Matching Input and Output Current on Transformers: A Common Question

  • Thread starter Thread starter harblargh
  • Start date Start date
  • Tags Tags
    Transformers
AI Thread Summary
The input current on a transformer does not need to match the rated output current to achieve that output, but the output voltage can vary based on the load and the transformer's regulation. For transformers with poor regulation, running below the rated load can lead to higher output voltage. The input current is calculated by dividing the output current by the turns ratio, with the relationship between input and output voltages following the inverse of this ratio. In a scenario with a 100V input and a 75V@50A output, the required input current would be 37.5A, assuming ideal efficiency. Understanding the power involved clarifies the relationship between input and output currents in transformers.
harblargh
Messages
3
Reaction score
0
Okay, now this may sound ridiculously stupid, but I know next to nothing about transformers and so I have a question:

Does the input current on a transformer have to match the rated output in order to get that output?

(i.e. A trans rated 5 amps putting out the full rated current on only 2.5 amps of input current.)
 
Engineering news on Phys.org
No, but depending on the regulation (essentially a measure of how efficient the transformer is) the output voltage maybe be dependent upon load--put another way, in order to get the stated output voltage given the correct input voltage, it must be run at or very near the rated load current. For transformers with good regulation, this is less important, but for those with relatively poor regulation, running it at much less than full rated load can result in a significantly higher output voltage. This may or may not be important, depending on the application's tolerance for voltage deviation.

However, to answer the question I think you're asking, the input current will be equal to the output current divided the turns ratio (the voltages will follow the inverse of this). For example, if your transformer has a load of 10 amps on the output and it's a 120-to-12 volt stepdown transformer, the input will be 10 A / (10/1) = 1 A, assuming an ideal transformer.
 
Oh. Thanks for the help.

The problem I just can't wrap my brain around is the actual flow of current. That's the issue.

For this hypothetical transformer, the rating is 50A with a matching load run though a VFD with a max load of 75V. Ergo, at maximum, the output would have to be 75V@50A.

What amperage would the 100V input would have to be to reach the maximum output rating?
 
Assuming decent regulation the loaded voltage ratio will roughly approximate the turns ratio (for a much more accurate figure, take the no-load voltage ratio). So, you can plug those figures into the equation in my previous post [input current = output current / (primary turns / secondary turns)] and that will be your answer.
 
Maybe it would help to look at the power involved.

Assuming a resistive load...

power in load = 75 volts times 50 amps = 3750 watts

The transformer is 100% efficient so...

3750 watts at 100 volts must mean a current in the primary of
(3750 watts / 100 volts) = 37.5 amps
since power = voltage times current.

This is the same answer as you get if you divide the output current by the turns ratio as above.
ie 50 amps /( 100 / 75 ) =37.5 amps.
 
Ah, I get it now. Thanks alot, I really appreciate it. :biggrin:
 
Very basic question. Consider a 3-terminal device with terminals say A,B,C. Kirchhoff Current Law (KCL) and Kirchhoff Voltage Law (KVL) establish two relationships between the 3 currents entering the terminals and the 3 terminal's voltage pairs respectively. So we have 2 equations in 6 unknowns. To proceed further we need two more (independent) equations in order to solve the circuit the 3-terminal device is connected to (basically one treats such a device as an unbalanced two-port...
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...
Back
Top