Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12EE201 Lecture 37 P. 1Average Power and Effective ValueConcept of effective value for currentIs(t) - periodic Ieff - dc+-R RVs(t)= tVscosAverage power absorbed by resistor is PaveAverage power absorbed is same as for case with periodic currentEffective value of periodic current (Ieff) is such that a dc current of value Ieff causes the same amount of power (average) to be absorbed in resistor.EE201 Lecture 37 P. 2 Pave,R = R*Ieff2 =TRdttiT02)(1Tttef fdttiTI00)(1222/1200)(1TtteffdttiTIIeff > 0Eq (1) impliesor,2/1200)(1TtteffdttvTV(2)(3)Similarly for the effective value of a periodic voltage,(1)EE201 Lecture 37 P. 3Example: Compute the effective value of the periodic voltage waveform sketched.1 2 3 4042v(t), Vt, sPeriodic waveform having effective value Veff = 2.309 V.SolutionV2eff = 14∫40v2(t)dt = 14∫10(2t)2dt + 14∫2122dt +14∫3242dt =163Therefore, Veff = 2.309 V.V2Example: Show thatEE201 Lecture 37 P. 4effeffeffeffaveIVRVRIP 22TavedttitvTP0)()(1 T TavedttiTRdttiTRP0 022)(1)(For periodic currents i(t) and voltage v(t)TavedttvTRP02)(11Pave=R*Ieff2Ieff2EE201 Lecture 37 P. 5 RVRIPeffeffave222222effeffaveVIP Pave=VeffIeffNote: Effective or rms values of sinusoidal currents or voltages are 0.707 times the maximum value of the waveform.tImFor dcIeff=ImRVPeffave2EE201 Lecture 37 P. 6tIImFor sinusoidal currents2meffII 3meffII For triangle current waveformsImImIIttFor square wavesIeff=ImEE201 Lecture 37 P. 7Similar relationships are valid for voltage waveforms of the same shapes as current waveforms above.Derivation of as function of effective (rms) quantitiesRecall, ivmmaveIVP cos2 ivmmaveIVP cos22effmVV2effmII2 iveffeffaveIVP cos;ButPaveEE201 Lecture 37 P. 8Dropping the ‘eff’ subscript and defining the difference in phase angle as Θz= Θv- ΘiPave = VI cos(Θv- Θi) = VI cos(Θz)Θz – angle representing extent voltage phasor leads current phasorExample: Find the rms value for the current waveform below.1234i(t)3 6 9 12 15t(s)Period T = 9sEE201 Lecture 37 P. 9TtteffdttiTI00)(1229022)(91dttiIeff )75(9104827912effI35effIIeff2 = 8 1/3 A= 2.88AEE201 Lecture 37 P. 10Example: The figure shows two types of household loads connected in parallel to a 110-V, 60-Hz source, vin(t) = 110√2 cos(120πt) V. Lamp 1 and lamp 2 have effective hot resistances of 202 Ω and 121 Ω, respectively. The impedance of the fluorescent light is Zfl(jω) = (56 + j 66) Ω.(a) Find the average power consumed by each light.(b) Find the average power delivered by the source.Fluorescent light+ Lamp 1 Lamp 2I1I2I3VinSolutionPart (a) For lamp 1, Z1(jω) = 202 0° Ω. Hence I1 = Vin/Z1 = 0.545 0° A. P1,av = V1,effI1,effcos(θz1) = 110 × 0.545cos(0°) = 59.9 WWhere V1,eff = |V1,eff| and similarly for I1,eff. This means that lamp 1 is a 60-W bulb.Similarly for lamp 2, Z2(jω) = 121 0° Ω. Hence I2 = Vin/Z2 = 0.909 0° A. P2,av = V2,effI2,effcos(θz2) = 110 × 0.909 cos(0°) = 100 WFinally, for the fluorescent light, Zfl(jω) = 56 + j66 = 86.56 49.7° Ω. Hence, I3 = Vin/Zfl = 1.27 -49.7° A. Pfl,av = V3,effI3,effcos(θz3) =110 × 1.27cos(49.7°) = 90.4 WEE201 Lecture 37 P. 11EE201 Lecture 37 P. 12Part (b)For this part we first compute Iin and compute the average power delivered by the source. Here by KCL,Iin = I1 + I2 + I3 = 0.545 0° + 0.909 0° + 1.27 -49.7° = 2.276 – j0.969 = 2.474 -23.06° ASo, the average power delivered by the source isPav = |Vin| × |Iin| cos(θv – θi) = 110 × 2.47cos(23.06°) = 250.3
View Full Document
Unlocking...