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The circuit characteristics...voltage supply is 117v, Impedance1 is 2k at 0degrees, Impedance2 is 1k at -45degrees and impedance3 is 5k at 30degrees. What is the apparent, real, reactive power and the power factor. It would be better if you can show the work in steps. thanks in advance!

2007-04-07 11:31:43 · 3 answers · asked by Anonymous in Science & Mathematics Engineering

volt of 117 is at 0degrees and the circuit is in series

2007-04-09 02:46:19 · update #1

3 answers

I think I ran into a similar problem . I worked for NASA at Gold Stone and there was a tremendous amount of third harmonic square waves in the neutral and with the neutral and ground tied together in many places causing square wave noise in our ground system . All connections between neutral and ground was separated except at the transformers . There was also other indications of a flattening on top of each half waveform ,indicating the transformer was reaching saturation but the normal meter reading said no.
What was happening is that the diodes in a power supply after first surge only conduct in about the top 30% of each half cycle. And at the NASA station almost all the load was power supplies ,so most of the load only happened when the diodes conduct. Our instantaneous power demand was over our transformer saturation load . Fluorescent light are very similar. With a load like that with the diodes off the load is inductive . with the diodes conducting the load was capacitance. Hope this helps Good luck.

2007-04-07 12:14:23 · answer #1 · answered by JOHNNIE B 7 · 0 0

Looking at the three impedances, they are all highly inductive, so that means there is a large reactive current flowing in each impedance, and hence a large total reactive power. You note the "answer" is 856 VAR. I find that impossibly small. Here are my calculations using phasors : Power = V^2 / Z for each load : ( 200 ∠ 0° )^2 / 8 ∠ 76° ( 200 ∠ 0° ) * ( 200 ∠ 0° ) / ( 8 ∠ 76° ) ( 200 * 200 / 8 ) ∠ ( 0° + 0° - 76° ) ( 5000 ∠ -76° ) rectangular form { (+1210) + j(-4850) } => very high inductive load ( 200 ∠ 0° )^2 / 6.5 ∠ 30° ( 200 ∠ 0° ) * ( 200 ∠ 0° ) / ( 6.5 ∠ 30° ) ( 200 * 200 / 6.5 ) ∠ ( 0° + 0° - 30°) ( 6,150 ∠ -30° ) rectangular form { (5330) + j(-3077) } => moderate inductive load ( 200 ∠ 0° )^2 / ( 4 ∠ 45° ) ( 200 ∠ 0° ) * ( 200 ∠ 0° ) / ( 4 ∠ 45° ) ( 200 * 200 / 4 ) ∠ ( 0° + 0° - 45° ) ( 10,000 ∠ ( -45° ) rectangular form { (+7070) + j(-7070) } add these together : { (+1210) + j(-4850) } { (+5330) + j(-3077) } { (+7070) + j(-7070) } ------------------------------- +13,610 + j(-15,000) So the reactive power is 15 kVAR, lagging (inductive) Convert to polar form : +13,610 + j(-15,000) = 20,300 ∠ 47.8° So total power = 20.3 KVAR Power factor = 13,610 / 20,300 = 0.67 Numbers have been rounded off to three significant figures and caculations were made with the rounded off numbers. I treat each load as if the other two loads do not exist, which means I ignore the different phase angles of the voltage of each phase. Then I sum them together. The real power matches your calculation. Mainly because I cannot remember how to work the calculations of unbalanced three phase loads and haven't found a webpage that shows how (yet). I do note that if the third term was this { (+7070) + j(+7070) } the sums would be { (+1210) + j(-4850) } { (+5330) + j(-3077) } { (+7070) + j(+7070) } ------------------------------- +13,610 + j(-857) See the magnitude of the j-component ? I suspect whoever did these calculations originally made a sign mistake somewhere. But dang if I can figure out where. It's been decades since I performed these calculations. And I can't find my notes.

2016-04-01 02:42:18 · answer #2 · answered by Anonymous · 0 0

We need to know if the impedances are series or parallel... V
Assume series and 117V at 0degrees
change the impedance to 2 + 2.5i + -.707i (rectangular form)
2+1.79i
(2.5i is 5 sin30) (-.707 is 1 sin-45)
Now take V=I*R so I = V/R which is 117/(2+1.79i) or
2+1.79i is SQRT(2^2+1.79^2) angle is (tan-1) (1.79/2) simple trig
117/2.68 at 41.8 degrees I=32.48-29.07i or 43.59 at -41.8 degrees
power factor equals cos of -41.8 = .745
P=VI pf or VI cos -41.8 (real) 117*43.59*.745
Q=VI sin -41.8 (reactive)
S = VI (apparent)

2007-04-07 15:46:04 · answer #3 · answered by Paul T 2 · 0 0

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