Power Factor Correction 

"My circuit will live happily ever after" :) Thank you! helpful video.

Your explanation was simply amazing. Thank you for clearing this up for me

Thank you, I had issues finding the right capacitance for the machine I built (there are ballasts in it so the PF is only 0.5)... can’t wait to remake all the calculations again !

Your channel is very good, important content and well taught. Congratulations for the initiative.

Good job! Had no idea what my teacher was trying to say, but this is great and obvious.

Wow 12 mins worked out 12 months. Again thanks a lot Sir.

You are explaining things like magic, thanks

i really wish you were able to post more videos your lessons are very straight forward and easy to follow it's so hard to find that for these topics especially amplifiers!!

Not sure why there are downvotes. This is a great video David!

That's was a amazing explanation

This video was so damn helpful. Thank you so much!

Amazing video, thank you for the summary!

watching your videos helps me brush off the.spider webs on my brain! thanks for the awesome content!

Straight to the point !!!!!! Awesome

Hi David, I like your lecture. Thanks

This is an awesome video thanks for making it really helped

Hi David, great video. Would it be possible to do a similar video but instead we are asked to improve the power factor from a certain value to another value but less than unity power? The example could be to improve the power factor from 0.6 to 0.9?

This is very useful, thank you.

this was very helpful thanks for posting

Thank You David , good Job

Best of the best of explanation for beginners

Very nice summary... when you find S dividing P by the power factor, it might be worth saying that P is the average power delivered to the load. I'm writing this because often, since power is not in the same phasor domain of voltage and current, it creates confusion.... especially (to me :)) if rms quantities are needed to find non rms amplitudes S = Vrms x Irms P(t) = P + S x cos(2 x omega x t = phi) expression of instantaneous power delivered to the load from the expression above, it is evident how P is the average of the overall power which goes to the load P = P(t) averaged over T/2 = S x power factor expression of average of instantaneous power delivered to the load Prms = rms value of the P delivered to the load = sqrt ( (P^2) + S^2/2 ) the expression of the Prms delivered to the load is not usually useful essentially this is the rms value of the instantaneous power which is a sinusoid plus its offset (P)

Great work,thanks lot...

amazing explanation.Thx

Other calculations have wanted an efficiency rating as on a motor's tag. This suggest I can use the resistance and perhaps derive the reactance and use those numbers to size a cap.

That was great, thank you!!

Nice example! Thanks

Good explaination...👍

i love you bro thank you so much for this video

awesome video sir! THank you!

Awesome. Thanks.

Very cool explaination. What about opposite PF, where there is an over capacitance creating what i assume would be a leading PF

Thank you very much

Hi sir i did not understand how does the final pf become =1 ? as i understand in the end of video P is 480 right and where is the new S that makes the pf=1?

Thank you so much

Is this example (semi-)correct? Imagine a tube(wire) and that you want to pass an x quantity of water(current) through it(ignore time), you can pass water at any speed(voltage) as long as it's volume doesn't overwhelm the tube due to its volume, then the water passes with virtually no resistance but if we increase the volume of the water or decrease the mass the tube can manage without getting overwhelmed, then the pressure of the water inside the tube increases, so then, the tube will have to deal with that pressure(wasted thermal energy) which is why there has to be a balance, right?

For me, I prefer Ql=480*96*sin53.13 then C equals to 14.52 microfarad, thank for great explanation.

It means power factor is always less tha 1 in both RLC series and parallel circuit?

We are assuming the phase angle of the source is 0 degrees in these power factor formulas, could you please explain how to correct the power factor of a true parallel RLC circuit when the phase angle is not 0, say 90 degrees? I am really struggling with this at the moment. PFC could really help with alternative energy solutions in the future and your approach to this problem would be greatly appreciated. Thank You!

The inductance is a motor. The R will go down if there is resistance to the motor shaft. - The motor has a load, draws more current. Inductance will be the same. Has the same number of turns for the windings. Therefore, the capacitance will be a factor of how much the motor is loaded?

Please explain why I sub t + I(RMS).

hi Mr.David. Thank you. can u make vidio again about this case. can u solve how much Power measured, I mean Power Real Power, Apparent Power, Reactive Power and current after compensation Power factor using capacitor parallel.

What will happen if we eliminate the 3 ohm resistor?as in practice,no resistors are used in series with the load.

How do we know that the 9 kVAR on the capacitor and the 9 kVAR on the inductor cancel each other out? Near 6:15.

On which programme do you use to write ?

Nice topic....

so if the load is capacitive, the PFC would be an inductive reactance

very helpful playlist and life saver for the exam, i want to ask why we added the capacitance in parallel not in series with the inductance and the resistance?

So I keep hearing reactive power is never consumed then I hear inductors consume reactive power and capacitors supply reactive power. I’m confused. By ‘consuming’ or ‘supplying’ reactive power is one simply trying to say-causing the current to lag or lead the voltage? Coz that would be much clearer.

Please I would like to know the screen writing tool you used for this video. Is it a pen of some kind??

sweet info, do you have a workbook? thanks

Thanks

Hi, good day, TX for the opportunity. I am busy building a rotary phase converter, naturally the output voltage of the 380v idler motor are not gonna be equal. Can i use this formula to correct the voltage on each leg by means of run caps.please help TX Basie

Sir how did you get the 5 ohms and 53.13 °. Thanks

Amazing video! One observation, though. At 10:40, are you sure Q = V^2/X? I once tried to prove that equation, but I concluded that was false. So, can you please prove it?

well-done

How did you decide to place the capacitor in parallel to the other circuit elements as opposed to in series? Of course it is helpful to do so to know the voltage across it, but for example how is its power factor correction ability altered if a capacitor was placed in series with the inductor?

I didnt understand the last part..why would it be pf=1...after getting the capacitance. You get pf=1..pls explain.

Also, PF = cos(phase angle).

Thanks 🐕

So power factor has to be corrected with a specific capacitor for a given load & pf?

Does using a power correction capacitor results in more kW being drawn from the grid just so we can sync the voltage and current? I mean it can't be free right? There're gotta be some kind of losses?

Great, with a PF of 1 we get full resonance with massive resonance voltages

THANKS A LOT BUT IM CONFUSED HERE WHY Qc should be = QL ??? HELP PLC

Since reactive power is I^2 * X, how come the capacitor's reactance isn't -4? (to offset Xl = 4)

Wow

I still dont understand if capacitor have impedance and current pass through it, doesn't it consume power too and wasted?

I've seen power factor correction circuits rather than just a simple parallel cap. What's that all about?

watch the short video I just uploaded on my channel if you want to be sure you understand Power Factor Correction.... always a good idea to see another example ...... just to be sure...... plus I got playlist of projects to build ..... and lots other videos on Electronics design theory.

Your video is total impedance formula from 3+j 4 ohm what means is value of j

how to convert 3+j 4 ohms = 5 ohms 53.13 degree angle

where is the 96A?

if only there was a chart for domestic appliances like fluorescent lights

how do I put omega function in calculator?

why did 53,15 degres become -53,15 degres?

@7:50 2pi x 10.6 does not equal 4 (ohms) regardless of the order of magnitude. What's going on here?

I have forgotten all these math algorithms!

3:33 42.1k/480=87 you wrote that but typed into your calculator p/v, 40k/480 =83

My brain still stuck and cant understand even watching the whole video.

why is too much voltage in the secondary instead of 13v dc I got 33v dc and capacitors overheating ? help me somebody

Just a quick question. when you calculated the reactive power of the inductor (96)^2 * 4. I noticed you used a current value of 96 amps. Isn't it true that once you add the capacitor the current through the inductor will no longer be the 96 amps and instead will be divided between the two branches. I guess I'm just confused on how the reactive power of the inductor will stay the same after connecting the capacitor.

Sam can also cry for love

I would plead that this be viewed by all the 'over unity' hobbyists who believe they have found greater output power than input power because they fail to account for the affects of V and I phase difference caused by inductor/capacitor in the circuit. The majority of these folks (1) use a multimeter (only) to measure output current and output voltage, then (2) multiply them together and believe they have discovered more output power than input power. A common example: Faraday's law shows that if you discharge an inductor very abruptly, the back EMF voltage spike can be quite high - since time (dt) is in the denominator: E = di / dt (Faraday's law) Using a MOSFET to very abruptly interrupt current flow in a coil creates a large back EMF spike (which is the reason 'snubber' diodes are popular!) and nearly all 'over unity' hobbyists then multiply this temporary large voltage spike by current flow and calculate an immense power value which greatly exceeds input power. I've tried explaining the issue to these guys, some of whom spend years, literally, not knowing Faraday's law and the difference between real and apparent power. Your video I believe would help them understand why reactive components change the power delivered to the load - it is about the clearest explanation I've found - good job on your part.

Hi, I sent an email. Thanks

S

4:02 real power, not apparent.

Its kind of dumb that you don't show the formula your using. Instead you just plug in the values without any regard that your audience might get lost

afterAddingCapacitor,TheCurrentInCoilIncreases.SoTheCoilCouldBurn,IfPowerFactorIsTooHigh?

Thank you so much

Wow