Good afternoon viewers.
I have simulated the 3 phase AC circuit to get the active power in Multisim 11.0 as below
As shown in the simulation I get an active power of -1.2kW with a power factor of 1.
However, when I used the formula of calculating active power with a watt-meter involved would be 3 multiplied by the reading from the watt-meter.
Therefore according to the simulation and formula, 3 x 1.2kW = 3.6kW
Now I try to calculate using another formula which is 3 multiplied with phase current multiplied with phase voltage multiplied with power factor.
From the simulation below, I got phase current as 20.0A
So I began calculating
3 x 20.0 x 20 x 1.00000 = 1.2kW!!!
This is not equal to the 3 x 1.2kW = 3.6kW.
Is it due to the error of either of my formulas?
Or is it that the Multisim 11.0 is advanced enough to multiply the value by 3 to give the total active power?
Thank you for your time and I hope someone is able to help me on this.
Solved! Go to Solution.
When requesting help, you should upload your file so people don't have to recreate it. Fortunately its a small one and I could use the practice. This is what I think you are trying to do.
This gives you the correct answer of 400 watts per phase.
Unfortunately simulated instrumentation allows you to make mistakes and sometimes doesn't point them out.
Your voltage measurement is across a 1 ohm resistor that is between the neutral and ground, in a balance y circuit the neutral current averages out to zero so the voltage here should be 0. How the watt-meter actually gave you any power reading is a mystery to me.
Also you are not measuring current through one of the load resistors, but between one output and ground, with the ground not connected directly to Neutral.
I'm not sure what the 1 ohm resistor is supposed to represent on the neutral, its quite high for a parasitic load estimation.
Your original circuit is below
Here you see that the voltage that the watt-meter is reading is zero volts RMS, and the current in the leg going through the watt-meter to ground is reading 60 amps RMS. That the watt-meter reads 1.2 kW you expect for the whole circuit is purely a coincidence, and a mystery to me....
Double the voltage of the 3 phase source and you should expect double the power.
Here you see the RMS current through the shorted leg is now 120 AMPS RMS, and the power is quadrupled which is expected.
After more thought, your neutral is actually connected to ground through a 1 ohm resistor and what should be a very high resistance voltmeter. The current going through the ammeter is not shorted as I first believed, it does bypass R15, but then has to flow back through R16 or R17. So this will lead to the higher RMS currents than expected if the ammeter current was isolated to flowing through R15 or one of the other legs.
One more probe added below
The AC voltage is back to 20 V and on the bottom right you see there is 98 amps peak to peak and 34.6 amps RMS flowing through the neutral wire while there is 60 amps RMS flowing through the ammeter. This means that 23.4 amps RMS of AC current is flowing through what should be a high impedance voltmeter as the only way the generator neutral is connected to ground is through the one ohm resistor and the voltmeter section of the watt-meter.
I'm open to someone more knowledgeable about AC circuits to further add to this, but I think this is an interesting coincidence and a bug in the multisim watt-meter that it can measure power with no voltage across it and also flow over 20 amps RMS through it with no measurable voltage across it.
Hope this helps and looking forward to an interesting answer to the odd watt-meter behavior.
My Multisim 14.1 files posted.
Couldn't let it go and the 1.2kW was too coincidental. In the first group of pictures I measured the voltage wrong at the wattmeter, I should have used a differential voltmeter as shown and now it indicates 20 VRMS.
And what the 1 ohm resistor was doing between the neutral and ground was bugging me. So I shorted the resistor below and the power didn't change
Still reads the estimated full power through all three legs.
After more thought I removed R15 and
Still the power is 1.2 kW, further it the voltage waveform from what I thought was the nuetral to the common point of the 3 phases matchs the L1 voltage waveform.
Still trying to figure out where all the electrons are going!!!
New file attached.
Hi, Dave thanks for helping.
Sure I will upload the file next time, thanks for informing. I just learnt how to do that hahaha.
I was actually trying to follow the diagram below
That is why the R18 was there in my circuit.
Sorry for troubling you.
Can I take it as a solution for this thread?
I think you can call it solved, it looks like the way you hooked up the ammetter part of the watt-meter it created a 4th path for current flow and a corresponding unbalance in the circuit. At first I assumed this was a short, but you tied the neutral to ground through the high impedance voltmeter so the current flow through the ammeter was not a short and controlled by the load resistors R16 and R17. Actually was not affecting the circuit. This unique connection scheme had the coincidental effect of reading the expected power for the whole circuit, but actually doubled the total power flow.
After further playing around with the circuits there is no mystery, everything works like expected now that I understand 3 phase power just a little bit better.
Got it, I made your first reply as a solution. Thanks for helping, have a nice day, wherever you are hahaha. See you around! 😉
By the way, if you replied a thread and someone adds in ideas or anything without tagging you, do you get notified? Or must you click on the specific thread?