1. REVIEW OF COMPLEX NUMBERS
Complex numbers are widely used to facilitate computations involving ac voltages and currents
j = (-1); j2 = -1
A complex number C has a real and imaginary part
C = a + jb a is the real part, b is the imaginary part
Can also use C = (a, b)
C = a + jb (rectangular form)
C = M θ (polar form)
C = Mcosθ + jMsinθ Re Im
θ =tan-1(b/a)
a =Mcosθ
b = Msinθ
M = (a2 + b2)
Complex Algebra and Phasors

2. ARITHMETIC OPERATIONS
(a + jb) + (c + jd) = (a + c) + j(b + d)
(a + jb) - (c + jd) = (a - c) + j(b - d)
Polar form:
(a + jb)  (c + jd) = (ac – bd) + j(bc + ad)
Complex conjugate C’ = a – jb =
CC’ = a2 + b2
Special case reciprocal 1/j:
Complex Algebra and Phasors

3. Complex Algebra and Phasors
A phasor is a mathematical representation of an ac quantity in polar form
A phasor can be treated as the polar form of a complex number, so it can be converted to an equivalent rectangular form
To represent an ac voltage or current in polar form, the magnitude M is the peak value of the voltage or current
The angle θ is the phase angle of the voltage or current
Examples:
The frequency of the phasor waveform does not appear in its phasor representation, because we assume that all voltages and currents in a problem have the same frequency
Phasors and phasor analysis are used only in circuit problems where ac waveforms are sinusoidal
Complex Algebra and Phasors

4. Complex Algebra and Phasors
We can also use phasors to convert waveforms expressed as sines or cosines to equivalent waveforms expressed as cosines and sines respectively Im
cos
cos (ωt - 30°)
=sin(ωt + 60°)
-30°
60° Re
sin
-sin
45°
-135°
-sin(ωt + 45°)
=sin(ωt - 135°)
-cos
Complex Algebra and Phasors

5. Complex Algebra and Phasors
WORKED EXAMPLE
Take the two voltage waveforms:
Converting to rectangular form and adding
The polar form of v1 + v2 is
Finally converting the polar form to sinusoidal form
Example: Find i1 – i2 if i1 = 1.5sin(377t + 30°) A and i2 = 0.4sin(377t - 45°) A
Complex Algebra and Phasors

6. PHASOR FORM OF RESISTANCE
The voltage v(t) = Vpsin(ωt + θ) V is in phase with the current i(t) = Ipsin(ωt + θ) A when across a resistor
By Ohm’s Law
Converting to phasors we have
We can regard resistance as a phasor whose magnitude is the resistance in ohms and whose angle is 0°
In the complex plane, resistance is a phasor that lies along the real axis
The rectangular form of resistance is R + j0
Example: The voltage across a 2.2kΩ resistor is v(t) = 3.96sin(2000t + 50°) V. Use phasors to find the current through the resistor. Draw a phasor diagram showing the voltage and current
Complex Algebra and Phasors

7. PHASOR FORM OF CAPACITIVE REACTANCE
The current through a capacitor leads the voltage across it by 90°
When v(t) = Vpsin(ωt + θ) V, i(t) = Ipsin(ωt + θ + 90°) A
Applying Ohm’s Law for capacitive reactance:
v(t) = XCi(t) or
Converting to phasor form
Capacitive reactance is regarded as a phasor whose magnitude is |XC| = 1/ωC ohms, whose angle is -90°
Capacitive reactance is plotted down the negative imaginary axis
The rectangular form is XC = 0 – j|XC|
Example: The current through a 0.25F capacitor is i(t) = 40sin(2104t + 20°) mA. Use phasors to find the voltage across the capacitor. Draw a phasor diagram showing the voltage and current
Complex Algebra and Phasors

8. PHASOR FORM OF INDUCTIVE REACTANCE
The voltage across an inductor leads the current through it by 90°
When i(t) = Ipsin(ωt + θ) A, v(t) = Vpsin(ωt + θ + 90°) V
Applying Ohm’s Law: v(t) = Xli(t) or
Converting so phasor form
Inductive reactance is regarded as a phasor whose magnitude is |XL| = ωL ohms with angle 90°
Inductive reactance is plotted up the imaginary axis
The rectangular form is XL = 0 + j|XL|
Example: The voltage across an 8mH inductor is v(t) = 18sin(2π106t + 40°) V. Use phasors to find the current through the inductor. Draw a phasor diagram showing the voltage and current.
Complex Algebra and Phasors