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**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

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**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

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**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

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**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

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**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

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**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

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**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 0.25F capacitor is i(t) = 40sin(2104t + 20°) mA. Use phasors to find the voltage across the capacitor. Draw a phasor diagram showing the voltage and current Complex Algebra and Phasors

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**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(2π106t + 40°) V. Use phasors to find the current through the inductor. Draw a phasor diagram showing the voltage and current. Complex Algebra and Phasors

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