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Using a step up transformer we could, for example, take a 10V source and get 20V in the secondary coil. This doesn't violate the law of conservation of energy because the current will be halved.

The answer I can't find anywhere is HOW the current gets halved? V=IR therefore it seems to me that the load attached to the secondary coil will determine the current passing through it. So to my understanding, if we place a 10Ω resistor on each coil, we will get 1A through the primary and 2A through the secondary. Quadruple (2Ax20V) the power for free - impossible!

Where am I going wrong?

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  • \$\begingroup\$ Look up law of induction. \$\endgroup\$
    – winny
    Jan 23 '17 at 20:16
  • \$\begingroup\$ What do you mean by 'place a 10 ohm resistor on each coil'? \$\endgroup\$
    – Chu
    Jan 23 '17 at 20:28
  • \$\begingroup\$ I mean attach each coil to a resistor so that the current passing through each coil also has to pass through the resistor \$\endgroup\$
    – CodeNovice
    Jan 23 '17 at 20:30
  • \$\begingroup\$ @CodeNovice: The current passing through one coil does not proceed to the other coil. Rather it loops through the primary coil and creates a magnetic field which induces current into the other coil. The ratio of turns for each coil and the quality of magnetic core material determines the step up or down ratio. \$\endgroup\$
    – jbord39
    Jan 23 '17 at 20:33
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    \$\begingroup\$ @pjc50 I think you may have steered me in the right direction. The impedance is probably what I have not been taking into consideration. I will look into it because it's not something I know a lot about. Thanks! \$\endgroup\$
    – CodeNovice
    Jan 23 '17 at 20:57
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I assume you have a simple AC transformer and not a complex switcing regulator. Especially the flyback regulator needs more complex explanations.

In step up transformer the input 10 V comes from an AC source. The output 20 V is connected to passive load that adapts its current intake to the voltage that is provided by the transformer.

You think it upside down. You think that something actively forces the output current to be a half of the input current and the input current is given at first.

The right causal chain is the following:

  • let the 10/20 tranformer have 10 V input voltage
  • 20 volts is present for the load
  • the load takes as much current as load's operating law states; for example a 10 Ohm resistor takes 2 A (otherwise it's not a 10 Ohm resistor)
  • The transformer takes 4 A from the 10 V supply, that's the output current as doubled (otherwise it's not a 10/20 transformer)

Of course in practice it's possible to misuse the transformer so that 2:1 law for the currents is not true. For example, connect the input to 10 V DC. No continuous 20 V output will be available because DC is not in the usable frequency band of the transformer. Transformer sinks maybe tens of amperes, gets hot, but still no output until someone cuts the input off. (A historical fact: Faraday found the induction by this experiment in 1831)

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  • \$\begingroup\$ Thanks for the reply, but it still makes no sense to me. Maybe I need to start from scratch. How does the transformer "take" (or 'draw'?) current from the 10V supply? I would have thought the the loads attached to the primary coil would determine the current that flows through it. \$\endgroup\$
    – CodeNovice
    Jan 23 '17 at 20:42
  • \$\begingroup\$ @CodeNovice Yes, you must start from the elementaries. Now you are jailed by a bad obsession. It's as serious as a real flat earth supporter has. May be he tries to understand the science, but it all stops, because he cän't figure out, how something flat can be spherical at the same time. \$\endgroup\$
    – user287001
    Jan 23 '17 at 20:55
  • \$\begingroup\$ Loads are attached to the secondary coil. The primary coil is just attached to a voltage source. You should think of a transformer as passing power (voltage times current), not just voltage or not just current. The power input to a transformer will equal the power out of the transformer (plus a little for losses in the transformer). \$\endgroup\$ Jan 23 '17 at 20:56
  • \$\begingroup\$ @PeterBennett, thanks for the reply. What seems clear is that I don't have enough underlying knowledge to even understand the answers! I think I'm trying to project DC circuit understanding (which is all I have) to an AC concept. \$\endgroup\$
    – CodeNovice
    Jan 23 '17 at 21:01
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The primary voltage produces a magnetisation current. That current produces an alternating magnetic flux. That alternating flux couples both primary and secondary coils (assume 100% coupling). From Faraday's law of induction, the secondary voltage induced is proportional to the number of turns so, twice as many secondary turns means twice the voltage.

It is easily shown that a secondary load current generates exactly the same magnitude (but with opposite polarity) ampere turns as the primary ampere turns due to that load i.e. there is no increase in the flux in the core. So with twice the primary turns, the secondary current is half the primary current "increase" due to the load.

I call it an "increase" because the starting point for the primary current is the magnetization current and that remains stable irrespective of load.

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  • \$\begingroup\$ From that we would have to conclude that in a loaded transformer the sum of primary and secondary flux is zero. ("exactly the same but with opposite polarity") \$\endgroup\$
    – neonzeon
    Jan 23 '17 at 23:04
  • \$\begingroup\$ @neonzeon - no, the sum of fluxes due to loading effects are zero. There is still the flux due to the primary voltage magnetizing the core and this is a fixed amount of flux. \$\endgroup\$
    – Andy aka
    Jan 24 '17 at 8:23
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It's quite easy to get confused by transformers - their math can become challenging, especially when you have a complex load.

In your example, a 10 Ω resistor over the secondary transforms to an equivalent 2.5 Ω resistor ("Apparent Load Impedance") on the primary side.

enter image description here

Your circuit model therefore simplifies to two resistors in series: The 10 Ω resistor you inserted in the primary and the 2.5 Ω ohms of "Apparent Load Impedance".

enter image description here

If you used a 40 Ω resistor in the secondary, the transformed impedance would be 10 Ω and the same power would be delivered to your 10 Ω primary resistor and the transformer/load combination.

enter image description here

Another way to think of the above is to realize you need 4x more resistance to get the same power from a 2x higher voltage.

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