A correction: you don't increase primary voltage until you reach primary rated current of 4.8A rather you increase primary voltage until full rated secendary current is achieved.
@altuber99_athlete
2 жыл бұрын
But when you reach the rated secondary current, you also reach the rated primary current. So aren’t both the same, with the benefit of measuring the primary current that you don’t need a huge ammeter or current transformer?
@RobiBue
2 жыл бұрын
"poteytoes, potahtoes". You could add a second ammeter to the secondary coil and read the "secondary rated current" while at the same time the "primary rated current" reaches the same _ratio_ level. In this case secondary = 167A, primary = 4.8A The only difference is, that on the secondary side you end up with 0V (thanks to the short circuit) while on the primary side you end up with 104V which enables you to get the Impedance %Z = V _input short_ / V _input rated_ so it makes more sense to get the _input rated Amps_ instead of the _output rated Amps_ note: at the end I switched to _input_ and _output_ for _primary_ and _secondary_ respectively since I am applying power to the _primary_ , ergo my _input_
@annavupie
5 жыл бұрын
Loving these videos. Hopefully you keep posting more problems so I can finish my program with good understanding of everything! I've struggled so much because the theory never seems to click right away.
@TheElectricAcademy
5 жыл бұрын
Anything you’d like to see in particular?
@hfe1833
2 жыл бұрын
how about for 3phase xformer?
@sethtaylor5938
Жыл бұрын
Actually in a 4160 three phase -Y- system the single phase voltage is 4160 divided by 1.732 or 2400 volts phase to neutral ground. 2400 is more appropriate in a “4160” distribution transformer. At First Energy we’d use 2400 in this example. In a three phase -Y- power bank in a 4160 system the phase to phase voltage is 4160. The phase to neutral is 2400. Our impedance calculations are based on a 2400 volt reference per transformer.
@shivamverma-uk7bh
4 жыл бұрын
Where is the 104 volts come from or is there any calculations to find that voltage or we have to do it practically
@Santana2163
3 жыл бұрын
Exactly what I want to know
@RobiBue
2 жыл бұрын
@@Santana2163 the 104 volts was _assumed_ (made up) for the sake of calculation. In the real world you would get that number from the volt meter reading the input voltage across the primary coil. if the voltage would be, lets say, 85V, then the Impedance would be 2% (or 2.0433%) on the other hand if it were 125V, then the impedance would be to 3%. The results of the short circuit test give you the means to calculate the impedance for the transformer at hand. The numbers used in the video were just arbitrary and for demonstration purposes only 😉
@ryanstephens6484
4 жыл бұрын
Very helpful! I have one question. How did you get the 104 volts?
@TheElectricAcademy
4 жыл бұрын
The 104 volts is the voltage it would take to reach maximum current in a short circuit situation. For the purpose of the video I had an assumed voltage.
@Theo0x89
5 жыл бұрын
How does the percent impedance relate to the actual impedance of the transformer?
@ranjan1436
5 жыл бұрын
why E- rated short ckt voltage is divided by E-rated secondary voltage? when per cent impedance is required to calculate short ckt secondary current, breaker size of the secondary side of the trafo.
@walterbrown8694
3 жыл бұрын
More years ago than I care to remember, (early '50s) when I began my career in electrical engineering, transformer ratings and electrical equipment ratings were marked % IZ , which of course meant % voltage. Eventually, the I in the markings was dropped because all the percentage calculations used 100 percent of the reference current, hence the marking as % Z. The percentage ratio is still the same for % Z as for % IZ. ( IZ (sc)/ IZ (rated) )
@ruthsariban1072
2 жыл бұрын
What do you do with a transformer?
@donberg01
4 жыл бұрын
Why use a percentage of Z & not the resistance in ohms? Every calculation done for R is measured in OHMS???
@sergioarrieta2571
3 жыл бұрын
Because it is based on nominal power of the transformer and at the rated voltage (per unit)
@RobiBue
2 жыл бұрын
%Z is used because when you're short-circuiting the transformer you're basically overloading at maximum current. 21.6Ω would be a bad value if the minimum Z given 4160V/4.8A =~ 867Ω. remember that if the transformer runs with a lesser load, then it also needs less amps so if let's say you run at half power (10KVA) you also run at half the max amps (2.4A) -> 4160V/2.4A = 1.73kΩ Impedance Z = sqrt *(* R^2 + ( *X **_inductive_** - X **_capacitive_** )* ^2 *)* X _inductive_ = 2πfL where f = frequency (60 or 50 Hz depending -- could be a different frequency as well) and L the Inductance (in Henry H) X _capacitive_ = 1 / (2πfC) where C is the capacitance (in Farad F) *X* (reactance) is measured in Ω (ohm)
@surendrakverma555
3 жыл бұрын
Good
@truthteller2700
10 ай бұрын
wrong explanation ! short the secondary and increase the voltage on primary till the current on secondary reaches is rated current. the voltage that causes the secondary rated current / rate primary voltage is Z%
@sinethembalusawana
6 ай бұрын
No that's a very bold statement for you to say. What you are say is not that different from what is explained here. When the secondary side reaches its rated current also the primary side reaches its rated current automatically. They reach their rated current at the same, I think it does not matter on which side you measure the current, still you would find the voltage at which the rated current appeared.
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