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The first figure below from Power Systems Analysis (Grainger/Stevenson/Chang) shows an ideal transformer. The authors write v1 as shown. Later on in the same chapter, however, the authors write v1 as a function of i2 while describing a practical transformer.

Why is v1 of the ideal transformer not also a function of i2? Is there a reason why i2 has no effect on v1 in an ideal transformer?

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\begin{align} \mathrm{Ideal:~} v_1&=e_1=N_1\frac{d\Phi}{dt} \end{align}

enter image description here enter image description here \begin{align} \mathrm{Non-ideal:~}v_1&=r_1i_1+L_{11}\frac{di_1}{dt}+L_{12}\frac{di_2}{dt} \end{align}

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2 Answers 2

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It's ideal so there are no losses, no magnetic or resistive losses, or other losses of any kind.

So as energy cannot be created or destroyed, observing only the ideal transformer, power in must equal power out.

So therefore, both primary and secondary ideally work with Faraday's law and Lenz's law. Voltage/current on primary creates flux and same flux creates voltage/current on secondary.

Ideal transformer also assumes infinitely high core permeability, infinite winding inductance, and zero net magnetomotoric force.

So all of the flux passes through both primary and secondary. No part of the flux is lost to core losses due to hysteresis or eddy currents. No energy is lost to resistance or heating. No stray capacitances between any windings are considered.

If there are losses, such as wire resistance, or magnetic losses, then the voltage does depend on current.

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  • \$\begingroup\$ I get that power in = power out for an ideal transformer. What I'm asking specifically is why v1 in the ideal transformer doesn't seem to be a function of i2 \$\endgroup\$ Feb 9, 2023 at 19:48
  • \$\begingroup\$ @artist_and_not_EE_by_training Because there are no losses that would cause voltage drop from current. \$\endgroup\$
    – Justme
    Feb 9, 2023 at 19:51
  • \$\begingroup\$ If you allow us to forget resistance (even for the non-ideal transformer), I think the idea of the second figure is that v1 is affected by the flux generated by the secondary current. Why is this phenomenon not also present in the ideal transformer? \$\endgroup\$ Feb 9, 2023 at 19:54
  • \$\begingroup\$ The same flux goes through primary and secondary coils because there are no magnetic losses either. \$\endgroup\$
    – Justme
    Feb 9, 2023 at 19:56
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    \$\begingroup\$ @artist_and_not_EE_by_training Because v1 comes from an ideal voltage source in the ideal example so nothing connected to it could ever cause it to deviate. Since the transformer on top of that is ideal, you don’t have any losses. i2 will just be transformed to primary side, causing i1 to flow. \$\endgroup\$
    – winny
    Feb 9, 2023 at 20:32
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More to the point, in this ideal model , winding resistance r1 does not exist which is the only source of input voltage drop measured at e2 from v1 source and i2 load.

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