Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13EE201 Lecture 36 P. 1Instantaneous and Average PowerInstantaneous power absorbed by a circuit element is p(t) = v(t) i(t) (unit:Watts)The average power absorbed between time T1 and time T2 is Pave[T1,T2] = 1/(T2-T1) p(t) dt Pave= p(t) dtwhere T is the period. T2 T1T 01TFigure 11.3EE201 Lecture 36 P. 2EE201 Lecture 36 P. 3 Example: Show that the average power absorbed by a 1 resistor in the circuit below is (V0)2 over any time interval [T1,T2]. 1+VoSource voltage-time relationshipVotVs_EE201 Lecture 36 P. 4 Pave[T1,T2]= V I dt V = V o I = V o/R = Vo Pave [T1,T2] = Vo2 dtT2T1 1(T2-T1) 1 (T2-T1) T1T2Pave[T1,T2] = tVo2T2-T1T2T1Pave[T1,T2] = Vo2T2-T1(T2-T1) =Vo2EE201 Lecture 36 P. 5 tVmTo2To(a)_+-Vs(t)R(b)Example: Compute the average power absorbed by the resistor R connected to an independent voltage source, as in figure (b), with the excitation shown in figure (a)VSolution:Compute the instantaneous power for 0 ≤ t ≤ Top(t) = {(2Vmt / To)2 0 ≤ t ≤ 0.5ToR 0 0.5To ≤ t ≤ ToAverage Power in SSS CircuitsAll voltages and currents have same frequency. Instantaneous power:p(t) = v(t)i(t) = Vmcos(t+v) Imcos(t+i) (1)EE201 Lecture 36 P. 6 Compute Pav. Observing that the fundamental period is To, we have Pave= 0.5To 0 1To4Vm2To2Rt2 dt = Vm2 6RUsing the trig identity,cos(x)cos(y) = 0.5cos(x-y) + 0.5cos(x+y)Eq(1) becomes Eq(2)p(t) = cos(v -i) + cos(2t+V +i) VmIm 2 VmIm2EE201 Lecture 36 P. 7 Integrate over time to find PavePave= cos(v - i)dt + 1TVmIm 2 cos(2t+V +i)dtT0VmIm 2 1TThe second integral equals zero,Pave = cos(v - i) (3)Vm Im 20TEqn. 3 is the average power absorbed by a circuit element in period T. Consider average power absorbed by resistors, inductors and capacitors.EE201 Lecture 36 P. 8ResistorVR= v(t) = R i(t) = R IRVoltage and current have same phase angle v = iPave= = =Vm Im Im2 R Vm 2 2 2 2RInductorVL= (jL) ILMultiplication by j causes a phase difference between voltage and current v - I = 90ºPave = Vm Im cos (90º) = 0 2EE201 Lecture 36 P. 9 CapacitorVC = (-j/C) ICAgain, multiplication by j causes a phase diff. between voltage and current.v - i= - 90ºPave = Vm Im cos (-90º) = 0 2Concept of superposition of average powerSuperposition is valid for two independent sources if 12 or if (1 - 2) = ± k (/2) R_+-+-v2v1v1(t) = Vm1 cos(1t + 1)v2(t) = Vm2 cos(2t + 2)(Odd integer)EE201 Lecture 36 P. 10Example: Find the instantaneous power for the circuit below assuming = 3rad/sec and the source voltage is a maximum value.+_ 0.1F+vL-2.550ºLoadis(t)VStep 1: Find current from sourceYL= (1/R) + jC = 0.4 + j0.3YL= 0.5 36.9 º mhosI S = YL Vs = (0.5 36.9 º)(50º) = 2.536.9º AEE201 Lecture 36 P. 11Step 2: Calculate instantaneous powerp(t) = 5cos(3t) * 2.5cos(3t +36.9 º)Or, using Eq(2)p(t) = 6.25cos(-36.9º) + 6.25cos(6t +36.9 º) p(t) = 5 + 6.25cos(6t +36.9 º) WEE201 Lecture 36 P. 12Vs = 10 sin(5t) = 10 cos(5t -90 )Vs =10 -90 Example: Determine the average power delivered by the source.+_1/ j1.1550.510sin(5t)ZinI =Vs 10 -90 10 -90 Zin 0.5+(1/j1.155) 0.5-j0.87 = =I =10 -901 -60= 10 -30EE201 Lecture 36 P. 13p(t) = Vs(t) I(t)p(t) = (10 -90) (10 -30)p(t) = [10cos (5t -90)] [10cos (5t -30)]Using Eq (2),p(t) = 50cos (60) + 50cos (10t -120)p(t) = 25 + 50 cos(10t -120 )Finding the average power absorbedT 01T Pave= p(t) dtT 01T Pave = [ 25+50 cos(10t -120)] dt Pave = 25 WStep 2: Calculate instantaneous and average
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