3-Lesson_Notes_Lecture_25 - More Examples on Power Factor...

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EE 204 Lecture 25 More Examples on Power Factor and the Reactive Power The power factor has been defined in the previous lecture with an example on power factor calculation. We present two more examples in this lecture. Example 1: A load absorbs 5 KW of power at a leading p.f. of 0.8 when a 60 Hz sinusoidal voltage source, with a peak voltage of 140 V is applied across it. Calculate: a) Assuming the load voltage V has a zero phase angle, calculate the load current I b) The load impedance c) Is the load impedance capacitive, inductive or purely resistive? I 140 0 o V = Figure 1 Solution: The 5KW power absorbed by the load is actually the average power absorbed by the load. It is common practice in power engineering to refer to the average power, simply as power. Using 0.5 cos PI V θ = In the above equation, we know: 5 5000 P KW W == p.f. 0.8 cos 140 VV = 5000 0.5 (140) 0.8 I 5000 89.286 0.5 140 0.8 IA × × 89.286 = is the amplitude of the load current
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[Note that 89.286 ii II A θ == we need to know i to completely find I ] Since p.f. 0.8 cos 1 cos 0.8 36.870 o Is 36.870 o =+ or 36.870 o =− ? Because the p.f. is leading 0 < 36.870 o 0 o v = (the voltage across the load has a zero phase angle) 36.870 o vi θθ = 36.870 0 o i −= 36.870 o i = 89.286 36.870 o i A 140 0 o V = 89.29 36.87 = o IA Figure 2 b) the load impedance 140 0 1.568 36.870 89.286 36.870 o o L o V Z I =
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This note was uploaded on 01/16/2010 for the course EE ee204 taught by Professor Profosama during the Spring '09 term at King Fahd University of Petroleum & Minerals.

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3-Lesson_Notes_Lecture_25 - More Examples on Power Factor...

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