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COMPLEX NUMBERS
Topic 12
COMPLEX NUMBERS
GEOMETRIC REPRESENTATION
In Mathematics, a1 is known as real component and b1
as imaginary component but in electrical engineering, these
are known as in phase (or active) and quadrature (or
reactive) components respectively.
CONVERSION BETWEEN FORMS
Example 1
1. Convert 5 ∠ 30° to rectangular form.
Solution:
a = C cos Ø
b = C sin Ø
= 5 (cos 30°)
= 5 (sin 30°)
a = 4.33
b = 2.5
C = 4.33 + j 2.5
Example 2
2. Convert 6.93 – j4 to polar form.
Solution:
C = √ (a2 + b2)
= √((6.93)2 + (-4)2)
C=8
C = 8∠− 30°
Ø = tan-1(b/a)
= tan-1(-4/6.93)
Ø = − 30°
PROPERTIES OF j
ADDITION & SUBTRACTION of COMPLEX NUMBERS
MULTIPLICATION & DIVISION of COMPLEX NUMBERS
APPLICATION
Example 1
In an alternating circuit, the impressed voltage is given by
V = (100 − j50) volts and the current in the circuit is I = (3 + j4) A.
Determine the real and reactive power in the circuit.
Solution:
Power will be found by the conjugate method. Using current
conjugate, we have
PVA = VI
= (100 − j 50) (3 + j 4)
= 300 + j 400 − j 150 + 200
= 500 + j 250
Hence, real power is 500 W and reactive power of VAR is 250.
Since the second term in the above expression is positive, the reactive
volt-amperes of 250 are inductive.
Example 2
In the circuit of below, applied voltage V is given by (0 + j10) and
the current is (0.8 + j 0.6) A. Determine the values of R and X and also
indicate if X is inductive or capacitive.
Solution:
V = 0 + j10 = 10 ∠90° ;
I = 0.8 + j0.6 = 1∠36.9°
As seen, V leads the reference quantity by 90° whereas I leads
by 36.9°. In other words ,
I lags behind the applied voltage by (90°−36.9°) = 53.1°
Hence, the circuit in the given figure is an R-L circuit. Now,
Z = V/I
= 10 ∠90°/1 ∠36.9°
= 10 ∠53.1°
Z = 6 + j8
Therefore,
R=6Ω
and
XL = 8 Ω (inductive)
ACTIVITY NO. 9
SUBMISSION - DECEMBER 20, 2021
Answer/solve the following problems.
1. Convert the following currents to polar form :
a) 6 − j 8,
b) − 6 + j8 ,
c) − j5.
2. Perform the following indicated operations :
a) (60 + j 80) + (30 − j 40) , b) (12 − j 6) − (40 − j 20) ,
c) (6 + j 8) (3 − j 4) ,
d) (16 + j8) ÷ (3 −j 4)
3. In a given R-L circuit, R = 3.5 Ω and L = 0.1 H. Find a) the current
through the circuit and b) power factor if a 50-Hz voltage
V = 220 ∠ 30° is applied across the circuit.
Excerpt from Electrical Technology by Theraja,
Basic Electricity by Gussow,
Electrical Circuits by Siskind and
https://www.slideserve.com/paloma-bradley/lesson-18-phasorscomplex-numbers-inac/?utm_source=slideserve&utm_medium=website&utm_campaign=auto
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