Calculating Amps and VA for Multiple Secondary Windings in a 28W Transformer

[Divendra Nath]

I have a transformer that has 120/240 on the primary,but the secondary has multi volt ,33v 0v,13.5v 0v,2.7v 0v 2.7v,9v 0v 9v.The power consumption is 28watts only.How you work out 28watts with all secondary windings to find amps so I can calculate for total VA,and wire gauge?.Some one out there is and has that knowledge to show me the right way to do this please.

 

[phinds]

I have a transformer that has 120/240 on the primary,but the secondary has multi volt ,33v 0v,13.5v 0v,2.7v 0v 2.7v,9v 0v 9v.The power consumption is 28watts only.How you work out 28watts with all secondary windings to find amps so I can calculate for total VA,and wire gauge?.Some one out there is and has that knowledge to show me the right way to do this please.

It depends on what current you are drawing from what output voltage. If you consume all 28 watts out of the 2.7v output, you'll get a very different current than if you consume all 28 watts out of the 33v output.

 

[Divendra Nath]

Thankyou for your kind response.So how it can be achieved .Do I have to share that 28watts with all secondary.So if I use 28watts with one winding it will be loaded and I can't use it with others.How should I do this?.

 

[phinds]

Thankyou for your kind response.So how it can be achieved .Do I have to share that 28watts with all secondary.So if I use 28watts with one winding it will be loaded and I can't use it with others.How should I do this?.

I have no idea what your load is going to be so have no possible answer for you.

 

[Tom.G]

So how it can be achieved .Do I have to share that 28watts with all secondary.So if I use 28watts with one winding it will be loaded and I can't use it with others.

Correct.

A transformer transfers power between windings through a magnetic field. The total power a transformer can handle is determined by the material and the physical size of both the core and of the wire used in the windings. Different core materials can support different strengths of magnetic field for a given size.

The wire sizes used will determine how much current they can carry before they overheat and burn the insulation, or even melt.

WATTS is the product of VOLTS times AMPS. W = V x A. So if you know any two of them you can find the third one.

In your example if you draw 28W from the 33V winding the current thru that winding will be W/V = A, or 28/33 = 0.85A. If you draw 28W from a 2.7V winding the current there would be W/V = A, or 28/2.7 = 10.4A.

In both cases the primary current (in a perfect transformer) will be W/V = A, or 28/240 = 0.12A. Since no one has come up with a perfect transformer yet, the primary current will be a little bit higher; around 10% to 20% higher is common for small transformers.

 

[jim hardy]

Do I have to share that 28watts with all secondary

Yes.
You can put 28 watts into the primary and apportion it to the secondaries as you see fit, provided each secondary is capable of handling 28 watts worth of current by itself.

Just like paying my monthly bills - if i have $28 available and i have four creditors, i can give $7 to each of them or divide it between them according to my whim.