power factor isimproved to 1. Find the reduction in current drawn from the generator.

Before the improvement:P = 23000 W = 230·I·0.5 I = 200 A

After the improvement:P = 23000 W = 230·I·1 I = 100 A 26

Unit 2. AC SINGLE-PHASE CIRCUITSPOWER FACTOR IMPROVEMENT

Inductive circuits: C in parallel

Capacitive circuits: L in parallel

+ L

R

I ef

efV

càrrega

C eq

eq

P=P'

S

'

Q < 0c

Q'=Q-Q S'

c

PfC2VQ

PfC)1/(2

VQ

PQQ

'tg2

2

C

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Unit 2. AC SINGLE-PHASE CIRCUITSPF IMPROVEMENT BENEFITS

Reduces reactive power and the amps absorbed by the load.

Active power driven by transformers is optimized.

Reduce voltage drop in the conductors.

Reduces power losses in the consumer’s conductors.

Reduces power losses in the conductors during transmission.

The conductor section can be minimized (capital savings).

The base price of electrical energy (kWh) increases if the PFis low.

The installation is used more efficiently.

Energy is used more efficiently.

Less power has to be generated (environmental benefits).28

Unit 2. AC SINGLE-PHASE CIRCUITSPF IMPROVEMENT

Electricity bill reduction

Kr = 17/cos2 – 21 %

The improvement of the PF of an existing electrical installation has many financial advantages and reduces the base cost of the kWh.

Kr is limited within: + 47 % (PF = 0.5) – 4 % (PF = 1)

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Unit 2. AC SINGLE-PHASE CIRCUITSPF IMPROVEMENT

Capacitor banks with automatic regulation can be used to adapt thecompensation to variable load.

Capacitor banks with automatic regulation are placed at the end of the LVinstallation or in the distribution board (cuadro de distribución), withconsiderable consumption of reactive power.

Capacitor banks with automatic regulation consist of several steps of reactivepower.

Automatic capacitor bank with several stepsAutomatic capacitor bank for PF correction30

Unit 2. AC SINGLE-PHASE CIRCUITSPF IMPROVEMENT

Types of power faction correction: Overalll Partial Individual.

Overall correction

This method is suitable for stable andcontinuous operated loads.

Partial correction

This method is suitable whendistribution of loads is unbalanced andwhen a distribution board (quadre dedistribució) feeds a considerable load.

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Unit 2. AC SINGLE-PHASE CIRCUITSPF IMPROVEMENT

Individual correction

This method is suitable when some loads are very heavy in relation to the totalload.

It is the most advantageous method.

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Unit 2. AC SINGLE-PHASE CIRCUITSQUESTIONS REGARDING PF IMPROVEMENTExample. A sinusoidal source (230 V, 50 Hz) feeds an inductiveload which absorbs 8 A and 1000 W. Calculate the PF, Q and Sin the load. Improve the PF to unity and calculate the associatedcost savings.

P = V·I·cos 1000 = 230·8·cos cos = 0.54 (i)

Q = V·I·sin = 230·8·0.84 Q = 1545 VAr

S = V·I = 230.8 S = 1840 VA PF = P/S = 0.54 (i) = cos

Capacitor needed to improve the PF to unity: Qc = -1545 VAr.

This results in C = 93 F

Increase in base price due to PF: Kr = 17/cos2 – 21 %

When PF = 0.54 Kr = 17/0,542 – 21 = + 37,3 % (increase)

When PF = 1.0 Kr = 17/1,0 – 21 = - 4 % (discount)

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 1A 230 V and 50 Hz AC power supply feeds a 5 KVA single-phase loadthat presents a PF = 0.6 lagging. Improve the PF to: a) 0.9(i) b) 0.9(c).

P=P'

S

'

Q < 0c

Q'=Q-Q S'

c PQQ

'tg C

a) cos’ = 0.9 ’ = 25.84º

This results in C = 153.26 F

b) When b) operates in a similar manner to a), the result is: C = 328.1 F

3000

Q4000 tg25.84

PQQ

'tg CC

Before improving the PF: P = S·cos = 3 kW, Q = S·sin = 4 kVAr

50C)(2230VAr 2547.16Q 2C

lagging = (i)

leading = (c)

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 2The consumption of 2 resistance is 20 W. Represent the powertriangle of the circuit in the figure.

P2 = 20 = I122 I1 = 3.162 A

VTOT = I1Z1 = 3.162·(22+52)1/2 = 17.03 V

I2 = VTOT/Z2 = 17.03/(12+12)1/2 = 12.04 A

PTOT = P1 + P2 = 20 + 12.042·1 = 165 W

QTOT = Q5 + Q1 = -3.1622·5 + 12.042·1 = 95 VAr

STOT = (PTOT2 + QTOT

2)1/2 = 190.4 VA

TOT = arctg(QTOT/PTOT) =29.93º

2

5

1

1 I1 I2

P

QS

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 3Determine the PF2 when the measured PFTOTAL = 0.90 (i).

cosTOTAL = 0.90 (i) TOTAL = +25.84º

P1 = 2000·0.8 = 1600 W Q1 = 2000·0.6 = 1200 VAr

P2 = 500·cos2 W Q2 = 500·sin2 VAr

S1 = 2kVA

PF1 = 0.8(i)

S2 = 500 VA

PF2 = ??

2

2

21

21

cos5001600sin50020015.84º2 tg

PPQQtg

TOT

-425.08=500sin2 - 242.16cos2

which results in: 2 = -24.07º PF2 =0.913(c) 36

Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 4From the following circuit, determine P, Q, S and PF in the load.

The equivalent impedance of the circuit is: Zeq = 7.5219º

PF = cos(19º) =0.945 (i)

I = V/Zeq = 2300º/ 7.5219º = 30.61-19º A

S = V·I* = 2300º·30.61+19º = 7039.2519º VA = 6655.75 (W) + j.2291.76 (VAr)

230 V 50 Hz 21

8

10

25

15

1-º19º96.30º34.51º0

eq

1331.00343.00781.0211

j15251

j1081

j0211

Z1

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 5. The mesh method (malles)From the following circuit, determine thecurrent in the 200 V voltage sources.

10 50

30

10

2000º V 2000º V

10

I1 I2

Loop1: Vi = IiZi 2000º = I1(10+j50+30+j10) - I2(30+j10)

Loop 2: Vi = IiZi -2000º = - I1(30+j10) + I2(10+30+j10)

Aj

jjjj

jj

º81.841 905.0

22002002000

1040103010306040

10402001030200

I

Ajj

jjjj

jj

º12.1862 616.4

2200200100002000

1040103010306040

20010302006040

I

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 6. Series RLC circuitDetermine a) the voltage and current in each element; c) P, Q and S in eachelement and the overall power factor; c) the resonance frequency (thefrequency at which the imaginary part of the equivalent impedance is null).

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 7. Parallel RLC circuitDetermine a) the voltage and current in each element; c) P, Q and S in eachelement and the overall power factor; c) the resonance frequency of thiscircuit.

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Unit 2. AC SINGLE-PHASE CIRCUITSEXAMPLE 8. The mesh method (malles)Calculate the impedance value needed to balance this AC bridge. Expressyour answer in both polar and rectangular forms. What type and size ofcomponent will provide this exact amount of impedance at 400 Hz?

http://powerelectrical.blogspot.com/2007/04/short-questions-and-solved-problems-in.html

197501975350

1000)·15,6910(350

1000)·0( º90 jjjXZ Lx

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Unit 2. AC SINGLE-PHASE CIRCUITS

How to balance a Wheatstone bridge?

http://utwired.engr.utexas.edu/rgd1/lesson07.cfm

Online resources:

The bridge is balanced when Va – Vb = 0

The same can be expressed as: Vab = 0

311 ZZ

VI

xZZ

VI

2

2

31

331 ··

ZZZVZIVac

x

xxbc ZZ

ZVZIV

2

2 ··

0)·(231

3

x

xbcacab ZZ

ZZZ

ZVVVV xZZZZ ·· 132

Balanced condition

1

32·Z

ZZZx

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Unit 2. AC SINGLE-PHASE CIRCUITSELECTRICAL MEASUREMENTS

Multimeters Voltage AC/DC Current AC/DC Resistance Others: capacitance, frequency, temperature, etc.

Current probes Voltage AC/DC (depending on models) Current AC/DC

Wattmeters Voltage AC Current AC Power : S, P, Q AC PF Frequency Single-phase or three-phase

Single-phase Three-phase

INSTRUMENTS FOR INDUSTRIAL MAINTENANCE

laboratory

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Unit 2. AC SINGLE-PHASE CIRCUITSELECTRICAL MEASUREMENTS

Three-phase

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Unit 2. AC SINGLE-PHASE CIRCUITSELECTRICAL MEASUREMENTS

INSTRUMENTS FOR INDUSTRIAL MEASUREMENT AND CONTROL

Useful for carrying out low-cost energy control checks on electrical consumption Can also be used to accurately control the consumption of any other physical

unit that has a meter with a digital impulse output.

Possible measurements: Voltage Current Hz Power (S, P, Q, PF) Energy Single-phase or three-phase

Circutor

Unit 2. AC SINGLE-PHASE CIRCUITSELECTRICAL MEASUREMENTS

LAB MEASUREMENTS: OSCILLOSCOPE

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