EE201 Lecture 36 P 1 Instantaneous and Average Power Instantaneous 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 T Pave T1 T2 1 T2 T1 p t dt 2 1 T 0 Pave T p t dt where T is the period T1 EE201 Lecture 36 Figure 11 3 P 2 EE201 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 Vo 1 Source voltage time relationship Vs Vo t EE201 Lecture 36 1 T2 I V o R Vo T2 Pave T1 T2 V Vo V I dt T2 T1 T1 1 Pave T1 T2 T1 Vo2 dt T2 T1 Vo2 Pave T1 T2 t T2 T1 T2 T1 Vo2 T2 T1 Vo2 Pave T1 T2 T2 T1 P 4 EE201 Lecture 36 P 5 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 V b a Vs t Vm R t To To 2 Solution Compute the instantaneous power for 0 t To p t 2Vmt To 2 R 0 0 t 0 5To 0 5To t To EE201 Lecture 36 P 6 Compute Pav Observing that the fundamental period is To we have 1 0 5T 4Vm2 Pave 0 To To2R o t2 dt Vm2 6R Average Power in SSS Circuits All voltages and currents have same frequency Instantaneous power p t v t i t Vmcos t v Imcos t i 1 Using the trig identity cos x cos y 0 5cos x y 0 5cos x y Eq 1 becomes Eq 2 VmIm VmIm p t cos v i 2 cos 2 t V i 2 EE201 Lecture 36 P 7 Integrate over time to find Pave Pave 1 T T 0 VmIm cos dt v i 2 T 1 V I m m cos 2 t dt V i T 0 2 The second integral equals zero Pave Vm Im 2 cos v i 3 Eqn 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 8 Resistor VR v t R i t R IR Voltage and current have same phase angle v i Pave Vm Im Im2 R 2 Vm 2 2 2R Inductor VL j L IL Multiplication by j causes a phase difference between voltage and current v I 90 Pave Vm Im cos 90 0 2 EE201 Lecture 36 P 9 Capacitor VC j C IC Again multiplication by j causes a phase diff between voltage and current v i 90 Pave Vm Im cos 90 0 2 Concept of superposition of average power Superposition is valid for two independent sources if 1 2 or if 1 2 k 2 Odd integer v1 v2 R v1 t Vm1 cos 1t 1 v2 t Vm2 cos 2t 2 EE201 Lecture 36 P 10 Example Find the instantaneous power for the circuit below assuming 3rad sec and the source voltage is a maximum value is t 5 0 V Load vL 2 5 0 1F Step 1 Find current from source YL 1 R j C 0 4 j0 3 YL 0 5 36 9 mhos I S YL Vs 0 5 36 9 5 0 2 5 36 9 A EE201 Lecture 36 P 11 Step 2 Calculate instantaneous power p t 5cos 3t 2 5cos 3t 36 9 Or using Eq 2 p t 6 25cos 36 9 6 25cos 6t 36 9 p t 5 6 25cos 6t 36 9 W EE201 Lecture 36 P 12 Example Determine the average power delivered by the source 10sin 5t 0 5 1 j1 155 Zin Vs 10 sin 5t 10 cos 5t 90 Vs 10 90 10 90 10 90 I Vs Zin 0 5 1 j1 155 0 5 j0 87 10 90 I 10 30 1 60 EE201 Lecture 36 P 13 Step 2 Calculate instantaneous and average power p 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 absorbed Pave 1 T 0 p t dt T 1 T Pave 0 25 50 cos 10t 120 dt T Pave 25 W
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