well, that's a very nice presentation, but you spend almost 20 minutes talking about Ohm's (and Kirchhoff's) law. We still don't know, and that's why I came here, how to determine the power of an unknown transformer (where nothing is labeled 12 V/1 A or something like that). What if we only know the open circuit voltage? How many VA is it able to deliver? Some people try to determine the power of an unknown transformer based on the size of the laminated core - I myself like to look at the shape of the secondary side signal on the oscilloscope under load. I increase the load continuously until the transformer saturates and the sine flattens out. Then I subtract about 20% from the load current and know APPROXIMATELY what load the transformer was designed for (of course, the core temperature in the windings should always be observed).
Wow! This is complicated even for me. I'm a 75year old CET. If yo don't know the secondary voltage or current, here is a simplier way to find out. 1) find out if the transformer is step-up or step-down. This is accomplished by measuring the voltage at the secondary. If you have 120v at the primary and 12v at the secondary, then the transformer is a 10:1 step-down. 2) get a large value potentiometer, somewhere around 10k. Put the pot in series with the primary. One side goes to the line in and the slider (center terminal) goes to the transformer. 3) put AC voltmeter across the same pot wires and plug it in. Adjust the pot until you have 1/2 the line voltage, then unplug from source and measure resistance across the pot. Use Ohms law to calculate the current through the pot using the 1/2 votage drop and the resistance measured. 4) multiply the calculated current times the turns ratio found in step #1. In our example it was 10. This gives you the current rating at the secondary. Why is this true? Because in any transformer, the power is the same on the primary and secondary. Therefore, if the voltage decreases, then the current increases proportionately. Have a blessed day!
There is no way to measure the amperage, only estimate, because it depends on how many amps/mm² were estimated for the wire calculation. The core area does not give the power of a transformer, it only shows the maximum power that the lamination can support. When nothing is written on the transformer we first have to assume the primary voltage (120V or other). Connect to 120V and measure the voltage on the secondary, divide the value by the square root of 2 and this will be the approximate value of the secondary at 120V. Short the secondary with a low value resistor until you find a lower value resistor that gets the secondary voltage minus 10%, then connect the ammeter in series or use the formula. It will have an approximate amperage value.
The best way is by labeling the device in the factory and not destroying that label. The second one is by knowing the core area, added to the aknowledgement of the coil gauge
Not that it's totally wrong according to one of the definitions, this is the first time I've heard the term filaments used by an electronics guy to define the input & output leads of a transformer.
No, this method is not simple neither practical at all. The simplest way is to calculate the core area which gives the power the transformer can deliver (in VA). Then, if you know (measure) the voltage and divide the VA value by the V found with the multimeter you have the A value. As simple as that!
@@uksuperrascalProfessor he is certainly not. His hands show that he is very young for a professor! Apart from that, i think that your method with the coil resistances is proper for working with DC relay coils, but not with transformers. With transformers, the impedances matter and these are combinations of resistances (R plus XL together), so the Ohms Law in AC becomes I=U/Z, where Z represent the impedance now. And i agree with you about the test procedure using an incandescent lamp in series with the primary winding (the one with the highest Ohmic resistance). If everything is OK with the transformer, the lamp (connected to the primary only) will not glow when the system is powered. But it will glow in full brightness if anyone of its windings has even only one single shorted turn. Besides this is a well known test method for testing a transformer for shorted turns...
@@USA-GreedyMenOfNoIntegrityWhen he needs almost 20 minutes against the calculating procedure which takes no more than 5 minutes, where do you see the simplicity?
@@USA-GreedyMenOfNoIntegritySo the criteria for the easiness of the method depends on the availability of the components on your bench. Nevertheless i think that the calculating method, given that it takes no longer than five minutes to find out the result, instead of almost twenty minutes spent on this video, it is much more easy at least as regards time...
NEVER connect any Unknown Unmarked Transformer Direct to AC Mains!!! - First Measure and Record the 3 dimensions of the Transformers L,D,H, To decide what is the Primary winding OR Secondary winding, Using your multi meter set to 200 OR lower Ohms, Measure and Record the resistance of the windings. The winding using the thinnest wire is probably the the primary, V/R=A If your Ohms reading is lower than 10 Ohm and the transformer is say no bigger than 5cm x 5cm x 5cm this winding is not the primary, Because 240V AC / 10R = 24 Amp, This transformer is way to small to Consume OR Deliver 24 Amp. So Thinnest wire, Ohms reading of say 35 Ohms, 240V / 35R = 6.857A This would be the Maximum Consumption and Deliverance AMPS this transformer could short time covert and isolate 240V AC mains. With this information about the transformer I would use a LED indicated, input switched and LED indicated bypass switched 50 Watt incandescent 240V lamp, Then a switched input with LED indication and LED indication switched output 240V to 240V Isolation Transformer with 100mA trip between my unknown Transformer. Because fuses are to slow to Burn out if there is a short circuit in the winding of the unknown Transformer, Connecting all before applying AC 240V.
Volt im eter ?.. It measures Volts, therefore it is a Volt meter. Very long winded to measure the internal impedance and thereby deduce the current rating.
5 minutes in and all he's done is confused the hell out of everybody over a simple 1 minute measurement and calculation, I can't bear to listen to it any longer, and what is a mul timiter, it's a multi meter.
Thank you very much. This is exactly what I needed.
well, that's a very nice presentation, but you spend almost 20 minutes talking about Ohm's (and Kirchhoff's) law. We still don't know, and that's why I came here, how to determine the power of an unknown transformer (where nothing is labeled 12 V/1 A or something like that).
What if we only know the open circuit voltage? How many VA is it able to deliver? Some people try to determine the power of an unknown transformer based on the size of the laminated core - I myself like to look at the shape of the secondary side signal on the oscilloscope under load. I increase the load continuously until the transformer saturates and the sine flattens out. Then I subtract about 20% from the load current and know APPROXIMATELY what load the transformer was designed for (of course, the core temperature in the windings should always be observed).
You make sense. The professor should know about saturation and heat.😊
I agree with your comments. It is very convoluted and does not address the aim of the video.
Very useful, thanks!
Thank you for sharing technical know how.
Wow! This is complicated even for me. I'm a 75year old CET. If yo don't know the secondary voltage or current, here is a simplier way to find out. 1) find out if the transformer is step-up or step-down. This is accomplished by measuring the voltage at the secondary. If you have 120v at the primary and 12v at the secondary, then the transformer is a 10:1 step-down.
2) get a large value potentiometer, somewhere around 10k. Put the pot in series with the primary. One side goes to the line in and the slider (center terminal) goes to the transformer.
3) put AC voltmeter across the same pot wires and plug it in. Adjust the pot until you have 1/2 the line voltage, then unplug from source and measure resistance across the pot. Use Ohms law to calculate the current through the pot using the 1/2 votage drop and the resistance measured.
4) multiply the calculated current times the turns ratio found in step #1. In our example it was 10. This gives you the current rating at the secondary.
Why is this true? Because in any transformer, the power is the same on the primary and secondary. Therefore, if the voltage decreases, then the current increases proportionately.
Have a blessed day!
This is excellent.thanks
Excellent. Thank you.
There is no way to measure the amperage, only estimate, because it depends on how many amps/mm² were estimated for the wire calculation.
The core area does not give the power of a transformer, it only shows the maximum power that the lamination can support.
When nothing is written on the transformer we first have to assume the primary voltage (120V or other).
Connect to 120V and measure the voltage on the secondary, divide the value by the square root of 2 and this will be the approximate value of the secondary at 120V.
Short the secondary with a low value resistor until you find a lower value resistor that gets the secondary voltage minus 10%, then connect the ammeter in series or use the formula.
It will have an approximate amperage value.
The best way is by labeling the device in the factory and not destroying that label. The second one is by knowing the core area, added to the aknowledgement of the coil gauge
The guage of wire on the secondary windings will give you a clue!
Not that it's totally wrong according to one of the definitions, this is the first time I've heard the term filaments used by an electronics guy to define the input & output leads of a transformer.
Example 20 AWG wire is rated to carry 3.5 amps @ 12 volts
Amperage? You mean current!
why you did not meseare the volage the current especially the power of the primiry since it is the primiry which determine the secondary parameters ?
Long life and prosperity
No, this method is not simple neither practical at all. The simplest way is to calculate the core area which gives the power the transformer can deliver (in VA). Then, if you know (measure) the voltage and divide the VA value by the V found with the multimeter you have the A value. As simple as that!
My comment above was a bit more long winded, But I agree with your comment, Also This Person is clearly Not a professor LOL.
Actually, it’s very simple because most of us have all these items in our lab and meters of all kinds.
@@uksuperrascalProfessor he is certainly not. His hands show that he is very young for a professor!
Apart from that, i think that your method with the coil resistances is proper for working with DC relay coils, but not with transformers. With transformers, the impedances matter and these are combinations of resistances (R plus XL together), so the Ohms Law in AC becomes I=U/Z, where Z represent the impedance now.
And i agree with you about the test procedure using an incandescent lamp in series with the primary winding (the one with the highest Ohmic resistance).
If everything is OK with the transformer, the lamp (connected to the primary only) will not glow when the system is powered. But it will glow in full brightness if anyone of its windings has even only one single shorted turn. Besides this is a well known test method for testing a transformer for shorted turns...
@@USA-GreedyMenOfNoIntegrityWhen he needs almost 20 minutes against the calculating procedure which takes no more than 5 minutes, where do you see the simplicity?
@@USA-GreedyMenOfNoIntegritySo the criteria for the easiness of the method depends on the availability of the components on your bench.
Nevertheless i think that the calculating method, given that it takes no longer than five minutes to find out the result, instead of almost twenty minutes spent on this video, it is much more easy at least as regards time...
Useful when u can't sleep
NEVER connect any Unknown Unmarked Transformer Direct to AC Mains!!! - First Measure and Record the 3 dimensions of the Transformers L,D,H, To decide what is the Primary winding OR Secondary winding, Using your multi meter set to 200 OR lower Ohms, Measure and Record the resistance of the windings. The winding using the thinnest wire is probably the the primary, V/R=A If your Ohms reading is lower than 10 Ohm and the transformer is say no bigger than 5cm x 5cm x 5cm this winding is not the primary, Because 240V AC / 10R = 24 Amp, This transformer is way to small to Consume OR Deliver 24 Amp. So Thinnest wire, Ohms reading of say 35 Ohms, 240V / 35R = 6.857A This would be the Maximum Consumption and Deliverance AMPS this transformer could short time covert and isolate 240V AC mains. With this information about the transformer I would use a LED indicated, input switched and LED indicated bypass switched 50 Watt incandescent 240V lamp, Then a switched input with LED indication and LED indication switched output 240V to 240V Isolation Transformer with 100mA trip between my unknown Transformer. Because fuses are to slow to Burn out if there is a short circuit in the winding of the unknown Transformer, Connecting all before applying AC 240V.
Greu tare explici
Eu am folosit viteza de redare X1.5.....
Multimeter isn’t pronounced like an altimeter. It is multi-meter. My opinion.
HI, MISTER ELECTROROBOTICS!!!💁🙆💁♀️🙆♀️💁🙆♂️🙋🙆🙋♂️
A 22 gauge wire at 12 volts can typically handle around 0.9 - 1.0 amps of current
Amp meter across a known resistor capable of handling the emf or clmp meter .
Your method = smoke + dead meter at best
Since the o/p voltage is greater, how do you calculate the transformer is 12v/1a?.
👍
I like where theres no music background, very clear
Volt im eter ?.. It measures Volts, therefore it is a Volt meter. Very long winded to measure the internal impedance and thereby deduce the current rating.
If you measure the output voltage and know the secondary winding gauge will get you in the ball park
Thanks
That's about a 20W transformer you have in your hand there.
What is el transformador?
What is a volttimameter
Amperage?
You mean current in Amps or milliamps!
Too. Much time consuming..use. ammeter. Thus it. !! Thanks..
Niveau polytechnique !
5 minutes in and all he's done is confused the hell out of everybody over a simple 1 minute measurement and calculation, I can't bear to listen to it any longer, and what is a mul timiter, it's a multi meter.
Turn the air off. No dinglish
You cannot, mfr should specify.
Total nonsense
Wasted my precious time.
Useless video yoo!
Crap AI
boys and girls?
What’s wrong with that? At least he’s not calling us names like felonious cat grabbing Trump does, “enemy within, deplorables”.
@@USA-GreedyMenOfNoIntegrity deplorables was from Clinton
@@drstrangelove09 yes, you are correct about deplorables. We were called radical left lunatics recently. So boys and girls is just fine. Great videos!
Yes. Trannies can be boys and/or girls.