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Could someone please explain why, when there is no secondary load, the applied voltage of the primary roughly equals the back EMF of the primary? Since the primary coil is an inductor I wondered exactly how the back EMF is always there and always cancelling the applied voltage when there is no secondary load. Take an inductor in a DC circuit for example: the back EMF of the inductor only cancels the applied voltage (source voltage) at $t=0$. I know transformers use AC but how can it continually match the applied EMF when there is no secondary load?

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closed as unclear what you're asking by hazzey, Fred, Chris Johns, Dave Tweed, Wasabi Feb 19 '16 at 1:05

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  • $\begingroup$ I'd really like to try to answer this question, but I really have no idea what you think "back EMF" is, and why you think it applies in the context of an inductor. Where did these ideas come from? $\endgroup$ – Dave Tweed Feb 18 '16 at 4:44
  • $\begingroup$ I think back EMF could be used in the inductive context to describe the imaginary force that results from a magnetic field not wanting to move. Usually its applied to motors to describe the loading of the magnetic field from the magnetic elements. $\endgroup$ – Voltage Spike Feb 18 '16 at 20:31
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The specific question is too confused to be answerable, so I'll try to explain some basics of transformers in general.

A transformer with open secondary is just a inductor. The concept of back EMF makes no sense here. What will happen is the current will lag the voltage by 90° in the ideal case.

Power delivered to the inductor is current times voltage. Note that this averages to 0 over any whole cycle when the current and voltage are sines with 90° phase shift between them. Energy is put into the inductor during half the cycle, then given back during the other half. The average power transfer is 0. The energy that is absorbed and then given back is temporarily stored in the magnetic field of the transformer core.

In reality, the inductor will not be purely inductive, but will also appear to have some resistance in series with its inductance. That resistance is the DC resistance of the wire the primary coil is made from. That resistance will dissipate power proportional to the square of the current thru it. In a "good" transformer or inductor, this resistance is small compared to the impedance due to the inductance over the frequency range the device is intended to work.

Looking at the voltage and current, the current will lag the voltage a little less than 90° due to this resistance. The average of the voltage x current over a whole cycle is now a little positive, which is the energy dissipated by the resistance each cycle.

Getting even closer to reality, there will also be some loss in storing then retrieving magnetic energy to/from the transformer core each cycle. This will also appear as a resistive component electrically. The resistance will appear in parallel with the primary inductance. Both together are in series with the DC resistance of the coil.

When a load is connected to the secondary, it looks a lot like the core gets more lossy to the circuit driving the primary. The primary looks more resistive, which accounts for the higher power it now draws at the same voltage. Like the core loss resistance, this additional resistance appears in parallel with the inductance of the primary, from the point of view of the circuit driving the primary. In a ideal transformer with ideal load, the secondary removes energy from the magnetic core exactly as the primary tries to put it there. The result is the primary current is now in phase with its voltage, and therefore looks purely resistive. This of course never happens exactly because there are always inevitable losses.

In the ideal case, a transformer primary looks like whatever impedance is connected to the secondary, divided by the square of the turns ratio. For example, let's say a transformer has a 1:3 turns ratio. You put 12 VAC into the primary, and ideally get 36 V out of the secondary. If there is 10 Ω on the secondary, then it is delivering (36V)2/10Ω = 130 W. This has to go into the primary, which means the primary current is 130W/12V = 10.8 A, which means the primary looks like a 12V/10.8A = 1.11 Ω resistor. Note that 1.11Ω/10Ω = 1/9, which is the square of the turns ratio.

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  • $\begingroup$ Just to be clear, the resistance of the primary winding itself is in series with the inductance, but the "transformed" resistance of the secondary load appears in parallel with the primary inductance. $\endgroup$ – Dave Tweed Feb 18 '16 at 19:26
  • $\begingroup$ @Dave: Good point. I have added clarification to the answer. $\endgroup$ – Olin Lathrop Feb 18 '16 at 22:56

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