Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Circuits IIEE221Unit 3Instructor: Kevin D. DonohueInstantaneous, Average, RMS, and Apparent Power, and, Maximum Power Transfer, and Power FactorsPower Definitions/Units:Work is in units of newton-meters or joules same as energy.1 joule is work done for 1 ampere passed through 1 ohm for 1 second, or work done by a force of 1 Newton applied over 1 meter. Power is a measure of the rate at which work is done and is in units ofjoules per seconds, 746 Watts in one horsepower.Power ConversionElectric motors and generators convert electrical power to mechanical power and vise versa.Other devices exist that convert electric power to light, heat, sound, …. and vise versaInstantaneous and Average PowerThe instantaneous power is the power absorbed by an element at an instance of time. In an electric circuit this is given by:where i(t) is the current through the element and v(t) is the voltage drop over the element.Average Power over some time interval T is given by:If the i and v product is periodic, then PAV can be reduced to the integration over a single period. For random/non-periodic signals T goes to infinity.)()()( titvtP TAVdttitvTP0)()(1Instantaneous Power Instantaneous power is simply computed by the product of 2 functions. For sinusoidal problems, trig identities or phasors can be used to simplify the products for easier average power formulae. Example: Write a Matlab script that plots the instantaneous power of a cosine voltage across an impedance load. Write the script so the voltage waveform and impedance can be easily changed and instantaneous power replotted.Matlab ScriptsA script is a series of Matlab command line instructions typed in a text file and stored with an *.m extension. This is referred to as an mfile. To run these commands change your current directory associated with the Matlab work space to the one containing the mfile. Then type the base name of the file in the Matlab command line. Or on the editor menu there is a green “play” arrow that will also execute the program.Matlab ScriptsFor this example comments are included in the script, and are identified by text following a “%” character. The functions plot and cos are used in this script. Type help plot or help cos to get a full description of these functions in the Matlab workspace.Matlab Scripts% This script will plot the instantaneous power absorbed% by an impedance load with a sinusoidal voltage over it.% Set the parameters of the Voltage signalf = 1000; % Frequency of signal in Hzph = 30; % Phase of signal in degreesA=2; % amplitude of signal%Set impedance valuezm = 8; % Impedance magnitudezp = -40; % Impedance phase in degrees% Create a time axis of 2 periods over which to plot the powertp = 1/f; % Determine periodt = 2*tp*[0:5000]/5000; % Make a 5001 point time axis (row vector) over 2 periodsMatlab Scripts% Create a vector of points for voltage v = A*cos(2*pi*f*t + pi*ph/180);% Create a vector of points for the current (adjust magnitude and phase% based on impedancei = (A/zm)*cos(2*pi*f*t + pi*(ph-zp)/180);% Take an element by element product to get powerp = v.*i;% Plot itplot(t,p)title(['Instantaneous Power '] )xlabel(['Seconds'])ylabel(['Watts'])Result0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2x 10-3-0.100.10.20.30.40.50.6Instantaneous PowerSecondsWattsWhat is the meaning of negative instantaneous power?Average Power in Periodic SignalsGiven a sinusoidal voltage and current in a device:Show:)cos()()cos()(iivvtAtitAtv)cos(21ivivAVAAPAverage Power in Periodic SignalsGiven a phasor representation of a voltage and current in a device:Show:iivvAIAVˆˆ )cos(21ˆˆRe21*ivivAVAAIVPConservation of Power In a given circuit the average power absorb (denoted by positive values) equals the power delivered (denoted by negative values).For a circuit with N elements the sum of all power is zero: NiiP10Note: These are all real or average power values.Passive Sign ConventionIt is assumed that positive charge entering the positive terminal of an element implies power absorbed by the element.Therefore, charge leaving the positive terminal of an element implies power supplied or delivered by the element.If the words absorbed or supplied are not given with a power value, power absorbed will be assumed. (A negative sign will imply power supplied).IVIVExample Average PowerFind the average power each of the elements given is(t) = 3cos(1000t)A isia25Ω80µF10mH5Ω10iaP25Ω=34.61WPL=PC=0P5Ω=21.13WPCCVS=-5.54WPis=-50.19WMaximum Power TransferGiven a Thévenin circuit with load ZL:Show that for a maximum power transfer to the load:And the maximum power transfer is: *ˆˆthLZZ ZthZLVth]ˆRe[8ˆ*2maxththZVP Example Maximum PowerFind the impedance of the load Z to result in the maximum power transfer, and find the resulting power. Assume vs(t) = 110cos(377t) VZ=2.87-75.88Ith=1.6583.12AVth = 4.71 159VPz=3.98W vsZ8Ω40µF7.5mH12ΩRoot Mean Square (RMS) ValuesThe RMS value of a periodic current or voltage is its DC equivalent value for delivering average power to a resistor.T TrmsAVRIdtTtiRdttR iTP222)()(1TrmsdttiTI )(12 T TrmsAVRVdtTtvRdtRtvTP222)(1)(1TrmsdttvTV )(12RMS Formula for SinusoidsThe RMS value for a sinusoid of any frequency or phase is its amplitude divided by square root of 2.   TTrmsTrmsTrmsTrmsdttjtjTAdtTAIdttjtjTAIdttjtjTAIdttATI))(2exp())(2exp(424))(2exp()0exp(2))(2exp(42))(exp())(exp()(cos1222222220242ATTAIrmsApparent Power and Power FactorApparent power (S) for sinusoidal waveforms is the product of the RMS voltage and current magnitudes without regard to their phase offsets. It is in units of Volt-Amps (VA). The cosine of their phase difference is the power factor (pf).)cos()()cos()(iivvtAtitAtv )cos()cos(21ˆˆRe21*ivrmsrmsivivAVIVAAIVPiivvAIAVˆˆ power)(complex )(expˆfactor)(power )cos(pfpower) (reactive )sin(power) (real