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I think I understand how a step-up transformer works, but don't understand how a step-down transformer converts high voltage AC to low voltage AC. Following the ratio $\frac{V_1}{V_2}=\frac{N_1}{N_2}$ I see why the emf in the second inductor can be smaller than the emf in the first inductor, but I don't understand how the voltage in the second wire can be smaller than the voltage across the whole first wire.

For example if there's one loop on the first inductor and on the second inductor ,then the voltages on the both wires are the same. If you add one more loop on the first inductor, now it has a two times greater flux and the second inductor has 0.5 of that flux because it has two times less loops. But how does it make to voltage in the second wire smaller than in the first? As I see it, the voltage across the second wire should be bigger, because the first inductor produces more flux so there's a greater emf on the second inductor. Please show me where is my confusion. Thanks.

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    $\begingroup$ I don't think that if you double the number of turns in the primary coil of an ideal transformer that the flux is doubled if the primary voltage is held fixed. I think that you're tacitly assuming that the electrical current through the primary coil would remain the same if you double the number of turns in the primary. $\endgroup$
    – user93237
    Commented May 15, 2017 at 19:58
  • $\begingroup$ @SamuelWeir, Then the rate of change of flux with respect to time drops as we add more loops? $\endgroup$
    – Snate
    Commented May 15, 2017 at 20:57
  • $\begingroup$ But they're ideally the same transformer. The distinction only comes about by deciding which winding to drive (which winding to consider the primary winding). If you drive the winding with the smaller number of turns, the transformer acts to step up the voltage. If you take the same transformer and 'flip it around' so that you're driving the winding with the larger number of turns, the transformer acts to step down the voltage. $\endgroup$
    – Hal Hollis
    Commented May 15, 2017 at 21:27
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    $\begingroup$ @Snate:"Then the rate of change of flux with respect to time drops as we add more loops?" - If you drive a current through the primary coil with an AC current source which forces the same sinusoidal AC current though the primary regardless of the loading conditions, then you're right in thinking that the magnitude of the total magnetic flux by the primary coil is proportional to the number of turns in the primary coil. But that's typically not how power is applied to a transformer. Usually the power source is more like a constant-voltage AC source. $\endgroup$
    – user93237
    Commented May 15, 2017 at 22:12

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now it has a two times greater flux and the second inductor has 0.5 of that flux because it has two times less loops.

This isn't true. Ideally, all of the flux $\phi$ threading the primary threads the secondary (no flux leakage).

However, if the secondary has fewer turns, it has less flux linkage $\lambda_s = N_s \phi$ than the primary $\lambda_p = N_p \phi$ (the unit of flux linkage is weber-turns)

The voltage across the primary is (again, ideally)

$$V_p = \frac{\mathrm{d}\lambda_p}{\mathrm{d}t} = N_p \frac{\mathrm{d}\phi}{\mathrm{d}t}$$

and the voltage across the secondary is

$$V_s = \frac{\mathrm{d}\lambda_s}{\mathrm{d}t} = N_s \frac{\mathrm{d}\phi}{\mathrm{d}t}$$

and so

$$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$

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