Why is the output current of a transformer fixed?

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Summary:

Im confused about how the output current of a transformer is fixed ie can be calculated by the law of conservation of energy only...(VI)in = (VI)out ?
If the secondary coil is part of a circuit, and the secondary voltage is defined by V1/n1 = V2/n2, then the current in the secondary circuit should be defined by I = V/R right?
So what happens when you vary the resistance of the secondary circuit, which law is obeyed?
Thanks

Main Question or Discussion Point

Im confused about how the output current of a transformer is fixed ie can be calculated by the law of conservation of energy only...(VI)in = (VI)out ?
If the secondary coil is part of a circuit, and the secondary voltage is defined by V1/n1 = V2/n2, then the current in the secondary circuit should be defined by I = V/R right?
So what happens when you vary the resistance of the secondary circuit, which law is obeyed?
Thanks
 

Answers and Replies

  • #2
phinds
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You have a fundamental misunderstanding. There is no limitation on the output current of a transformer other than that the transformer can't put out more power on the output than is available on the input (minus inefficiencies).
 
  • #3
berkeman
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Im confused about how the output current of a transformer is fixed ie can be calculated by the law of conservation of energy only...(VI)in = (VI)out ?
The output power is equal to the input power, minus the losses due to resistive losses and other losses. For most transformers working in their passbands, they are pretty close to 95% efficient or better.

So Vin*Iin = Vout*Iout minus some efficiency losses. The actual ratios of Vin and Vout depend on the winding ratios of the transformer (same for the currents, but inversely so).

Hope that helps. Ask more questions if things still don't make sense, but please do a little reading about transformer modeling first in case that answers your questions. Thanks :smile:

https://en.wikipedia.org/wiki/Transformer
 
  • #4
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Thank you very much for your help.
I am still a little confused why the answer to the attached problem is 15A however. Whilst Law of Conservation of Energy is obeyed, Ohms Law seems to be largely disobeyed (ie I would have predicted 5000A based on Ohms Law...ie 10000V/2ohms)?

1575246151559.png
 
  • #5
berkeman
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Assume ideal transformers with no losses. What it the output power consumed at the output voltage? So assuming the same input power to that final distribution transformer, what is the current input at that higher input voltage to that distribution transformer?
 
  • #6
berkeman
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Upon further review, just use the power of 150kW at 10,000V to figure out the current in the transmission line. As a more accurate answer, you would nee to know the load power and factor in the line resistance of the TL.
 
  • #7
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Upon further review, just use the power of 150kW at 10,000V to figure out the current in the transmission line. As a more accurate answer, you would nee to know the load power and factor in the line resistance of the TL.
Thank you
As I said, I understand why 15A is the right answer if only Law of Conservation of Energy is used but I dont understand how it is the answer if Ohms Law is used.
 
  • #8
berkeman
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Thank you
As I said, I understand why 15A is the right answer if only Law of Conservation of Energy is used but I dont understand how it is the answer if Ohms Law is used.
Can you show the Ohm's Law equations that you think are a problem? There's probably just some transformer ratio issues that are confusing you. Thanks. :smile:
 
  • #9
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As I stated,
I = V/R = 10000/2 = 5000A
Thank you
 
  • #10
phinds
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As I stated,
I = V/R = 10000/2 = 5000A
Thank you
You are confusing what is clearly stated as the POWER LINE resistance with the load resistance.
 
  • #11
berkeman
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As I stated,
I = V/R = 10000/2 = 5000A
Thank you
The 2 Ohms is the series resistance of the TL, which contributes to losses. It has nothing directly to do with the load resistance, That's a common confusion.

To look at this problem more accurately, you would need to know the load power (150kW?) and the transformer losses. That way your could solve for all of the voltages and currents exactly.
 
  • #12
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So what happens when you vary the resistance of the secondary circuit, which law is obeyed?
When you vary that resistance, the currents in both the primary and secondary circuits change.
So both laws are obeyed.
 
  • #13
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You are confusing what is clearly stated as the POWER LINE resistance with the load resistance.
@pkc111 this is the key. Both Ohms law and energy conservation must be used in this problem. Ohms law for the power line and energy conservation to figure out the load.
 
  • #14
sophiecentaur
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Summary:: Im confused about how the output current of a transformer is fixed ie can be calculated by the law of conservation of energy only...(VI)in = (VI)out ?
If the secondary coil is part of a circuit, and the secondary voltage is defined by V1/n1 = V2/n2, then the current in the secondary circuit should be defined by I = V/R right?
So what happens when you vary the resistance of the secondary circuit, which law is obeyed?
Thanks

So what happens when you vary the resistance of the secondary circuit, which law is obeyed?
All those laws apply but you need to be using the appropriate ones and in the right order.
Your secondary volts (assumption about the Voltage Ratio of the transformer) will be applied to the Load and TL Resistances in series. That determines the secondary current. The Secondary VI (plus some losses) will be reflected in a different pair of V and I in the primary circuit. The VI product will be the same (ideally).
Cause and effect (if you want to introduce them) are always in the direction of Energy Source - Energy Sink but it doesn't mean your calculations have to follow that order. You can write down an Equation but that doesn't necessarily say anything about the direction of Energy Flow. But it gives the right answer.
 
  • #15
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Wind turbine may be rated 150 kW but, as you can see, its blades are motionless, it is generating no power...
Hence the line current is zero...
:wink: :wink: :wink:
 
  • #16
gleem
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Assuming your physics knowledge is limited to general physics then you might not know all the intricacies of using Ohms law for AC power. In the our case of the power line load for the secondary of the step up transformer is the primary for the step down transformer as well as the resistance of the transmission line. So the output voltage of the step up transformer must be split among the elements in that circuit which includes the 2 ohms for the transmission line and the primary windings of the step down transformer. While the windings may have small DC resistance they have rather large AC resistance (called inductive reactance). This reactance inhibits AC current like a resistance (and thus produces a voltage drop) but does not dissipate any energy. You do not have enough info to determine this reactance and therefore cannot assume all the 100,000 volts is dropped across the 2 ohms of the transmission line. So you can simply use conservation of energy to determine the current in the secondary of the step up transformer.

Remember that energy is stored in the magnetic fields of the windings of the transformers and is used or lost when the current passes through an actual resistance (not a reactance)
 
  • #17
sophiecentaur
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you might not know all the intricacies of using Ohms law for AC power.
For resistive loads, it's exactly the same. RMS Volts and Current are always assumed unless told otherwise. First pass through the solution for a transformer doesn't need to consider Inductance - just the basic Transformer equation. There's no need to make things any more difficult until you have to.
cannot assume all the 100,000 volts is dropped across the 2 ohms of the transmission line.
I should hope not. V2/R would be very high!!!!
 
  • #18
gleem
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First pass through the solution for a transformer doesn't need to consider Inductance - just the basic Transformer equation.
therefore cannot assume all the 100,000 volts is dropped across the 2 ohms of the transmission line.
Yes I know but one has to be aware that the inductance takes most of the potential drop of the 100,000 volts of the secondary of the step up transformer. The OP did not consider this and wondered why 15 amps was the correct answer rather than 5000 amps which he calculated using Ohms law assuming the 100,000 volts is applied only to the transmission line thus raising the question in post #4.
 
  • #19
sophiecentaur
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The OP did not consider this and wondered why 15 amps was the correct answer rather than 5000 amps which he calculated using Ohms law assuming the 100,000 volts is applied only to the transmission line thus raising the question in post #4.
I don't quite know where you are coming from here. The secondary winding is a source of induced emf. It's the Load Resistance that governs the secondary current (include the extra 2 Ohms if you want to). If you short out the secondary then that's not a mode that most transformers are happy with.
 
  • #20
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Im confused about how the output current of a transformer is fixed
You're confused, and the answers are not helping because we don't really understand your question.

I propose that you should try again. Think carefully about what you know about transformers, and which part is confusing, and restate your question.
 
  • #21
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You have my sympathy that you are dealing with trying to solve such a poorly constructed problem. Or, perhaps there is other information that you haven't shown us. Without knowing how much power the town is consuming, there is no answer to this question.
 
  • #22
sophiecentaur
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We need a response from @pkc111 before we can progress this. He / she feels that The R=V/I should fit in somewhere but you have to use the correct R in each of the circuits.

An additional concept that may or may not help is to point out that, if the transmission line is supplying 150kW at 10kV, then it must be 'seeing' a load R that's given by the line volts and the power. The idea that transformers also 'Transform Resistance' is not intuitive but the load resistance presented by all the town's lights etc. can be found from the local 250V supply volts and the supply power. This is not the same (much lower, in fact) than the load resistance that the transmission line 'sees'.

(I haven't included formulae or answers because @pkc111 can work them out as a useful exercise.) Ignore the 2Ω initially; any system will be designed with a low enough line resistance (thick enough wire) to keep the losses low 'enough'.
 
  • #23
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I see that the OP has not returned to PF since posting this thread. He hasn't seen any of the replies. That means we are wasting out time.

Thread closed.
 

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