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?
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.
In short yes, but no. Capacitive loads and inductive loads produce sin waves that are both out of sync with a given resistive load. The reason they neutralize each other is because inductance's wave lags behind the resistive load 90° where as capacitance's wave is 90° ahead, thus the two loads are a full 180° off of each other and thus directly nullify. Thus it is indeed yechnically possible to have a circuit with a low power factor due to a capacitive load. Why I say "technically" is because I don't personally know of any electrical gear / application that innately produces an unwanted and intrinsic capacitive load, and in fact none may well even exist. Inductance, however, is present in the the supply of every single electronically driven motor simply due to mere physical methods by which they operate. Thus, PFC is generally understood as adding capacitors to circuits with inductive loads rather than it being versa-vice. It's a solution to an inherent problem that can't be solved otherwise whereas the obvious solution to a circuit with too high of a capacitive load would be to remove the capacitor you added.
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 !
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?
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?
bro, because the after adding the capacitor the reative power will be cancelled out as both(capacitor and inductor) will be having same values of Reactive power, and the impedance of the circuit will become purely Resistive, and you know for a purely Resistive network the power factor is UNITY!
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
My first impression was the same, but then I realized the voltage drop across the inductor is reduced because of the resistance in series on that arm of the circuit. It's not the same as a simple LC circuit. So, you need to account for that extra resistor, the reactive power will be reduced because the resistor eats up some of the voltage drop, reducing current that will appear on that arm of the line. He conveniently used current, which builds in a correction automatically when calculating power with current , Q=I^2*R.
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)
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!!
@@ElectronXLab thank you for the reply I look forward to it! I've been doing everything I can going to tutoring every week and watching youtube videos and practicing as much as possible. My professor gives 0 partial credit so it has been rough...
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.
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!
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.
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.
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?
You might be referring to active power factor correction circuits which directly measure the voltage and or current and use a switching circuit to actively correct for low power factor. The main advantage is that power factor will be corrected even as the load changes. The main disadvantage is that the circuit is much more complex.
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.
Dude the title is power factor correction. If you know what that means you gotta be able to look up the formulae yourself. Would you rather not hear some with knowledge to share? Maybe they don't have video editing software but still decide to share their know how. You want them to do the work for you?
An ideal capacitor doesn't consume power. It's just like an electron storage container, you can fill it up with electrons to store energy and then release all the electrons to release energy
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?
The pizza lol. Well, this is really the power equation P=V*I but we know I=V/R so we sub that in to the power equation and we get P=V*(V/R) or P=V^2/R. He’s dealing with imaginary power so he’s using Q instead of P and X instead of R. Hope this helps.
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?
No, counterintuitively adding the new component will result in lower losses (for the same amount of real power to the load). The power factor correction means the apparent power will be less and so the the peak current will be less. The lower peak current means there will be lower resistive losses in the wires.
@@ElectronXLab Say I have a machine that consumes 10kwh and 10kvarh as registered on the electric meter per hour, and therefore has a power factor of 0.71. If I were to correct it to 0.9, the meter would now register 10kwh and 5kvarh per hour?
Because that's impossible. The minus in this case is technically an imaginary number and only serves to represent that it has the opposite effect to an inductor.
Because the current in the C leg is not the same as the current through R and L.That makes capacitor's reactance different. If C was in series with R & L then Yes that would've been the case.
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?
You could do a similar calculation to see what capacitor you need placed in series. Practically speaking, it is far easier to place a capacitor in parallel to the circuit than in series with it.
@@ElectronXLab thank you! Also, when placing the capacitor in parallel with the other load, there would be a division of current between the capacitor and the resistor-inductor pair- would this result in less reactive power being consumed by the inductor, since Q=I^2 * X(inductor)?
@@justinbieber19010 The capacitor is in parallel with the RL pair and also in parallel with the voltage source. This means (assuming an ideal voltage source ) that the voltage across the RL branch will stay the same
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?
@@souvikdas2009 if u put in series, then u change current to machine or circuit that it needs...so the only option to place in parrallel so u dont affect the original machine or circuit requirments.
if you're trying to convert AC to DC with a transformer and rectifier, you multiply the RMS Ac volts by about 1.4, so to charge a 12v battery for example, you only need about a 9v AC transformer/rectifier combo.
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.
Not sure why there are downvotes. This is a great video David!
Thanks Oscar. I appreciate the comment
You are explaining things like magic, thanks
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?
"My circuit will live happily ever after"
:)
Thank you! helpful video.
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.
so if the load is capacitive, the PFC would be an inductive reactance
In short yes, but no.
Capacitive loads and inductive loads produce sin waves that are both out of sync with a given resistive load. The reason they neutralize each other is because inductance's wave lags behind the resistive load 90° where as capacitance's wave is 90° ahead, thus the two loads are a full 180° off of each other and thus directly nullify. Thus it is indeed yechnically possible to have a circuit with a low power factor due to a capacitive load.
Why I say "technically" is because I don't personally know of any electrical gear / application that innately produces an unwanted and intrinsic capacitive load, and in fact none may well even exist. Inductance, however, is present in the the supply of every single electronically driven motor simply due to mere physical methods by which they operate. Thus, PFC is generally understood as adding capacitors to circuits with inductive loads rather than it being versa-vice. It's a solution to an inherent problem that can't be solved otherwise whereas the obvious solution to a circuit with too high of a capacitive load would be to remove the capacitor you added.
@@XTYRMIN8Zvery good answer.
Good job! Had no idea what my teacher was trying to say, but this is great and obvious.
Best of the best of explanation for beginners
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 !
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?
Your explanation was simply amazing. Thank you for clearing this up for me
Amazing video, thank you for the summary!
That's was a amazing explanation
Your channel is very good, important content and well taught. Congratulations for the initiative.
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?
This is an awesome video thanks for making it really helped
Straight to the point !!!!!!
Awesome
watching your videos helps me brush off the.spider webs on my brain! thanks for the awesome content!
I didnt understand the last part..why would it be pf=1...after getting the capacitance. You get pf=1..pls explain.
bro, because the after adding the capacitor the reative power will be cancelled out as both(capacitor and inductor) will be having same values of Reactive power, and the impedance of the circuit will become purely Resistive, and you know for a purely Resistive network the power factor is UNITY!
i love you bro thank you so much for this video
My brain still stuck and cant understand even watching the whole video.
now i understand already. Thanks for it xd
It means power factor is always less tha 1 in both RLC series and parallel circuit?
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
That is a very complicated way to say LC tank resonates at 1/sqrt(LC)
Whole thing could take 1 minute doing it that way.
Sir how did you get the 5 ohms and 53.13 °. Thanks
Nice example!
Thanks
For me, I prefer Ql=480*96*sin53.13 then C equals to 14.52 microfarad, thank for great explanation.
My first impression was the same, but then I realized the voltage drop across the inductor is reduced because of the resistance in series on that arm of the circuit. It's not the same as a simple LC circuit. So, you need to account for that extra resistor, the reactive power will be reduced because the resistor eats up some of the voltage drop, reducing current that will appear on that arm of the line. He conveniently used current, which builds in a correction automatically when calculating power with current , Q=I^2*R.
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)
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!!
Thanks for your kind comments. I'm actually going to get back into making videos again soon. Stay tuned
@@ElectronXLab thank you for the reply I look forward to it! I've been doing everything I can going to tutoring every week and watching youtube videos and practicing as much as possible. My professor gives 0 partial credit so it has been rough...
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.
What will happen if we eliminate the 3 ohm resistor?as in practice,no resistors are used in series with the load.
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.
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!
Also, PF = cos(phase angle).
This video was so damn helpful. Thank you so much!
5:34 how does this calculation work on a calculator 10x2 + 9x2 becomes 13.45??? 😩
The calculation is square root of 10*10 + 9*9...sqrt(10^2+9^2) = sqrt(181) = 13.45
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.
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 explain why I sub t + I(RMS).
Hi, I sent an email. Thanks
Sam can also cry for love
Thank you very 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?
Awesome. Thanks.
Thank you so much
I've seen power factor correction circuits rather than just a simple parallel cap. What's that all about?
You might be referring to active power factor correction circuits which directly measure the voltage and or current and use a switching circuit to actively correct for low power factor. The main advantage is that power factor will be corrected even as the load changes. The main disadvantage is that the circuit is much more complex.
how to convert 3+j 4 ohms = 5 ohms 53.13 degree angle
@@wfvideos4099 thanks a lot sir
How?
3:33 42.1k/480=87 you wrote that but typed into your calculator p/v, 40k/480 =83
why did 53,15 degres become -53,15 degres?
Very cool explaination. What about opposite PF, where there is an over capacitance creating what i assume would be a leading PF
if only there was a chart for domestic appliances like fluorescent lights
sweet info, do you have a workbook? thanks
Nice topic....
Thank you so much
how do I put omega function in calculator?
2 x Pi x f
Hi David, I like your lecture. Thanks
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.
Your video is total impedance formula from 3+j 4 ohm what means is value of j
J is an imaginary number, in maths it’s called i. do some more YouTubing 👍
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
Dude the title is power factor correction. If you know what that means you gotta be able to look up the formulae yourself. Would you rather not hear some with knowledge to share? Maybe they don't have video editing software but still decide to share their know how. You want them to do the work for you?
I still dont understand if capacitor have impedance and current pass through it, doesn't it consume power too and wasted?
An ideal capacitor doesn't consume power. It's just like an electron storage container, you can fill it up with electrons to store energy and then release all the electrons to release energy
On which programme do you use to write ?
Wow 12 mins worked out 12 months. Again thanks a lot Sir.
this was very helpful thanks for posting
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?
The pizza lol. Well, this is really the power equation P=V*I but we know I=V/R so we sub that in to the power equation and we get P=V*(V/R) or P=V^2/R. He’s dealing with imaginary power so he’s using Q instead of P and X instead of R. Hope this helps.
I have forgotten all these math algorithms!
Good explaination...👍
Thank You David , good Job
afterAddingCapacitor,TheCurrentInCoilIncreases.SoTheCoilCouldBurn,IfPowerFactorIsTooHigh?
amazing explanation.Thx
@7:50 2pi x 10.6 does not equal 4 (ohms) regardless of the order of magnitude. What's going on here?
I think you missed the 60.
2pi x 60 x 10.6 = 4 Ohms
Well that's embarrassing!
This is very useful, thank you.
THANKS A LOT BUT IM CONFUSED HERE WHY Qc should be = QL ??? HELP PLC
THANKS A LOT I GOT IT NOW YOU A LIFE SAVER
Great work,thanks lot...
where is the 96A?
well-done
thanks!
why can't you clear your throat when it becomes filled with mucus?
So power factor has to be corrected with a specific capacitor for a given load & pf?
Unless it's the opposite type what i think is called a "leading" rather than this "lagging" current.
awesome video sir! THank you!
Glad you liked it!
Please I would like to know the screen writing tool you used for this video. Is it a pen of some kind??
For this video I think I used a Wacom tablet. For more recent videos, I use a OneNote on a MS Surface tablet
Great, with a PF of 1 we get full resonance with massive resonance voltages
Does that mean we can now supply the flux capacitor and send us to the future?
Wow
Wow
S
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?
No, counterintuitively adding the new component will result in lower losses (for the same amount of real power to the load). The power factor correction means the apparent power will be less and so the the peak current will be less. The lower peak current means there will be lower resistive losses in the wires.
@@ElectronXLab Say I have a machine that consumes 10kwh and 10kvarh as registered on the electric meter per hour, and therefore has a power factor of 0.71. If I were to correct it to 0.9, the meter would now register 10kwh and 5kvarh per hour?
@@impactodelsurenterprise2440 it would cost less to run if you increase the PF toward 0.99
Since reactive power is I^2 * X, how come the capacitor's reactance isn't -4? (to offset Xl = 4)
Because that's impossible. The minus in this case is technically an imaginary number and only serves to represent that it has the opposite effect to an inductor.
Because the current in the C leg is not the same as the current through R and L.That makes capacitor's reactance different. If C was in series with R & L then Yes that would've been the case.
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?
You could do a similar calculation to see what capacitor you need placed in series. Practically speaking, it is far easier to place a capacitor in parallel to the circuit than in series with it.
@@ElectronXLab thank you! Also, when placing the capacitor in parallel with the other load, there would be a division of current between the capacitor and the resistor-inductor pair- would this result in less reactive power being consumed by the inductor, since Q=I^2 * X(inductor)?
@@justinbieber19010 The capacitor is in parallel with the RL pair and also in parallel with the voltage source. This means (assuming an ideal voltage source ) that the voltage across the RL branch will stay the same
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?
It's easier to physically place the capacitor there. You don't have to break the circuit and you can do it right at the power supply.
@@ElectronXLab So, will the apparent power change if we put capacitor in series with resistance?
@@souvikdas2009 if u put in series, then u change current to machine or circuit that it needs...so the only option to place in parrallel so u dont affect the original machine or circuit requirments.
Thanks 🐕
That was great, thank you!!
Yeah
4:02 real power, not apparent.
why is too much voltage in the secondary instead of 13v dc I got 33v dc and capacitors overheating ? help me somebody
I'm not sure what you are asking. There is no secondary in this example.
@@ElectronXLab ok. thanks too much weed.!
if you're trying to convert AC to DC with a transformer and rectifier, you multiply the RMS Ac volts by about 1.4, so to charge a 12v battery for example, you only need about a 9v AC transformer/rectifier combo.
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.
The voltage across the R-L branch of the circuit will not change, so the current will not change either.