EE201 Lecture 37 P 1 Average Power and Effective Value Concept of effective value for current Is t periodic Vs t Vs cos t R Average power absorbed by resistor is Pave Ieff dc R Average power absorbed is same as for case with periodic current Effective 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 T 1 2 i t Rdt T 0 Pave R R Ieff2 1 Ieff 0 Eq 1 implies I eff 2 t 0 T 1 T 2 i t dt t0 or I eff 1 T i t dt t0 t 0 T 1 2 2 2 Similarly for the effective value of a periodic voltage Veff 1 T v t dt t0 t 0 T 2 1 2 3 EE201 Lecture 37 P 3 Example Compute the effective value of the periodic voltage waveform sketched v t V Periodic waveform having effective value Veff 2 309 V 4 2 0 t s 1 2 Solution V2eff 1 4 3 4 4 0 v2 t dt 1 4 1 0 2t 2dt 1 4 Therefore Veff 2 309 V 1 4 2 1 22dt 3 2 4 dt 2 16 3 V2 EE201 Lecture 37 P 4 Example Show that 2 Veff 2 Pave I eff R Veff I eff R For periodic currents i t and voltage v t T 1 Pave v t i t dt T 0 T 1 T 2 R 2 Pave i t dt R i t dt T 0 T 0 Pave R Ieff2 Pave T 1 1 2 v t dt R T 0 Ieff2 EE201 Pave Veff R 2 Lecture 37 2 Pave RI eff 2 P 5 2 Veff 2 R 2 Pave I eff Veff 2 Pave VeffIeff Note Effective or rms values of sinusoidal currents or voltages are 0 707 times the maximum value of the waveform For dc Im Ieff Im t EE201 Lecture 37 For sinusoidal currents I P 6 Im 2 I eff Im t I Im For triangle current waveforms I eff Im 3 t Im I For square waves Ieff Im t EE201 Lecture 37 P 7 Similar relationships are valid for voltage waveforms of the same shapes as current waveforms above Derivation of Pave as function of effective rms quantities Recall Pave Pave But Vm I m cos v i 2 Vm I m cos v i 2 2 Vm Veff 2 Im I eff 2 Pave Veff I eff cos v i EE201 Lecture 37 P 8 Dropping the eff subscript and defining the difference in phase angle as z v i Pave VI cos v i VI cos z z angle representing extent voltage phasor leads current phasor Example Find the rms value for the current waveform below i t Period T 9s 4 3 2 1 t s 3 6 9 12 15 EE201 I eff 2 1 t0 T 2 i t dt T t0 Lecture 37 I eff 2 P 9 9 1 2 i t dt 90 1 1 2 I eff 27 48 0 75 9 9 Ieff2 8 1 3 A I eff 5 3 2 88A EE201 Lecture 37 P 10 Example The figure shows two types of household loads connected in parallel to a 110 V 60 Hz source vin t 110 2 cos 120 t V Lamp 1 and lamp 2 have effective hot resistances of 202 and 121 respectively The impedance of the fluorescent light is Zfl j 56 j 66 a Find the average power consumed by each light I1 Vin I2 I3 Lamp 1 Lamp 2 Fluorescent light b Find the average power delivered by the source EE201 Lecture 37 P 11 Solution Part a For lamp 1 Z1 j 202 0 Hence I1 Vin Z1 0 545 0 A P1 av V1 effI1 effcos z1 110 0 545cos 0 59 9 W Where V1 eff V1 eff and similarly for I1 eff This means that lamp 1 is a 60 W bulb Similarly for lamp 2 Z2 j 121 Hence I2 Vin Z2 0 909 0 0 A P2 av V2 effI2 effcos z2 110 0 909 cos 0 100 W Finally for the fluorescent light Zfl j 56 j66 86 56 49 7 Hence I3 Vin Zfl 1 27 49 7 A Pfl av V3 effI3 effcos z3 110 1 27cos 49 7 90 4 W EE201 Lecture 37 P 12 Part 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 A So the average power delivered by the source is Pav Vin Iin cos v i 110 2 47cos 23 06 250 3 W
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