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DrROOO
modified 5 years ago

✔ Lagging Power Factor Correction

5
4
214
04:52:02
Power factor of a circuit represent its power delivery efficiency. It reflects the amount of current needed to deliver a specific amount of power. Power factor can be corrected by adding an equal and opposite reactance in parallel with the original load. Assuming we are dealing with RMS values of voltages and currents instead of the peak values shown here. --- Switch is OFF (current = 1.41A) --- Z_total = 85.078 Ω ∠45.152° = 60 + j60.319 Ω I_total = 120 / 85.078 = 1.410 A P (consumed) = 1.410² × 60 = 119.286 W S (apparent) = 1.410² × 85.078 = 169.144 VA power factor = 119.286 / 169.144 = 0.705 --- Switch is ON (current = 0.994 A) --- Z_total = 120.639 Ω ∠0.038° = 120.64 - j0.573 Ω I_total = 120 / 120.639 = 0.994 A P (consumed) = 0.994² × 120.64 = 119.365 W S (apparent) = 0.994² × 120.639 = 119.366 VA power factor = 119.365 / 119.366 = 0.9999916 ------------------------------ Notice that in the first case 1.41 A is needed to deliver a true power of approximately 119 W, While only 0.994 A is needed in the second case to deliver the exact same amount of true power. Also notice that the power factor has a direct impact on the phase angle between voltage and current (impedance phase angle), higher power factor means smaller phase shift between a circuit's voltage and current, and lower power factor means a larger shift between them.
published 5 years ago
PrathikP
5 years ago
Someone has made a copy of your circuit http://everycircuit.com/circuit/6730052988043264 you should report it to muzemaze
DrROOO
5 years ago
@PrathikP It seems to be deleted now. Thank you very much bro. I'm also looking for your constructive feedback about the circuit as usual.
PrathikP
5 years ago
Maybe use the motor to show a practical example?
PrathikP
5 years ago
Your post is straightforward, doesn't need any improvements.

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