revision of phasors

7/29/2019 Revision of Phasors

http://slidepdf.com/reader/full/revision-of-phasors 1/22

Revision of Phasors

• A phasor is a complex number

• A phasor is a way of writing down a sinusoidal signal

a

jb X

+ - x of complex numbers

A complex number can be

In Cartesian form: a + jb

In Polar form: X



a

jb X

22 ba X

a

btan 1

θ

θcos X a θ sin X b

a+ jb X

To + and - complex numbers

Complex numbers should be in Cartesian form

Add/subtract real and imaginary parts separately

Eg. What is 3 - j7 + 431 ?

+ 431 = 3.43 + j2

3 - j7 + 3.43 + j2 = 6.43 – j5



To x and of complex numbers

If in Cartesian form:

Multiply ad bc jdbac jd c jba Divide:

22

d c

ad bc jbd ac

jd c jd c

jd c jba

jd c

jba

2222

d c

ad bc j

d c

bd ac

jd c

jba

If in polar form:

1θ X 2θY 21 θθ XY x =

212

1θθ

θ

θ

Y

X

Y

X Divide:

Multiply



Representation of a sinusoidal signal

In a linear circuit, all signals at the same frequency = 2 f

But signals will have different magnitude and have different

phase displacements to each other

Eg. The following are two signals in a circuit

10B

80

A



10B

80

A

If we say that A goes through zero at t =0

Then A is

And B is

t sinω10 010

80ω4 t sin 804

The "magnitude" becomes the magnitude of the

complex number

The phase displacement becomes the angle of the

complex number



10B

80

A

Actually, the magnitude of the phasor (complex number)is, by convention, the RMS value of the signal

(peak/root 2).

The phasor for A is thus

The phasor for B is thus or

(the angle can be either degrees or radians)

41832 .. or

0077.

80832.



A

B

10

The phasors can be put on an Argand diagram:

0077.

80832.



10

A B

60

Question 1

Go from "waveform on a scope" to "phasor" to "argand

diagram". E.g. you see B on a scope

Write down its phasor and put it on the Argand diagram

A is 0077. B is 12010



Adding sinusoidal currents or voltages etc.

You can now add sinusoidal signals together. You must add

them as phasors (complex numbers). NEVER JUST ADDTHEIR MAGNITUDES.

Consider two AC currents and , both of peak 10A, flowing

in the circuit below:

Ammeter 1

Ammeter 2

Ammeter 3

?

2 I ~

1 I ~



101 I ̂ Ammeter 1

Ammeter 2

Ammeter 3

?

102 I ˆ

Q3 If ALL ammeters are set to "AC", what do

Ammeters 1 and 2 read?

Answer:

What does Ammeter 3 read?

Answer:

Q2. If ALL the ammeters are set to "DC", what will they

read? Answer: All 0A

Both 7.07A

Don’t know. Not told phase of sinusoids



Question 4

Let and be as in the example above: 1 I ~

2 I ~

10

80

00771 . I ~ 8083.2~2 I

What will Ammeter 3 read? Answer:

I 1 = 7.07 + j0

I 2= 2.83{cos(-80)+jsin(-80)} = 0.49 - j2.78

I 1+ I 2 = 7.56-j2.78 = 8.05

-20. It will read 8.05A



2 I ~

1 I ~

10

Question 5

Let1 I

~2 I

~and be as follows:

What will Ammeter 3 read? Answer : 0 A !!



R, C and L in AC Circuits

In steady state AC circuits, R, L and C all have an

IMPEDANCE.

This complex for inductors and capacitors:

Resistor (purely real)Inductor (purely imaginary)

Capacitor (purely imaginary)

You will normally be given the values of R , L, C and = 2 f , where f = 50Hz (60Hz in the US).

The quantity =2f is called the angular frequency of the

AC in radians per second.

R X R L j X L ω

C j / X C ω1 C / j ω



Why should L go to jL in AC circuits?

Consider Faraday's Law for a coil:dt

t di L Rt it v

)()()(

Let the current be t sin I ˆ )t ( i ω

dt

t sin I ̂ d Lt sin I ̂ R )t ( v

ωω

t cos I ̂ Lt sin I ̂ R )t ( v ωωω

90ωωω t sin I ̂ Lt sin I ̂ R )t ( v

j I ~

L j I ~

RV ~ 0ω01

I ~

L j I ~

RV ~

ω I ~

X I ~

RV ~

L

change to phasors:



Worked Example R1

A 50Hz voltage of 220V rms is applied to

a series combination of R = 10 and L = 38mH.

(a) What is the rms value of the current flowing?

(b) What is the rms voltage across R and L?

(a) Impedance of R = 10

Impedance of L =

Combined series Impedance

v

R

L

= 10 + j12 L jX R Z jXL

R

Z

Impedance Phasors:

2 x 50 x 0.038 = 12



I ~

X RV ~

L 1210

0220

j X R

V ~

I ~

L

2.501.142.500

6.15

220

2.506.15

0220~ I

v

R

L

iNote that a reactive impedance (with

+j component), causes the current

to lag the voltage (i.e. current has a

-j component)

A useful mnemonic is C I V I L:

In a Capacitive circuit I leads V ;

whereas V leads I in an inductive circuit (L)



LV L

RV (b) What is the rms voltage across R and L?

I ~ L j I ~ RV ~V ~V ~ L R ω

2.501412.501.1410~~

I RV R

Voltage and Current Phasors:

V

-j

I

IR

-50.2°



LV L

RV

2.501699012.501.1412~~

j I L jV L

8.39169

~ LV

-j

V

Voltage and Current

Phasors:

jIXLI

IR

Note that although

This is because VR and VL are sinusoids and NOT in phase

L R V ~V ~V ~ , 220 141 + 169 !!

Note: 141

2

+ 169

2

= 220

2



Ammeter 1

Ammeter 2

Ammeter 3

?

Worked example R2

1 I ~

2 I ~

t sin )t ( i ω31 02

31 I ~

t cos )t ( i ω42 902

4

2

I

~

Let

Let

t cost sin )t ( i )t ( i ω4ω321

θω

21

t sin X )t ( i )t ( iResult will be sinusoid. Must find

magnitude and phase



t cost sin )t ( i )t ( i ω4ω321

θω

21

t sin X )t ( i )t ( it cos sin X cost sin X )t ( i )t ( i ωθθω21

3θ cos X 4θ sin X Hence: and

222222 43θθ sin X cos X 5X 252 , X

3

4

θ

θ cos X

sin X

3

4θ

1 tan

Not a good way of adding sinusoids!!

53ω521 t sin )t ( i )t ( i A / I ~

253



Much better to use Phasors:

02

3

1

I ~

902

4

2 I ~

3

4

2

5 13

tan I ~

And the answer is obtained almost immediately

02

31 I ~

902

42 I

~

Have and



Revision of Phasors

• A phasor is a complex number

• A phasor is a way of writing down a sinusoidal signal

• Animations on WebCT

a

jb X

http://hermes.eee.nott.ac.uk/teaching/cal/h51elc/elc0001.php




revision of phasors

Documents