Unformatted text preview:

1EE2315Lecture9Phasor CircuitAnalysis,EffectiveValueandComplexPower:Watts,VAR’sandVolt‐AmperesEffectiveValueofaSinusoid(1/2)Average Power:i(t) 10 2In our example:Also:The effective value is also called the Root Mean Square value or rms value.EffectiveValueofaSinusoid(2/2)R‐CCircuitExample(1/6)Capacitive Reactancevs(t)40 88.42 F+ vR -+vc-i3Using rms phasor for voltage source.R‐CCircuitExample(2/6)12040 -j30 + VR -~+VC-~I~Calculate Real Power:And Reactive Power:Apparent power is the product of voltage and current of the source.Also:R‐CCircuitExample(3/6)4Power Factor is the ratio of real power to apparent power:Power Factor is also the Cosine of the anglebetween the load voltage and the load current:If the load current leads the load voltage, the power factor is leading; if it lags the load voltage,the power factor is lagging.R‐CCircuitExample(4/6)Phasor Diagram of Voltage and CurrentCurrent leads voltage.R‐CCircuitExample(5/6)36.87oI~~V120V2.4A5The Power Triangle showing leadingpower factor.R‐CCircuitExample(6/6)CalculatingComplexPower(1/2)2SPjQIRjX  *2*22,I c jd I c jdthen I I c d I  *SPjQIIRjX  RjXI~+ V -~cjdI~6CalculatingComplexPower(2/2)VRjXI*SPjQVI From now on, we use the above method to calculate complex power.LaggingPowerFactorExample(1/4)v(t)5 22.97 mHi7Calculate complex power directly:LaggingPowerFactorExample(2/4)1200°5 j8.66 I~Power Factor:Power Factor is LaggingPhasor Diagram of Voltage and CurrentLaggingPowerFactorExample(3/4)60o1200° V12-60° Aimagreal8Power Triangle for lagging Power FactorLaggingPowerFactorExample(4/4)1440 VA720 W1247 VAR'simagrealPhasorPowerExample(1/4)480V(rms)0.5 j2 40 j30 -j150 +V-~I~480 07.94 20.08(56.75 20.75)VIAj(40 30)( 150)(56.25 18.75)40 30 150pjjZjjj9PhasorPowerExample(2/4)3.550.949 94.9%3.74PF or Lagging*(471 1.65 )(7.94 20.08 )3.55 1.18RLCSVIVAkW j kVAR (56.25 18.75) 471 1.65VI j    PhasorPowerExample(3/4)*(471 1.65 )(9.42 38.52 )3.55 2.66RL RLSVIVAkW j kVAR 471 1.659.42 38.52(40 30)RLVIAjCapacitorVA R ’s:224711.48150cVQ


View Full Document

UT Arlington EE 2315 - Lecture Notes

Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?