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I know that transformer is a constant flux device but there is one point I don't understand.

As you know, when secondary of transformer is connected to a load, current starts to flow. This current produce its own flux (let's say F2) which opposes the main flux (F1). Then E1 (back-EMF primary side) decreases and extra current starts to flow in primary side and this current cancels F2 flux out and the F1 kept constant in transformer. Everything is OK by now but the point I don't understand is when F2 flux is canceled out by extra current in primary side, what happens to E1?

  1. Does it go to beginning value (before secondary side current to flow)? If yes, so current in primary side should also goes to beginning value (by canceling extra current) right?

    Or,

  2. E1 stays decreased (at the point after secondary current flux decreases it)? If it is like this how is that possible? I mean there is constant flux in the core but E1 is smaller. Shouldn't it be the same since flux is same?

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  • \$\begingroup\$ the flux varies SLIGHTLY. View the transformer as a negative-feedback system with high gain, but not infinite gain. \$\endgroup\$ Jul 24 '19 at 17:55
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Before you connect a secondary load, magnetization current is flowing in the primary and this sets up the flux in the core. That flux remains constant as an RMS value with any load changes.

Any secondary load takes current and this creates a flux that is cancelled by the extra flux created by the primary winding due to that secondary load current. In other words load currents produce opposing fluxes that cancel and, all that remains, is the original magnetization flux.

This means that back emfs or induced voltages remain as they always were when there was no secondary load. Small print: this explanation assumes leakage inductance and winding losses are negligible.

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Using the approximations you appear to be using in your question (like neglecting winding resistance, which is fine) ...

1) E1 goes back to the value it had.

However, the flux in the core, constant flux as you assume, is due to not just the primary current, but the algebraic (signed) sum of the primary ampere turns (AT) and the secondary AT.

The E1, back EMF as seen at the primary, is due to the total flux in the core.

When there's no secondary load, the primary AT, I'll call the value ATmag as it's the AT required to magnetise the core, causes the core flux, and the core flux causes the E1.

When there's a secondary load, secondary AT opposes most of the primary AT, the difference being the ATmag that causes the core flux.

The same core flux generates the same voltage in both primary and secondary as it did before the secondary was loaded.

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  • \$\begingroup\$ Thanks for answer. As I understand, E1 goes back to value which it was before current flowing secondary. But how this extra current flows when E1 comes to initial value ? \$\endgroup\$
    – d.alex
    Jul 24 '19 at 19:55
  • \$\begingroup\$ Because we made the asssumption that the winding resistance was zero. With a real transformer, with finite resistance, the extra current indeed needs a small voltage drop, which is provided by the IR drop of the primary winding. \$\endgroup\$
    – Neil_UK
    Jul 25 '19 at 4:25
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Flux is constant but inductance of primery has changed. I mean when you use V= N.d/dt flux. N ( number of turns has changed!!!) Or if you use V= L. di/dt also L has changed. If we connect secondary of transformer to a load because of producing reverse flux the inductance ( unit is Henry) of primary decreases.

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