06-postal math tool polar notation

8/3/2019 06-Postal Math Tool Polar Notation

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AMIE( I ) STUDY CIRCLE(REGD.)

A Focused ApproachPolar Notation

POLAR NOTATION

An alternate method of expressing sinusoidally varying

quantities is the polar notation whereby the magnitude

and phase angle is expressed as A. The magnitude is

A and the angle is as shown in figure.

B- is the vector OQ.

Angles are measured positive in the anticlockwise

direction.

COMPLEX NOTATION(J FORM)Consider the phasor OP = 300. This is shown as a

horizontal line of magnitude 3 units as in figure. This

vector if rotated through 900(anti clockwise) becomes

OQ = 3900 and can be represented as j3. Thesymbol j represents an operator whereby the phasor

300 is rotated through an angle 900(anticlockwise).

The phasor, OA can therefore be considered to have

two components a and jb and can be written as phasor

OA = a + jb.

a and b are the horizontal and vertical components of OA and incidentally fix the vector.

1 btana

=

Also note that j2

= -1 i.e. j = -1 which is complex.

Two or more electrical quantities can be easily added or subtracted using the j notation. For example, if two

currents are a + jb and c + jd, then their sum is (a+c) = j(b + d). For multiplication and division of two

quantities, the polar form is more convenient i.e.

A x B = AB +

andA A

B B

=


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A Focused ApproachExample

Express the following in j form(complex form)

(i) 1230 0

(ii) 12-300

Solution

(i) 12300 = 12cos(300) + jsin(300) = 10.4 + j6

(ii) 12-300 = 12 cos(-300) + jsin(-300) = 12cos300 - jsin300 = 10.4 - j6

[cos(-) = cos and sin(-) = -sin]

Example

Express in polar form

(i) -10 - j10

(ii) -10 +j7.5

(iii) 8 - j6

(iv) 25.4 - j1.2

Solution

The respective vectors are shown in following figure.

To express ( -10 - j10) in polar form, we need to find the magnitude OP and the angle

2 2OP ( 10) ( 10) 14.14= + =

1 1 010tan tan (1) 45

10

= = =


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A Focused ApproachBecause OP lies in third quadrant the angle will be 180 + 45 = 2250 from OX(anticlockwise direction) or -1350

from OX(clockwise direction).

= 2250

Hence -10 -j10 = 14.142250 or 14.14-1350

Similarly, other vectors evaluated in polar form are

2 210 j7.5 ( 10) (7.5) 12.5 + = + =

1 1 07.5tan tan ( .75) 36.9

10

= = =

Because OQ = -10 + j7.5 lies in second quadrant the angle will be -180 - 36.9 = -216.90

from OX(clockwise)

i.e. 143.10

from OX(anticlockwise)

= 143.10

Hence -10 +j7.5 = 12.5143.10 or 12.5-216.90

Similarly, 2 2 1 1 06

8 j6 8 6 tan 10 tan ( .75) 10 36.98

= + = =

and 2 2 1 01.2

25.4 j1.2 (25.4) (1.2) tan 25.43 2.725.4

= + =

Example

Perform the following operations. Express in polar form.

(i) 5300

+ 8-300

(ii) (3 + j4) + (-7 - j8)

(iii) 10450 - 860

0

(iv) (8 +j6) x (-10 - j7.5)

(v)0

0

10 30

5 60

(vi)4 j3

3 j4

+

(vii) 10300x 5-60

0

Solution

(i) 5300 = 5(cos30 + jsin30) = 4.33 + j2.5


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A Focused Approach8-300 = 8(cos30-jsin30) = 6.93 - j4

530 + 8-30 = (4.33 + j2.5) + (6.93 - j4) = 11.26 - j1.5 = 11.36-7.60

(ii) (3 + j4) + (-7 - j8) = -4 - j4 = 5.6572250

(iii) 10450 - 8600 = (7.07 + j7.07) - (4 + j6.93) = 3.07 + j0.14 = 3.073 2.60

(iv) (8 + j6) x (-10 - j7.5) = 8(-10) - (8)(j7.5) +(j6)(-10) + j(6)(-j7.5)

= -80 - j60 -j60 -j2(45) = -80 - j120 + 45 [j

2= -1]

= -35 - j125 = 2 2( 35) ( 120) 125 + =

angle = 1 1 0120

tan tan (3.428) 73.7335

= =

now -35 - j125 lies in third quadrant hence angle from OX will be 180 + 73.73 = 253.730

from

OX(anticlockwise direction) or -106.270 from OX(clockwise).

Hence (8 + j8)(-10-j7.5) = 125253.730 or 125-106.270.

(v)0

0

0

10 302 [30 ( 60)] 2 90 2(cos90 jsin90) j2

5 60

= = = + =

(vi) ( )2 2

4 j3 (4 j3)(3 j4) 12 j16 j9 12 j25 10 j25

3 j4 (3 j4)(3 j4) 3 (j4) 25 25

= = = =

+ +

= 2 2 1 01 25

0 ( 25) tan 1 90 j125 0

+ = =

(vii) 0 0 010 30 x 5 60 50 [30 ( 60)] 50 30 50(cos30 jsin30) 43.3 j25 = + = = =


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A Focused ApproachASSIGNMENT

Q.1 Express in polar form

(i) -12 +j6

(ii) 700 + j200

(iii) -69.4 - j40

Answer: 20126.90, 726160, 802100

Q.2 Express in rectangular(complex or j) form

(i) 12300

(ii) 20-450

(iii) 0.022100

Answer: 10.4 + j6, 14.14 - j14.14, -0.0173 - j0.01

Q.3 Perform following operations

(i) (3 - j2)(1 - j4)

(ii) (105.310 ) + (4 + j2)

(iii) (5 + j5) / (1 - j1)

(iv) (3200)(8-650)

Answer: -5 - j10, 10 + j10, j5, 24-450

06-postal math tool polar notation

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