Transformer current rating


by brewcrew975
Tags: power engineering, rated current, transformer
brewcrew975
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#1
Apr7-09, 12:29 PM
P: 6
When a transformer is rated 225 VA, Vprimary=120 V and Vsecondary=9KV, Isecondary=25 mA,
does this mean that on the secondary side the transformer can only supply 25 mA, no more, no less than rated current. Thanks
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berkeman
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#2
Apr7-09, 01:00 PM
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Quote Quote by brewcrew975 View Post
When a transformer is rated 225 VA, Vprimary=120 V and Vsecondary=9KV, Isecondary=25 mA,
does this mean that on the secondary side the transformer can only supply 25 mA, no more, no less than rated current. Thanks
Welcome to the PF. It means no more than 25mA at 9kV can be supplied, without overheating the transformer.
Pumblechook
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#3
Apr7-09, 01:04 PM
P: 359
It can supply any current up to about 25mA and maybe a bit more.

The secondary voltage will be higher with a lower current drawn. Maybe quite a bit higher.

The classic Vin/Vout = Primary/Secondary turns only applies to ideal transformers with no losses which don't actually exist.

I think generally the rating means the secondary voltage will be the rated secondary voltage at that rating... 9kV at 25 mA (225 VA).

Unloaded or lightly loaded the transformer may kick out 11 -12 kV.

The ultimate rating will be where the transformer becomes too hot .. That point is debatable.

brewcrew975
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#4
Apr7-09, 05:41 PM
P: 6

Transformer current rating


OK. I understand the rated conditions, my other question is I have is the the same transformer, 225 VA, Vprimary=120 V and Vsecondary=9KV, Isecondary=25 mA, says that when the secondary side is shorted it will have a current of 25 mA, but what if I place say a 1 ohm resistor in place of the short to represent a load. A short basically represents "ideally" no resistance and would result max current, so would the 1 ohm load then draw 9000 A, but it has more resistance than the short? I know this is impossible but can someone please explain what would happen to the voltage and current on secondary side. Thanks in advance, I ask on here because my power systems professor is hard to understand and I don't like asking him detailed questions.
berkeman
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#5
Apr7-09, 07:03 PM
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Quote Quote by brewcrew975 View Post
OK. I understand the rated conditions, my other question is I have is the the same transformer, 225 VA, Vprimary=120 V and Vsecondary=9KV, Isecondary=25 mA, says that when the secondary side is shorted it will have a current of 25 mA, but what if I place say a 1 ohm resistor in place of the short to represent a load. A short basically represents "ideally" no resistance and would result max current, so would the 1 ohm load then draw 9000 A, but it has more resistance than the short? I know this is impossible but can someone please explain what would happen to the voltage and current on secondary side. Thanks in advance, I ask on here because my power systems professor is hard to understand and I don't like asking him detailed questions.
You are neglecting the winding resistance. That's a pretty weird spec, though. They want you to short the 9kV side of the transformer? I wouldn't do that.

Can you post a link to the actual spec?
bitrex
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#6
Apr8-09, 03:43 AM
P: 196
I certainly hope this 125VA 9kV transformer is a hypothetical one for the moment, not one you're planning on messing around with just yet. I'm not sure exactlywhat would happen if one put a 1 ohm resistor across a transformer like that to "test" it, but I'm pretty sure it would result in fuses and circuit breakers being tripped at best and smoke and flames and things melting and people getting hurt or killed at worst. You're correct that drawing 9000 A through the resistor is impossible, but that transformer will be darned if it won't try. It won't just go up to the maximum current printed on the side and say "oh, that's my maximum" and sit there; something down the line will give. My guess is that on a 9kV 35 mA transformer the secondary winding resistance will be high enough that the building circuit breaker wouldn't trip immediately and the secondary will behave like the coils in a toaster oven - for as long as it takes to toast itself.

If it were absolutely necessary to test such a transformer with a resistor dummy load it would have to be some kind of a series combination because a) if a resistor fails short circuit it won't necessarily be catastrophic and b) it'll may be a bit difficult to obtain a 360k resistor rated for 200+ watts and 9kV

It's unfortunate that your professor is hard to understand as power electronics is an area where misunderstandings can have serious consequences. We'd like to see you back here.
The Electrician
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#7
Apr8-09, 05:16 AM
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Quote Quote by brewcrew975 View Post
OK. I understand the rated conditions, my other question is I have is the the same transformer, 225 VA, Vprimary=120 V and Vsecondary=9KV, Isecondary=25 mA, says that when the secondary side is shorted it will have a current of 25 mA
A transformer that has a rating like that is undoubtedly a current limited transformer of the type used to power neon tube signs. See: http://en.wikipedia.org/wiki/Neon_sign_transformer

If that is what it is, then with a 1 ohm load the current won't be very different from what it is with a short on the ouput. You'll have 25 mA in the 1 ohm resistor, and the voltage across it will be 25 mV.
bitrex
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#8
Apr8-09, 04:23 PM
P: 196
I'd forgotten about neon sign transformers and their integral current limiting. I thought he might have been dealing with some kind of plate transformer like some of the ones shown here: http://www.surplussales.com/Transformers/HvLvTr-1.html


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