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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 P IV θ = In the above equation, we know: 5 5000 P KW W = = p.f. 0.8 cos θ = = 140 V V = 5000 0.5 (140) 0.8 I = × 5000 89.286 0.5 140 0.8 I A = = × × 89.286 I A = is the amplitude of the load current
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