# transformer

Discussion in 'General Electronics Discussion' started by GeoffC, Mar 19, 2013.

1. ### GeoffC

39
4
Mar 11, 2013
Please can anyone help with this problem?
I have a transformer which is a 240v primary 15-0-15v secondary. I want to find out what continuous working load the transformer can work at.
I know the resistance of the primary 80 ohms. So can I use ohms law to work the current load? – ie volts^2/80ohms=720W. So, Square root of 720W/80= 3Amps. This answer seems to be on the high side, the transformer is heavy, and is a handfull - (in practical terms). Have I gone off track with ohms law. I want to choose a ‘voltage Regulator LM78/79nnn’ to work within the limits of the transformer?

Thanks

2. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Ohm's law is not very helpful here. The transformer has a high inductance and therefore an AC resistance that differs wildly from the DC resistance.
You will need a datasheet to find the rating of the transformer (unless there's a hitherto unfound label somewhere on the transformer.

A somewhat risky alternative is to apply a load to the output of the transformer and increase the load until the transformer's core temperature rises to e.g. 40° above ambient temperature.

3. ### Electrobrains

259
5
Jan 2, 2012
You can also roughly estimate the VA rating of a standard iron sheet transformer by comparing it to the size and weight of known, similar transformers.

If your transformer has only one primary winding and one center-tapped 30V secondary winding, you can be quite sure that the transformer was dimensioned so that all possible stored energy in the iron core would be available on that secondary winding.

Also, you can measure the thickness of the secondary (and primary) copper wire and compare it to other known transformers. There are standard formulas used by professional transformer builders, that specify the wire thickness based on the nominal current.

4. ### GeoffC

39
4
Mar 11, 2013
Thanks for this, I understand now, I thought my answer was too good to be true. I will use it with a12v 1amp voltage regulator (LM7812). I am sure it will handle 1amp. Many Thanke. Regards, Geoff. (GeoffC).

5. ### Miguel Lopez

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63
Jan 25, 2012
Try to measure dimensions of the central column of the core (in centimeters), and then calculate the effective area of that column (in square centimeters).

Squaring that value will give you an approximate idea of how many watts your transformer can manage. This value will always be under the real one.

6. ### Electrobrains

259
5
Jan 2, 2012
Interesting!
Do you mean squaring the square cm's?
P (Watt) > A^2 (cm^4)

Last edited: Mar 20, 2013
7. ### Miguel Lopez

252
63
Jan 25, 2012
Yes, usually the number (in cm^4) is around 75% of the actual value, but it can vary with the geometrics of the central column. It is more accurate if the cross section tend to be square. It fails a lot in a core type with rectangular cross section; in this case, the value will be around 40% of the actual power. In shell type transformers with square cross section, it is reliable.

This is a mnemonic that a professor taught us in the college. I have used it till now with good results.