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In a transformer a sinusoidally varying voltage is applied to the primary coil of the transformer which give rise to a time varying flux which induce another sinusoidally varying EMF in the primary coil. The induced EMF has a 180° phase difference with the applied voltage and hence opposes it due to which the voltage across the coil should decrease leading to a decrease in the max value of the time varying flux. So the decrease in flux should then also decrease the max value of induced EMF.

How can the EMF across the primary then continue to be sinusoidal when the max value of sinusoidally varying out of phase induced EMF change with time? Shouldn't it be represented by a sinusoidal curve whose max value change with time?

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  • \$\begingroup\$ it's not 180 phase difference. The induced voltage is \$ \dfrac{d\Psi}{dt}\$ if \$\Psi=\Psi_{max}cos(\omega t)\$ then \$u=U_{max}sin(\omega t)\$ in both windings, primary and secondary. \$\endgroup\$
    – Marko Buršič
    Commented Jun 21, 2016 at 8:18
  • \$\begingroup\$ the flux is in phase with the current which lags the applied voltage by 90 degree(if we assume the primary winding is purely inductive). And as you have proved above the induced emf further lags the flux by 90 degree . So there is a net 90+90=180 degree phase difference between the applied and induced emf. \$\endgroup\$
    – Siddharth Prakash
    Commented Jun 21, 2016 at 8:42
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    \$\begingroup\$ No, it doesn't add 90+90. Both voltages: secondary and primary are in phase. Both voltages lags the flux by 90deg. You sholud twist the logics: the voltages are induced due to the alternating flux, of course the flux is due to the applied voltage on the primary, but if you want to understand the transformers basics you should skip that fact. Not an easy machine this transformer, right? \$\endgroup\$
    – Marko Buršič
    Commented Jun 21, 2016 at 9:08

2 Answers 2

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If you force a voltage in the primary (voltage source connected to the primary), then this is the voltage you'll have in the primary ("nature" can't change that). Then the magnetic flux will be such that you get this voltage in the primary (so that the Faraday law holds true). And finally the current will be such that you get this magnetic flux.

This is the variables causal relationship in your configuration (voltage source attached to the primary):

Voltage(t) -> Flux(t) -> Current(t)

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The current flowing in the primary is 90m degrees out of phase with that of the applied voltage, The changing current induces an emf which opposes the supply voltage and this leads to a limit in the amplitude of the primary current. This assumes no resistance in the circuit. The changing magnetic field, which theoretically would be sinusoidal induces a voltage in the secondary winding and any load on the secondary would produce a current which would change the magnetic balance. the EMF's would change and the primary current would rise to provide the power to balance out the change in magnetic fields.The changing magnetic field would induce currents in the core of the transformer which would cause it to heat up. This is known as IRON loss.Laminating the core provides multiple parallel paths for this induced current, the total current is the same but because the current is split into parallel paths the power loss is reduced because of the square effect.The parallel paths are effectively insulated from each other.Letting the laminations go rusty was one way of achieving this insulation.There are other losses involved and I would recommend you to read up a good text book on the subject. Viv

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  • \$\begingroup\$ "The changing current induces an emf which opposes the supply voltage and this leads to a limit in the amplitude of the primary current." So we agree upto that point. Now let's focus only on the primary winding. What happens after that. If the current through the primary is reduced then the opposing induced emf would also dicrease which then further increases the current and so on..So instead of pure sinusoidal curve shouldn't we get a sinusoidal curve with oscillating max value \$\endgroup\$
    – Siddharth Prakash
    Commented Jun 21, 2016 at 8:07

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