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I couldn't find the answer to this anywhere so, how can you tell what the power draw, in amps, of a transformer is knowing the input voltage and turns on primary and secondary.

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  • \$\begingroup\$ The input power will depend primarily on the output power consumed by the load. \$\endgroup\$ – Peter Bennett May 6 '16 at 4:21
  • \$\begingroup\$ Can the input power go over the maximum load for the transformer, saying it is drawing too much, making it overheat? \$\endgroup\$ – Zamen May 6 '16 at 4:26
  • \$\begingroup\$ Yes, it can.... \$\endgroup\$ – Peter Bennett May 6 '16 at 4:44
  • \$\begingroup\$ At full design power Ip~= Vp/Zload effective ~= Vp/( Zload x (Ns/Np)^2) . At "idle" the transformer draws "magnetising current. This is largely "wattless" current as the load is mainly inductive = reactive, so actual watts dissipated mainly relate to I^2Rcopper copper losses and core magnetisation hysteresis losses. When loaded the load can exceed the design load if you connect a load larger than the designed one :-) - ie yes - you can breal it if you do silly things. \$\endgroup\$ – Russell McMahon May 6 '16 at 12:09
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If the secondary is unloaded then you just have an inductor, the primary. As an inductor for (say) a power transformer, it will have an inductance of maybe 10 henries. At 50 Hz, 10 henries has an impedance of 3142 ohms and, on a 230V AC 50 Hz supply, will take 73.2 mA RMS.

This is current but it certainly doesn't imply any power because current through an inductor (in an AC scenario) lags the voltage applied by 90 degrees and this makes the average power zero.

However, there are some losses that consume power such as "eddy current" losses. A power transformer core is made from laminations and laminations are used to stop the big iron core acting as a shorted turn. Laminations are insulated from each other but each laminate will have a small eddy current circulating and this represents a small power loss.

Core saturation is another problem - there are only so-many ampere-turns than can be tolerated before core saturation losses become intolerable.

As for the rest of the power dissipated in a transformer, this is usually just the copper losses i.e. the current flowing through the wires causes heat but, this is only usually noticeable when a secondary load is taking output current.

how can you tell what the power draw, in amps, of a transformer is knowing the input voltage and turns on primary and secondary.

Well, you need to understand what power is being taken by the load - this can be reflected back to the primary and is part of the story. Those amps flowing through both windings produce copper losses so measure the wire resistance for each winding and, knowing the secondary current you can predict that loss. For eddy current losses, measure them with an open circuit secondary - apply only enough primary voltage to make a decent estimation then scale up for the full primary voltage (this avoids confusion with saturation losses).

Once you have eddy current losses predicted at the full primary voltage you can calculate power loss due to core saturation - measure the total power (secondary open circuit) and subtract predicted eddy current losses.

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All transformers draw a little current (a percent of full power or so) when there isn't a load connected. This is the energy that is lost magnetising and reverse magnetising the transformer core. This can be really hard to model and is often measured empirically instead. But as this is usually only a few percent of full rated current, most people don't worry about it that much.

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  • \$\begingroup\$ Current and power should not be used interchangeably in this context, since AC power is a complex subject. An unloaded transformer is basically a reactive load; active power is only a small fraction of apparent power. In other words, losses in an unloaded transformer are far less than the product of voltage and current. \$\endgroup\$ – dmitryvm May 6 '16 at 9:07
  • \$\begingroup\$ Yes, you are absolutely correct, serves me right for writing that late at night \$\endgroup\$ – Sam May 6 '16 at 22:26

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