AC Systems and Phasors › uploads › production › document › ... · 2015-06-10 · AC Systems...

Basic Engineering

AC Systems and Phasors

F Hamer, A Khvedelidze & M Lavelle

The 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

http://www.plymouth.ac.uk/mathsmailto:protect �egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def mailto:[email protected]:protect �egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def

Table of Contents

1. Introduction (AC Systems)2. From Phases to Phasors3. Adding Phasors (Introduction)4. Final Quiz

Solutions to ExercisesSolutions to Quizzes

The full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.

http://www.plymouth.ac.uk/mathaid

Section 1: Introduction (AC Systems) 3

1. 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:

t

V

Vm

−Vm

0 Vpeak to peak

This function can be described by

V = Vm cos(2πft + 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.


Since 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:

t

V

Vm

−Vm

0

t

V 2

V 2m0

Voltage (volts)against time

Voltage squared (volts squared)against time

Now take the square root of the average of V 2. This defines the rootmean square or RMS voltage (units are

√volts2 =volts). It can be

shown thatVRMS =

Vm√2


Example 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 domestic 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 6

2. From Phases to PhasorsExample 2 Consider the two AC voltages below:

t

VVm

−Vm

0

The 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.


Quiz Which of the statements about the curves below is correct?

t

V

Vm

−Vm

0

(a) The blue curve leads the red one by 30◦.

(b) The red curve has a negative phase shift of 30◦ with respect tothe 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 tothe red curve.


Exercise 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◦.


Example 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:

t

V

Vm

−Vm

0Vm

Vm 6 30

In 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 ofthe red phasor has the same length as the reference phasor.• The red voltage leads the blue voltage by 30◦: the red phasor isrotated anti-clockwise by a relative angle of 30◦ (written Vm∠30◦).


Exercise 2. Re-express each case below as a phasor diagram.

(a)t

V

Vm

−2Vm

0Relative angle 30◦.Use blue curve asreference.

(b)t

V

Vm

−Vm

0Relative angle 30◦.Use red curve asreference.

(c)t

V

Vm

−Vm

0

Read off relativeangle from graph.Use blue curve asreference.


The 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.

t

V

Vm

−Vm

0t

V

Vm

−Vm

0

Adding in phase(adds)

Adding out of phase(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) 12

3. 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.

t

V

Vm

−Vm

0Vm

Vm√

2Vm 6 45

Adding two AC voltages Same result using phasors

This 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.


Vm

Vm VR 6 θ

θVm

VmVR

θ

From the right angled triangle, the angle θ is given by

tan(θ) =VmVm

= 1 , ⇒ θ = tan−1(1) = 45◦ .

while the amplitude of the resultant voltage, VR, is the length of thehypotenuse: by Pythagoras’s theorem

V 2R = V2m + V

2m , ⇒ VR =

√2V 2m =

√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


Example 5

Here we add: 340∠0 + 340∠135 (volts)

N.B. the red phasor points left at180− 135 = 45◦ above the horizontal.

340 V

3406 135

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 = 260V

and the phase is

θ = tan−1(

240100

)= 67◦ .

This result is shown on the next page.


The result, 340∠0◦ + 340∠135◦ = 260∠67◦, is shown below:

340 V

3406 135◦3406 67◦

The 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 16

4. Final Quiz

Begin Quiz

t

V100 V

−100 V

0340 V

3406 60◦

1. In the graph, what is the RMS voltage of the red curve?(a) 141 V (b) 282 V (c) 200 V (d) 71 V

2. 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◦ V

End Quiz

Solutions to Exercises 17

Solutions 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(2πft− 180)and the diagram for the voltages is

t

V

V2

−V1

0

A phase difference of 180◦ means that the two voltages are completelyout of phase.Click on the green square to return

�


Exercise 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(2πft + 90)

and the diagram is

t

V

V2

−V1

0

Click on the green square to return�


Exercise 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.

t

V

V2

−V1

0



Exercise 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(2πft− 45)and the diagram is

t

V

V2

−V1

0

Click on the green square to return �


Exercise 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(2πft− 60)and the voltages are

t

V

V2

−V1

0



Exercise 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).

t

V

V2

−V1

0



Exercise 2(a)

The blue reference voltageleads the red voltage by 30◦

and the peak voltages are re-lated by Vred = 2Vblue.

t

V

Vm

−2Vm

0

Therefore 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 6 30



Exercise 2(b)

The blue voltage lags thered reference voltage by 30◦

and the peak voltages arethe same Vred = Vblue.

t

V

Vm

−Vm

0

Therefore in the phasor diagram the blue phasor is rotated clockwiseby 30◦ relative to the red one and they have the same length:

Vm

Vm 6 − 30



Exercise 2(c)

From the graph, the redvoltage leads the blue ref-erence voltage by 90◦ whilethe peak voltages are equal.

t

V

Vm

−Vm

0

Therefore in the phasor diagram the red phasor is rotated anti-clockwiseby 90◦ relative to the blue one and the length are the same:

Vm

Vm 6 90



Exercise 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 is

tan(θ) =512≈ 0.42 , ⇒ θ = tan−1(0.42) ≈ 23◦ .

The result 12∠0◦V + 5∠90◦V = 13∠23◦V is drawn below

126 0◦

56 90◦136 23◦

θ



Exercise 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) =−43≈ −1.33 , ⇒ θ = tan−1(−1.33) ≈ −53◦ .

The result 3∠0◦V + 4∠− 90◦V = 5∠− 53◦V is drawn below

36 0◦

46 − 90◦

56 − 53◦

θ



Exercise 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 sin(60)) = (1,

√3) . Evaluating the

sum 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) =√

32≈ 0.87 , ⇒ θ = tan−1(0.87) ≈ 41◦ .

The result 1∠0◦A + ∠60◦A = 2.65∠41◦A is drawn below

16 0◦

26 60◦ 2.656 41◦

θ



Exercise 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

26 0◦

26 30◦3.866 15◦


Solutions to Quizzes 30

Solutions to QuizzesSolution to Quiz:

If the RMS voltage is 400 kV the peak voltage can be found as follows.Since

VRMS =Vm√

2,

therefore the peak voltage is

Vm =√

2× VRMS= 1.41× 400 kV= 566 kV ,

End Quiz


Solution to Quiz:

t

V

Vm

−Vm

0

In 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


Solution to Quiz:

To add two voltages, reference V1 and V2 , whose phasors are 3∠0◦Vand 4∠90◦ V we find first the relative angle

tan(θ) =V2V1

=43≈ 1.33 , ⇒ θ = tan−1(1.33) ≈ 53◦ .

The length of the resulting voltage, VR, is

V 2R = V21 + V

22 = 3

2 + 42 = 25 V 2 ,

∴ VR =√

25 = 5 V .

The result is thus:

3∠0◦V + 4∠90◦V = 5∠53◦V .

End Quiz


Solution to Quiz:

Drawing the diagram with two voltages 12∠0◦ and 5∠180◦ V

126 0◦

56 180◦ 7 6 0◦

we see that the resulting sum reads

12∠0◦ + 5∠180◦ = (12− 5)∠0◦ = 7∠0◦ V .

End Quiz

Table of Contents1 Introduction (AC Systems)2 From Phases to Phasors3 Adding Phasors (Introduction)4 Final Quiz Solutions to Exercises Solutions to Quizzes

eqs@qz:phasors: 1: 1: Off2: Off3: Off4: Off

2: 1: Off2: Off3: Off4: Off

3: 1: Off2: Off3: Off4: Off

qz:phasors: Score:eq@Mrk:qz:phasors:

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