24v transformer

[lor]

I took a resistance reading across the 240 v side of a transfomer (on a boiler) .It read 0.2 ohms .
Why does the fuse not blow as this is like a short circuit isnt it ?

If im wrong then how do you test a transformer .

ps im a boiler man not an electrician

thanks alot

 

[Adam_151]

Most of the impedance will be reactive, not resistive... a transformer wants to have a low resistance otherwise it'll get warm and be inefficent

If you want to know what current its taking, then hook up an ammeter in series with it

What are you trying to test though, if you are trying to prove that the Primary is not open circuit then you've just done it ,you might want to check the secondary as well, and do an IR between all connections of the primary together and all connections of the secondary. You might also want to check the secondar voltage with it power up

 

[Lectrician]

Most of the impedance will be reactive, not resistive... a transformer wants to have a low resistance otherwise it'll get warm and be inefficent

If you want to know what current its taking, then hook up an ammeter in series with it

What are you trying to test though, if you are trying to prove that the Primary is not open circuit then you've just done it ,you might want to check the secondary as well, and do an IR between all connections of the primary together and all connections of the secondary. You might also want to check the secondar voltage with it power up

Not sure what you mean by the RED bit!?!






A transformer coil is a coil of wire. This wire is not particularly tiny - is has a fair CSA. It is also copper - very low resistance. The resistance measured with a low ohm meter at low volts (9v typically from a meter) will be low.

When you power up this coil with 240v mains, there will be an initial surge caused by this near 'short circuit' until the coil starts to produce a back EMF, which then starts to 'choke' the current in the circuit (limits it). This is why me refer to these items in terms of IMPEDANCE rather than RESISTANCE. This is also why the inductors in flourescent fittings are often reffered to as 'chokes'........choking back the current, and presenting a 'high resistance' to the mains supply, even though the relative resistance of the coil is actually very low.

 

[Adam_151]

I'm not sure what the issue is? You've just backed up and expanded upon the bit in red there, haven't you?

 

[Lectrician]

No, I still do not understand the bit in red



There is no easy way to test to see if a transformer is down, other than doing an insulation resistance test to earth, however, in a boiler the transformer is likely to be an electronic double insulated type.

 

[ban-all-sheds]

No, I still do not understand the bit in red

It puzzles me too - surely the heating you get depends on the impedance of the primary (which will be related to resistance - more turns = more resistance and more reactance), not just its resistance?

And the impedance of the primary is what it is, and depends on the application of the transformer - some of them (e.g. signal transformers, valve output transformers) can be quite high.

 

[NotHimAgain]

The heat produced in a transformer is mainly as a result of 'Copper Losses' and 'Iron Losses'.

Copper losses are due to power dissipated by the windings.

Cu Loss = I²R watts, where: I is the winding current (primary or secondary as appropriate) and R the winding resistance.

Iron losses are due to eddy currents induced in the iron core of the transformer - this is why it is laminated.

When designing a transformer the appropriate number of turns of wire are determined for both primary and secondary to give the required voltage change. The windings would generally be made of the thickest grade of copper wire that is economic for the particular transformer. This is to reduce the resistance (and hence the copper losses) to the minimum economic value.

 

[ricicle]

.

Iron losses are due to eddy currents induced in the iron core of the transformer - this is why it is laminated.

You havn't mentioned the other iron loss

 

[NotHimAgain]

.
You havn't mentioned the other iron loss [/quot

No I can't spell Hysterisis and I even worse at magnetostriction (hum)

 

[ricicle]

.
You havn't mentioned the other iron loss [/quot

No I can't spell Hysterisis and I even worse at magnetostriction (hum)

 

[Adam_151]

No, I still do not understand the bit in red

The OP was supprised that the resistance of the coil was low, I pointed out that was how it should should be, as the vast majority of the impedance of the coil comes from reactance, not resistance

It puzzles me too - surely the heating you get depends on the impedance of the primary (which will be related to resistance - more turns = more resistance and more reactance), not just its resistance?

The heat disapated in the transformer due to electrical losses is as someone else meationed, P=I²R, where R is just the resistance, of course current through the coil will depend on impedance, but its only resistance that heats (if you consider a coil that has no resistance at all - a perfect coil that doesn't actually exist, and all the impedance of that is due to reactance, then the heating effect will be null)

 

[briwire]

crikey he only asked a simple question.

the I=V/R only applies to DC.

Mains is 50Hz, AC. When AC flows in a coil, the "resistance" is called impedance, as it varies with frequency. So a coil connected to 240v 50Hz would take more current than 240v 100Hz.

If you measure with your meter you are reading pure resistance which would only be applicable with 0 Hz or DC.

Baisically, if you are reading a resistance of up to a 1000ohms, it should be OK.