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Phasor algebra

The following are the methods of representing vector quantities :

1. Symbolic notation

3. Exponential form

2. Trigonometrical form

4. Polar form.

 

A vector as shown in Fig. 28 may be described in the above forms as follows :

1. Symbolic notation :

E = a + jb

2. Trigonometrical form :

E = √ a2 + b2 ( cos θ + sin θ )

= √ a2 x b2 ( cos θ + j sin θ )

3. Exponential form :              E =  a2 x b2 e + jθ

= a2 + b2 e+ fθ

4. Polar form :                         E = a2 + b2 < θ

= a2 + b2 < + θ

Significance of operator j. The letter j used in the above expressions is a symbol of an operation. It is used to indicate the counter-clockwise rotation of a vector through 90°. It is assigned a value of √ – l· The double operation of j on a vector rotates it counter-clockwise (CCW) through 180° and hence reverses its sense because, j x j =P = √ (-1)2 = 1.

In general, each successive multiplication of j, rotates the phasor further by 90° as given below (Refer Fig. 29)

j = √ -1

… 90° CCW rotation from OX-axis

j2 = √ (-1)2 = 1.

… 180° CCW rotation from OX-axis

J3 = √ (-1)3 = – √ -1 = – j

… 270° CCW rotation from OX-axis

J4 = √ (-1)4 = + 1

… 360° CCW rotation from X-axis

It should also be noted that,

1 / j = j / j2 = j / -1 = – j
 

Example 14. Write the equivalent potential and polar forms of vector 6 + j8. Also illustrate the vector by means of diagram.

Solution. Refer Fig. 30.

Magnitude of the vector = √ 62 + 82 = 10 , tan θ = 8 / 6

θ = tan -1 ( 8 / 6)  53.1°

Exponential form = 10 e J 53.1 (Ans.)

The angle may also be expressed in radians.

Polar form                      = 10 L 53.r. (Ans.)

The vector is illustrated by means of diagram in Fig. 30.

 

Example 15. A vector is represented by 30e-i2rt1 3• Write down the various equivalent forms of the vector and illustrate by means of a vector diagram, the magnitude and position of the above vector.

Solution. Refer Fig. 31. Draw the vector in a direction making an angle of 2π / 3  = 2 x 180 / 3 = 120° in the clockwise direction (since the angle is negative).

(i) Rectangular form :

a = 30 cos (- 120°) = – 15

b = 30 sin (-120°) =-25.98

. . Expression is = (- 15 – j 25.98). (Ans.)

(ii) Polar form = 30 <  – 120 (Ans.)

Addition and subtraction of vector quantities :

For addition and subtraction of vector quantities rectangular form is best suited. Consider two voltage phasors represented as:

V1 = a1 + jb1 and V2 = a2 + jb2

Addition . V + V1 + V2 = ( a1 + jb1) + ( a2 + jb2) = ( a1 + a2 0 + j ( b1 + b2)

The magnitude of the resultant vector            V = ( a2 + a2)2 + ( b1 + b2) 2

The position of V with respect to X-axis is θ = tan -1 ( b1 + b2 / a1 + a2 )

Subtraction.  V = V1 – V2 = (a1 + jb1)- (a2 + jb2) = (a 1 – a 2) + j (b1 – b2)

The magnitude of the resultant vector V = (a1 – a2)2 + (b1 – b2)2

The position of V with respect to X-axis is θ = tan -1 ( b1 – b2 / a1 – a2 )

Multiplication and division of vector quantities :

It the vectors are represented in the polar exponential form, their multiplication and division becomes very easy and simple.

Consider two voltage phasors represented as

V1 = a 1 + jb1 = V1< θ1, where θ 1 = tan-1 (b1 / a1)

V2 = a2 + jb2 = V2 2 , where θ2 = tan -1 ( b2 / a2)

Multiplication. When the phasor quantities are represented in polar form, while multiplying their magnitudes are multiplied and their angles added algebraically,

Division. In this case, the magnitudes of phasor quantities (expressed in polar from) are divided and their angles subtracted algebraically

Example 16. Perform the following operation and express the final result in polar form : 10 L30° + 16 L- 30°.

Solution.     10 < 30° = 10 (cos 30° + j sin 30°) = 8.66 + j5

16 < 30° = 16 [cos (- 30°) + j sin (- 30°)] = 13.86 – j8

10 < 30° + 16L- 30° = (8.66 + j5) + (13.86 – j8) = 22.52-j3

= (22.52 + 32) tan -1 ( – 3 / 22 . 52)

= 22.72 tan -1 ( – 3 / 22 . 52 ) = 22.72 < – 7.6

 

Example 17. Subtract the following given vectors from one another.

A= 15 + j26 and ii =-19.75 – j7.18.

Solution. A-B = C = (15 + j26) – (- 19.75-j7.18) = 34.75 + j33.18

Magnitude of C =  34.75 2 + 33.18 2 = 48

Slope of C = tan -1 (33.18/34.75) = 43.68°

C = 48 < 43.68. (Ans.)

 

Example 18. Perform the operation C and express the final result m polar form for the

vectors given below

A = 10 + j10; B = 15L- 120°; C = 5 + j0°.

 Solution. Rearranging vectors A and in polar form, we have

 

Example 19. The instantaneous values of two currents i1 and i2 are given as:

i1 = 5 sin (wt + π / 4 ) and I 2 = 2.5 cos ( wt – π / 2)

Find the r.m.s. value of i1 + i2 using complex number representation.

Solution. Given :     i 1 = 5 sin (wt + π / 4)

i 1 = 2.5 cos (wt – π / 2) = 2.5 sin [ 90 + ( wt – π / 2)] = 2.5 sin wt

I 1(max ) = 5 (cos 45° + j sin 45°) = (3.53 + j3.53)

I 2(max) = 2.5 (cos 0 + j sin 0°) = (2.5 + j0)

The maximum value of resultant current is

I max= (3.53 + }3.53) + (2.5 + j0) = 6.03 + }3.53 = 6.987 L30.34°

R.M.S value = 6.987 / √2 = 4.94 A.

 

Conjugate complex numbers

Two numbers are said to be conjugate if they differ only in the algebraic sign of their quadrature components. Accordingly, the numbers (a+ jb) and (a-jb) are conjugate.

• The sum of two conjugate numbers gives in-phase or active component and their difference

gives quadrature or reactive component.

i.e.                               (a+ jb) +(a-jb) = 2a (i.e., active component), and

(a+ jb)- (a- jb) = j2b (i.e., reactive component).

The resultant is the sum of two vertical components only.

• The resultant arising out of the multiplication of two conjugate numbers contains no quadrature

component.

i.e.,                              (a+ jb) x (a- jb) = a2 -j2b2 = a2 + b2

The conjugate of a complex number is used to determine the apparent power of an A. C. circuit in complex form.

 

Power and roots of vectors / phasors

The powers and roots of vectors can be found conveniently in polar form. If the vector are not, in polar form, these should be converted into polar form before carrying out the algebraic operations, as mentioned below.

Powers. Consider a vector phasor quantity represented in polar form as A =A < θ,

Then                            (A)n =A n < (n x θ)

Example. Suppose it is required to find cube of the vector 4< 12o

Then,                           (4< 12°)3 = (4)3 < (3 x 12°) = 64 < 36°

Roots. Consider a vector (phasor) quantity represented in polar form as (A )lin = (A)lln L9/n.

Example. Suppose it is required to find cube root of 125 L60o

Then,                           (125 L60°)1/3 = (125)1/3 L60°/3 = 5<.20°.

 

The 120″ operator

In case of 3-phase work, where voltage vectors are displaced by 120° from one another (Fig. 32) it is convenient to use an operator, which rotates a vector/phasor through 120° toward or backwards without altering its length. This operator is ‘a’ and any operator which is multiplied by ‘a’ remains unaltered in magnitude but is rotated in CCW (counter-clockwise) direction by 120°.

a= 1 L120″

In cartesian form,        a = cos 120° + j sin 120•

= 0.5 + j 0.866

Similarly, a 2 = 1 L 120° x 1 L120o = 1 L240° = cos 240° + j sin 240° = – 0.5 – j 0.866

Hence the operator ‘a’ will rotate the in CCW by 240° which is the same as rotating the vector in CW(clock-wise) direction by 120°.

a2 = 1 L-120°

Similarly                                  a3 = 1 L360° = 1