Phasor Circuit AnalysisPhasor DiagramPhasor Diagram ExampleCurrent DivisionVoltage DivisionIn-Phase and Out-of-PhasePhase Lead and Phase LabEampleKevin D. Donohue, University of Kentucky 1Phasor Circuit AnalysisPhasor Diagrams, Voltage and Current DivisionKevin D. Donohue, University of Kentucky 2Phasor DiagramPhasors are denoted by vectors in 2-D space. Phasor diagrams graphically illustrate the summation of complex valuesConsider the following summation of complex numbers resulting from setting up a KVL or KCL equation:36.664.036.636.67)180/29.84 sin(4.6)180/29.84 cos(4.6)180/45 sin(9)180/45 cos(9729.844.645907jjjj++−=++−+−=∠+−∠=∠°°ππππ°−∠ 45907∠RE°∠ 29.844.6IMKevin D. Donohue, University of Kentucky 3Phasor Diagram ExampleDraw Currents is, ir, il, icin a phasor diagram to show is = ir+ il+ icwhere is(t) = sin(1000t) A isirilic1Ω1mF0.5mHKevin D. Donohue, University of Kentucky 4Current DivisionBy substituting impedance in for resistance (or admittance for conductance), the current division formula can be generalize to:sIˆ1ˆZ2ˆZNZˆ………………kZˆkIˆNkksNkkskYYYYYIZZZZZIIˆˆˆˆˆˆˆ1ˆ1ˆ1ˆ1ˆ1ˆˆ2121+++++=+++++=LLLLKevin D. Donohue, University of Kentucky 5Voltage Division sVˆ1ˆZ2ˆZNZˆ...kZˆ−+kVˆ...By substituting impedance in for resistance, the voltage division formula can be generalize to:NkkskZZZZZVVˆˆˆˆˆˆˆ21+++++=LLKevin D. Donohue, University of Kentucky 6In-Phase and Out-of-PhaseVoltages and currents are considered in phase if the phase angle between their phasor quantities is zero, otherwise they are out of phaseby and amount equal to their phase difference.Which quantities are in phase? °∠°−∠°∠°−∠°∠°∠ 45ˆ 45ˆ 270ˆ 90ˆ 0ˆ 45ˆ221221IIIVVVKevin D. Donohue, University of Kentucky 7Phase Lead and Phase LabOut of phase voltages and currents are considered to either lead or lag each other. If the phase of quantity 1 is subtracted from the phase of quantity 2 and the result is positive, then quantity 1 leads quantity 2 or equivalently quantity 2 lags quantity 1.GivenDescribe the phase difference betweenÎ v1and i1Î v2and i2Î v3and i3°∠°−∠°∠°−∠°∠°∠ 45ˆ 45ˆ 270ˆ 90ˆ 0ˆ 45ˆ321321IIIVVVKevin D. Donohue, University of Kentucky 8EampleFind R such that the v1leads i1by 15 degrees, if v1= cos(400t) V 312.5µ F20 ΩRv1i1Show R = 15