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What is the phase difference between the induced emf in the secondary winding of a transformer and the current through the secondary circuit in the case where a pure resistance is connected across the secondary winding? Or are they in phase with each other?

I have work out this much so far - The secondary induced emf, say E2 is lagging the flux phi in the core by 90 degrees. The current I2, according to Lenz's law, should be in such a direction so as to oppose the cause producing it, i.e, the sinusoidally varying flux phi. So The current I2 will have to lag E2 by 90 degrees and also phi by 180 degrees, so that the flux phi2 produced by I2 will be lagging phi by 180 degrees as well, and phi will thus be opposed by phi2. So I think that I2 will lag E2 by 90 degrees. Am I right so far?

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    \$\begingroup\$ There is no phase difference between the windings for a regular transformer. \$\endgroup\$ – Eugene Sh. Aug 3 '16 at 17:58
  • \$\begingroup\$ @Eugene Sh In what other types of transformers would there be a phase difference? \$\endgroup\$ – user118685 Aug 3 '16 at 18:04
  • \$\begingroup\$ ecmweb.com/archive/basics-transformers \$\endgroup\$ – Eugene Sh. Aug 3 '16 at 18:04
  • \$\begingroup\$ For Resistive load there is no phase difference between Induced voltage and Current. If you connect any 'R' and 'L' load the current 'I2' lags the Induced voltage 'E2' \$\endgroup\$ – Photon001 Nov 2 '16 at 6:57
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With a pure resistor connected directly to a winding, the current through the resistor and the winding are equal. The voltages across the resistor and the winding are equal. (Either the voltage or the current may be viewed as opposite in sign.) There is no phase difference between the voltage and current on a pure resistor. Therefore there is no phase difference on the winding (from the perspective of a source).

You reasoning is for an inductor not a transformer. For example, an ideal transformer is assumed to have infinite inductance and develop zero net flux. Regardless of whether the transformer is ideal or not, the relationships in the first paragraph hold and is the easy way to answer your main question.

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  • \$\begingroup\$ "For example, an ideal transformer is assumed to have infinite inductance and develop zero net flux." - There is always a net flux. But this net flux will always be maintained constant. The constancy of the net flux of the transformer core is one of the main working principle of the transformer. \$\endgroup\$ – user118685 Aug 4 '16 at 4:13
  • \$\begingroup\$ @user118685 "this net flux will always be maintained constant" then how does this net flux move away from a constant zero to a non-zero constant? \$\endgroup\$ – rioraxe Aug 4 '16 at 8:33
  • \$\begingroup\$ First when a sinusoidal voltage is applied across the primary winding, a current will set up the flux phi in the core. At this time assume that secondary is open. Then if a purely resistive load is connected across the secondary winding, a current I2 will be formed in the secondary circuit, given by I2 = E2/R. But this current will be in such a direction that it will produce a flux phi2 whose physical orientation will be against that of flux phi. So this will cause a reduction in the net flux in the core and that will be {phi-phi2}. Since the net flux is now lesser, the primary induced voltage \$\endgroup\$ – user118685 Aug 4 '16 at 9:02
  • \$\begingroup\$ will also reduce. Due to this applied voltage on primary will be higher than the induced voltage across the primary. This will cause the current to rise above I1 to I1a and this will increase the flux from phi to phi1. But phi1 = -phi2. Now the net flux in the core will be phi+phi1+phi2, = phi+phi1-ph1= phi. So the net flux phi will be a non zero constant. \$\endgroup\$ – user118685 Aug 4 '16 at 9:10
  • \$\begingroup\$ If the statement is -- an ideal transformer is not possible in the real world and therefore there will be magnetic flux -- that I agree. But we are specifically talking about an ideal transformer here. \$\endgroup\$ – rioraxe Aug 5 '16 at 0:21

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