REVIEW OF COMPLEX NUMBERS

1. Im b = Msinθ M = (a2 + b2) θ =tan-1(b/a) a =Mcosθ Re 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 + jba 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 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) • Polar form: • Complex conjugate C’ = a – jb = • CC’ = a2 + b2 • Special case reciprocal 1/j: Complex Algebra and Phasors

3. PHASORS (1) • 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. PHASORS (2) • 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. 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