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Complex Algebra and Phasors 1 REVIEW OF COMPLEX NUMBERS Complex numbers are widely used to facilitate computations involving ac voltages and currents j = (-1); j 2 = -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 Re Im θ =tan -1 (b/a) a =Mcosθ b = Msinθ M = (a 2 + b 2 )

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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 = a 2 + b 2 Special case reciprocal 1/j:

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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

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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 Re sin -cos -sin 60° -30° cos (ωt - 30°) =sin(ωt + 60°) -135° 45° -sin(ωt + 45°) =sin(ωt - 135°)

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Complex Algebra and Phasors 5 WORKED EXAMPLE Take the two voltage waveforms: Converting to rectangular form and adding The polar form of v 1 + v 2 is Finally converting the polar form to sinusoidal form Example: Find i 1 – i 2 if i 1 = 1.5sin(377t + 30°) A and i 2 = 0.4sin(377t - 45°) A

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Complex Algebra and Phasors 6 PHASOR FORM OF RESISTANCE The voltage v(t) = V p sin(ωt + θ) V is in phase with the current i(t) = I p sin(ωt + θ) A when across a resistor By Ohms 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

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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) = V p sin(ωt + θ) V, i(t) = I p sin(ωt + θ + 90°) A Applying Ohms Law for capacitive reactance: v(t) = X C i(t) or Converting to phasor form Capacitive reactance is regarded as a phasor whose magnitude is |X C | = 1/ωC ohms, whose angle is -90° Capacitive reactance is plotted down the negative imaginary axis The rectangular form is X C = 0 – j|X C | Example: The current through a 0.25 F capacitor is i(t) = 40sin( t + 20°) mA. Use phasors to find the voltage across the capacitor. Draw a phasor diagram showing the voltage and current

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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) = I p sin(ωt + θ) A, v(t) = V p sin(ωt + θ + 90°) V Applying Ohms Law: v(t) = X l i(t) or Converting so phasor form Inductive reactance is regarded as a phasor whose magnitude is |X L | = ωL ohms with angle 90° Inductive reactance is plotted up the imaginary axis The rectangular form is X L = 0 + j|X L | Example: The voltage across an 8mH inductor is v(t) = 18sin(2π 10 6 t + 40°) V. Use phasors to find the current through the inductor. Draw a phasor diagram showing the voltage and current.

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