phasors

Basic Engineering AC Systems and Phasors F Hamer, A Khvedelidze & M LavelleThe aim of this package is to provide a short selfassessment programme for students who wantto understand AC systems and the basic ideaof phasors.c 2006 chamer, akhvedelidze, [email protected] Revision Date: October 16, 2006 Version 1.0

Table of Contents1. Introduction (AC Systems)2. From Phases to Phasors3. Adding Phasors (Introduction)4. Final Quiz Solutions to Exercises Solutions to Quizzes The full range of these packages and some instructions, should they be required, can be obtained from our web page Mathematics Support Materials.

Section 1: Introduction (AC Systems) 31. Introduction (AC Systems)This package discusses AC (alternating current) systems. We restrictourselves to sinusoidal oscillations (see the package on Waves). Thediagram shows the voltage measured at a fixed point as a function oftime: V Vm t Vpeak to peak 0 −VmThis function can be described by V = Vm cos(2πf t + φ)Here Vm is the maximum voltage (−Vm is the minimum); f is thefrequency and φ is a phase. Vm is also called the peak voltage oramplitude. The peak to peak voltage is 2Vm.

Section 1: Introduction (AC Systems) 4Since AC voltage varies with time between ±Vm we need a way todescribe its typical value. The average voltage is zero (the positiveand negative voltages cancel out). But the square of the voltage is,see the diagrams below, always positive: V V2 Vm t Vm2 0 0−Vm t Voltage (volts) Voltage squared (volts squared) against time against timeNow take the square root of the average o√f V 2. This defines the rootmean square or RMS voltage (units are volts2 = volts). It can beshown that V√m 2 VRMS =

Section 1: Introduction (AC Systems) 5Example 1 An important example is Vm = 340 V and f = 50 Hz.This means that every second the voltage supply undergoes 50 cycles.Its period is T = 1/f = 1/50 = 0.02 s. This is the voltage andfrequency of domestic electricity supply in the UK.Since in dom√estic electricity supply Vm = 340 V the RMS voltage isVRMS = 340/ 2 = 240V to the nearest volt.Note that AC voltmeters display the value of the RMS voltage. TheRMS values are also used for comparisons with DC systems.Quiz Electricity in transmission lines is typically transmitted at highervoltages to reduce power loss. If the RMS voltage in a SuperGridtransmission line is 400 kV what is the peak voltage in the line? (a) 283 kV (b) 340 V (c) 566 kV (d) 400 kV

Section 2: From Phases to Phasors 62. From Phases to PhasorsExample 2 Consider the two AC voltages below: V t Vm 0−VmThe peaks or troughs of the red curve are measured at earlier timesthan their blue equivalents. The red curve is ahead of the blue curveby one quarter of a cycle (1cycle = 360◦ or 2π). As the frequencies arethe same, this phase difference stays constant. Here is some notation: • the red curve leads the blue curve by 90◦. • it has a positive phase shift of 90◦ with respect to the blue curve. • the blue curve lags the red curve by 90◦. • It has a negative phase shift of 90◦ with respect to the red curve.

Section 2: From Phases to Phasors 7Quiz Which of the statements about the curves below is correct? V Vm t 0−Vm(a) The blue curve leads the red one by 30◦.(b) The red curve has a negative phase shift of 30◦ with respect to the blue curve.(c) The red curve lags the blue one by 30◦.(d) The blue curve has a negative phase shift of 30◦ with respect to the red curve.

Section 2: From Phases to Phasors 8Exercise 1. In each case below sketch two AC voltages, V1 and V2,which have the same frequency against time for the given relativephases. In your sketch assume that V1 has twice as large a peakvoltage as V2.Hint: draw, say, V1 first and then add V2 with the requested phaseshift and smaller amplitude.(a) V1 lags V2 by 180◦.(b) V1 leads V2 by 90◦.(c) V2 lags V1 by 90◦.(d) V1 lags V2 by 45◦.(e) V1 has a negative phase shift with respect to V2 of 60◦.(f) V1 has a positive phase shift with respect to V2 of 90◦.

Section 2: From Phases to Phasors 9Example 3 Phasors are an efficient method of analysing AC circuitswhen the frequencies are the same. (The amplitudes do not need tobe.) Let us draw the example of the last quiz in phasor form: V Vm Vm 30 0 t Vm−VmIn the phasor diagram: everything is defined relative to thereference phasor (which is always chosen to point to the right).Here we have chosen the blue voltage as our reference. • The two voltages have the same amplitude. Therefore the arrow of the red phasor has the same length as the reference phasor. • The red voltage leads the blue voltage by 30◦: the red phasor is rotated anti-clockwise by a relative angle of 30◦ (written Vm∠30◦).

Section 2: From Phases to Phasors 10Exercise 2. Re-express each case below as a phasor diagram. V Vm Relative angle 30◦. Use blue curve as(a) 0 t reference.−2Vm V Vm Relative angle 30◦. Use red curve as(b) 0 t reference. −Vm V Read off relative angle from graph. Vm t Use blue curve as reference.(c) 0 −Vm

Section 2: From Phases to Phasors 11The result of adding two AC quantities depends on their relativephase. This is a typical wave interference phenomenon (see the pack-age on Waves). Below two voltages (red and blue lines) are addedand the resulting voltage (black line) is shown.VV Vm Vm t 0 0−Vm t −Vm Adding in phase Adding out of phase (adds ) (cancels )If all signals have one, fixed frequency, adding them produces a signalwith the same frequency but a different amplitude and a phase shift.We will not prove this result here.

Section 3: Adding Phasors (Introduction) 123. Adding Phasors (Introduction)Phasors can be used to combine AC voltages or currents.Example 4 Below two voltages (red and blue curves), which are outof phase by 90◦ are added. The result (black curve) has the samefrequency but a larger amplitude and its peaks lie exactly betweenthe other two.V √ Vm 2Vm 45 Vm t Vm 0−VmAdding two AC voltages Same result using phasorsThis result can be obtained directly by adding the two voltages in thephasor diagram. They are added in the same way that vectors areadded. This is calculated in detail on the next page.

Section 3: Adding Phasors (Introduction) 13Vm VR θ VR Vm θ θ Vm VmFrom the right angled triangle, the angle θ is given bytan(θ) = Vm = 1 , ⇒ θ = tan−1(1) = 45◦ . Vmwhile the amplitude of the resultant voltage, VR, is the length of thehypotenuse: by Pythagoras’s theorem √VR2 = Vm2 + Vm2 , ⇒ VR = 2Vm2 = 2Vm √The result is thus: Vm∠0◦ + Vm∠90◦ = 2Vm∠45◦ .Quiz What is the result of adding 3∠0◦ + 4∠90◦ (volts)?(a) 7∠90◦ V (b)5∠53◦ V (c) 5∠1◦ V (d) 12∠60◦ V

Section 3: Adding Phasors (Introduction) 340 135 14 340 VExample 5 Here we add: 340∠0 + 340∠135 (volts) N.B. the red phasor points left at 180 − 135 = 45◦ above the horizontal.Calculate the components along the axes. These are horizontally: 340 − 340 cos(45) = 100 Vand vertically: 0 + 340 sin(45) = 240 VThus (by Pythagoras’s theorem) the resultant peak voltage is VR = 1002 + 2402 = 260 Vand the phase is θ = tan−1 240 = 67◦ . 100This result is shown on the next page.

Section 3: Adding Phasors (Introduction) 15The result, 340∠0◦ + 340∠135◦ = 260∠67◦, is shown below:340 135◦ 340 67◦ 340 VThe arrows show the geometrical way to add phasors. Phasors point-ing in opposite directions tend to cancel each other.Exercise 3. Here are some exercises on adding phasors. In each casedraw a phasor diagram and thus calculate the sum.(a) 12∠0◦ + 5∠90◦ (volts) (b) 3∠0◦ + 4∠ − 90◦ (volts)(c) 1∠0◦ + 2∠60◦ (amperes) (d) 2∠0◦ + 2∠30◦ (volts)Quiz What is the result of 12∠0◦ + 5∠180◦ (volts)?(a) 7∠180◦ V (b)17∠90◦ V (c) 13∠85◦ V (d) 7∠0◦ V

Section 4: Final Quiz 164. Final Quiz 340 60◦Begin Quiz V 100 V 0 t 340 V −100 V1. In the graph, what is the RMS voltage of the red curve? (a) 141 V (b) 282 V (c) 200 V (d) 71 V2. Which statement about the blue curve below is wrong? (a) It has a positive phase shift (b) It leads the red curve (c) It has a negative phase shift (d) The red curve lags it.3. What would be the result of adding the phasors in the diagram? (a) 680∠60◦ V (b) 524∠24◦ V (c) 340∠30◦ V (d) 589∠30◦ VEnd Quiz

Solutions to Exercises 17Solutions to ExercisesExercise 1(a)Since the AC voltage V1 lags V2 by 180◦ and its peak is twice as largeas the V2 peak, the voltage V1 is given by V1 = 2V2 cos(2πf t − 180)and the diagram for the voltages is V V2 0 t −V1A phase difference of 180◦ means that the two voltages are completelyout of phase.Click on the green square to return

Solutions to Exercises 18Exercise 1(b)The AC voltage V1 leads V2 by 90◦ and its peak is twice as large asthe V2 peak therefore the voltage V1 is V1 = 2V2 cos(2πf t + 90)and the diagram is V V2 0 t −V1Click on the green square to return

Solutions to Exercises 19Exercise 1(c)If V2 lags V1 by 90◦, then V1 leads V2 by 90◦. This is therefore justa different way of expressing the question of part (b) of the presentexercise. V V2 0 t −V1Click on the green square to return

Solutions to Exercises 20Exercise 1(d)The AC voltage V1 lags V2 by 45◦ and its peak is twice as large as theV2 peak, therefore the voltage V1 is V1 = 2V2 cos(2πf t − 45)and the diagram is V V2 0 t −V1Click on the green square to return

Solutions to Exercises 21Exercise 1(e)The AC voltage V1 has a negative phase shift with respect to V2 of60◦ and its peak is twice as large as the V2 peak therefore the voltageV1 is given by V1 = 2V2 cos(2πf t − 60)and the voltages are V V2 0 t −V1Click on the green square to return

Solutions to Exercises 22Exercise 1(f )If V1 has a positive phase shift with respect to V2 of 90◦, then V1leads V2 by 90◦ and therefore this is just a different way of expressingthe question of part (b). V V2 0 t −V1Click on the green square to return

Solutions to Exercises 23Exercise 2(a)The blue reference voltage Vleads the red voltage by 30◦ Vmand the peak voltages are re- 0 tlated by Vred = 2Vblue. −2VmTherefore in the phasor diagram the red phasor is rotated clockwiseby 30◦ and the arrow of the red phasor is twice as long as the bluereference phasor: Vm 2Vm 30Click on the green square to return

Solutions to Exercises 24Exercise 2(b) V t The blue voltage lags the Vm red reference voltage by 30◦ 0 and the peak voltages are the same Vred = Vblue. −VmTherefore in the phasor diagram the blue phasor is rotated clockwiseby 30◦ relative to the red one and they have the same length: Vm Vm − 30Click on the green square to return

Solutions to Exercises 25Exercise 2(c) V t From the graph, the red Vm voltage leads the blue ref- 0 erence voltage by 90◦ while the peak voltages are equal. −VmTherefore in the phasor diagram the red phasor is rotated anti-clockwiseby 90◦ relative to the blue one and the length are the same: Vm 90 VmClick on the green square to return

Solutions to Exercises 26Exercise 3(a)The sum of two phasors 12∠0◦V and 5∠90◦ V is the sum of two vectorsVR = V1 = (12, 0) , V2 = (0, 5) (see the package Introduction toVectors). The length of the resulting vector V1 + V2 = (12, 5) is √ VR = 122 + 52 = 169 = 13 V ,and its angle relative to V1 istan(θ) = 5 ≈ 0.42 , ⇒ θ = tan−1(0.42) ≈ 23◦ . 12The result 12∠0◦V + 5∠90◦V = 13∠23◦V is drawn below 5 90◦ 13 23◦ θ 12 0◦Click on the green square to return

Solutions to Exercises 27Exercise 3(b) The sum of two phasors 3∠0◦V and 4∠ − 90◦ V is asum of two vectors V1 = (3, 0) , V2 = (0, −4) . The length and therelative angle of the resulting voltage VR = V1 + V2 = (3, −4) are √ VR = 32 + 42 = 25 = 5 V .tan(θ) = −4 ≈ −1.33 , ⇒ θ = tan−1(−1.33) ≈ −53◦ . 3The result 3∠0◦V + 4∠ − 90◦V = 5∠ − 53◦V is drawn below θ 3 0◦ 4 − 90◦ 5 − 53◦Click on the green square to return

Solutions to Exercises 28Exercise 3(c)The sum of two currents 1∠0◦A and 2∠60◦ A is a√sum of two vectorsA1 = (1, 0) and A2 = (2 cos(60), 2 s√in(60)) = (√1, 3) . Evaluating thesum AR = A1 + A2 = (1 + 1, 0 + 3) = (2, 3) we find the lengthand the relative angle of the resulting current as √√ √AR = 4 + 3 = 7 ≈ 2.65A .tan(θ) = 3 ≈ 0.87 , 2 ⇒ θ = tan−1(0.87) ≈ 41◦ .The result 1∠0◦A + ∠60◦A = 2.65∠41◦A is drawn below 2 60◦ 2.65 41◦ θ 1 0◦Click on the green square to return

Solutions to Exercises 29Exercise 3(d)Since the amplitudes of the voltages 2∠0◦ and 2∠30◦ are the same, thesum must point halfway between them (here ∠15◦). The amplitudedepends upon the size of the angle between the phasors we are adding.The sum of the vectors V1 = (2, 0) and √ V2 = (2 cos(30), 2 sin(30)) = ( 3, 1) √is V1 + V2 = (2 + 3, 1) and the resulting amplitude is VR = √ (2 + 3)2 + 12 ≈ 3.86 V .The result 2∠0◦V + 2∠30◦V = 3.86∠15◦V is drawn below 2 30◦ 3.86 15◦ 2 0◦Click on the green square to return

Solutions to Quizzes 30Solutions to QuizzesSolution to Quiz:If the RMS voltage is 400 kV the peak voltage can be found as follows.Since V√m 2 VRMS = ,therefore the peak voltage is √ Vm = 2 × VRMS = 1.41 × 400 kV = 566 kV , End Quiz

Solutions to Quizzes 31 tSolution to Quiz: V Vm 0 −VmIn the diagram above the blue curve has a negative phase shift of 30◦with respect to the red curve because the peaks or troughs of the redcurve are measured at earlier times than their blue equivalents.Equivalent ways of stating this are: • the blue curve lags the red curve by 30◦ , • the red curve leads the blue curve by 30◦ . End Quiz

Solutions to Quizzes 32Solution to Quiz:To add two voltages, reference V1 and V2 , whose phasors are 3∠0◦Vand 4∠90◦ V we find first the relative angletan(θ) = V2 = 4 ≈ 1.33 , ⇒ θ = tan−1(1.33) ≈ 53◦ . V1 3The length of the resulting voltage, VR, is VR2 = √V12 + V22 = 32 + 42 = 25 V 2 ,∴ VR = 25 = 5 V .The result is thus: 3∠0◦V + 4∠90◦V = 5∠53◦V . End Quiz

Solutions to Quizzes 33Solution to Quiz:Drawing the diagram with two voltages 12∠0◦ and 5∠180◦ V 5 180◦ 7 0◦ 12 0◦we see that the resulting sum reads 12∠0◦ + 5∠180◦ = (12 − 5)∠0◦ = 7∠0◦ V . End Quiz