Can a Transformer Rated at 225 VA Only Supply 25 mA on the Secondary Side?

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Discussion Overview

The discussion revolves around the operational characteristics of a transformer rated at 225 VA with a primary voltage of 120 V and a secondary voltage of 9 kV, specifically focusing on the implications of its rated secondary current of 25 mA. Participants explore the behavior of the transformer under various loading conditions, including short circuits and resistive loads.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that the transformer can supply up to 25 mA at 9 kV without overheating, but may deliver higher voltages at lower currents.
  • Others argue that the transformer’s behavior under load, such as with a 1 ohm resistor, could lead to dangerously high currents, potentially resulting in circuit breakers tripping or catastrophic failures.
  • A participant mentions that the transformer might be a current-limited type, similar to those used for neon signs, suggesting that the current would remain around 25 mA even with a 1 ohm load.
  • Concerns are raised about the safety of testing such a transformer with low resistance loads, emphasizing the risks involved.
  • Some participants note that the secondary winding resistance could affect the actual current drawn, and that the transformer would not simply limit itself to the rated maximum current.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the transformer under load, particularly regarding the implications of connecting a 1 ohm resistor. There is no consensus on the outcomes of such tests, and safety concerns are highlighted by multiple participants.

Contextual Notes

Participants acknowledge the complexities involved in transformer operation, including the effects of winding resistance and the nature of current-limiting transformers. The discussion reflects a range of assumptions about the transformer's design and intended use.

Who May Find This Useful

This discussion may be of interest to those studying electrical engineering, power systems, or anyone involved in practical applications of transformers and their safety considerations.

brewcrew975
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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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brewcrew975 said:
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.
 
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.
 
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.
 
brewcrew975 said:
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?
 
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 :eek:

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. :smile:
 
brewcrew975 said:
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.
 
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 :eek:
 

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