0
\$\begingroup\$

The "transformer equation" is as follows, $$\frac{N_s}{N_p}=\frac{V_s}{V_p}$$ What is assumed when this equation is used?

enter image description hereenter image description here

(Primary coil on the left, secondary on the right, in both cases)

Is transformer 1 an ideal transformer, or transformer 2 (I have seen both in books and I'm not sure whether in the first picture, the other field lines are just not shown because they are irrelevant). It would make sense that transformer 1 is ideal and is 100% efficient because no work is done by the magnetic field in the air, but wouldn't a magnetic field through "ideal air" do no work regardless? How does this relate to flux linkage? Surely the flux linkage in both coils must be the same for the equation to hold, so in transformer 2, conceptually, less magnetic field lines go through the secondary coil than the primary coil, therefore the equation would not hold.

\$\endgroup\$
  • \$\begingroup\$ Bro, please insert a good picture. Draw it in a paper and then capture a photo. It looks like cartoon 😂😂 \$\endgroup\$ – Sadat Rafi Apr 26 at 14:18
  • 1
    \$\begingroup\$ Transformer 2 is not a transformer, it's two inductors next to each other. A transformer uses a magnetic core to guide the flux and minimize leakage flux (i.e. flux that doesn't go through the other coils in the transformer). \$\endgroup\$ – Hearth Apr 26 at 14:24
  • 1
    \$\begingroup\$ None of them is Ideal transformer. Because ideal transformer doesn’t exist. All transformer has some losses. They are hysteresis loss, eddy current loss, heat loss ( due to winding resistance) and magnetic leakage. But when someone is at primary level of transformer these losses are ignored for easy analysis. \$\endgroup\$ – Sadat Rafi Apr 26 at 14:25
  • \$\begingroup\$ And ' Np / Ns = Ep / Es ' is not transformer emf equation. \$\endgroup\$ – Sadat Rafi Apr 26 at 14:27
  • \$\begingroup\$ I recommend you to study " Electric Machinery Fundamentals - by Stephen J Chapman, Chapter 1 (Topics 1.4,1.5, 1.6,1.7) and chapter 2. \$\endgroup\$ – Sadat Rafi Apr 26 at 14:32
2
\$\begingroup\$

The most basic and important assumption required for the equation to hold is that all of the flux generated in one coil (primary or secondary) passes through the other coil. For this assumption to be true, generally a magnetic core is needed to trap and guide the flux from one coil to the other. Without the core, much of the magnetic flux generated by one coil does not pass through the other coil.

After that, there are many issues that lead to deviation from the ideal behavior, such as various forms of losses.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

The transformer equation is for an ideal transformer. A real transformer behaves like an ideal transformer with added components as shown in the equivalent circuit shown below. The added components account for the flux linkage between the primary and secondary being less than 100% as depicted in your transformer 2. It also accounts for the winding resistance and core losses.

It is the winding resistance and core losses that make the transformer less than 100% efficient, not the incomplete flux linkage.

Incomplete flux linkage results in leakage reactance, Xp and Xs in the equivalent circuit. There is no energy lost in those components, but there is a voltage drop across them that is proportional to the current. That means that the expected secondary voltage from the transformer equation could be reduced significantly as the load current increases. Some leakage reactance is unavoidable, but good designs minimize that to the extent required for the type of use intended.

For the equivalent circuit presented below, the transformer equation holds for the ideal transformer part of the circuit: $$\frac{N_s}{N_p}=\frac{E_s}{E_p}$$

enter image description here

For additional information about the above figure, see my answer to: Equivalent ciruit parameters of a 1kVA transformer.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ You might spell out the difference between resistive energy losses = less than 100% efficient, and leakage inductance, which despite having no losses still makes it a bad general purpose transformer (though good for microwave ovens, neon signs and welders). \$\endgroup\$ – Neil_UK Apr 26 at 14:41
  • \$\begingroup\$ @Neil_UK; thank you. I revised my answer. \$\endgroup\$ – Charles Cowie Apr 26 at 14:58
1
\$\begingroup\$

What is assumed when this equation is used?

  1. All the magnetic flux produced by the primary couples to all other primary turns i.e. there is no leakage flux that can give rise to leakage inductance that gives rise to formula errors when there is a load or magnetization inductance (which there always will be).
  2. All the flux generated by the primary couples to all the turns on the secondary - this ensures that the voltage in the secondary is truly dependent on the turns ratio and the voltage ratio

However, meeting the above (and satisfying the equation) can be done theoretically with what we sometimes refer to as a non-ideal transformer. But only if that non-ideal transformer has zero leakage components (see below for the listings in the equivalent circuit). It can have magnetization current and, it can have core losses and the basic equation still works - this is because core losses and magnetization current are parallel to the ideal "inner transformer".

Transformer equivalent circuit showing leakages and magnetization inductance and losses: -

enter image description here

Picture from here

  • \$L_P\$ is the primary leakage inductance

  • \$R_P\$ is the primary copper loss

  • \$R_C\$ is the core losses due to eddy currents and hysteresis

  • \$L_M\$ is the magnetization inductance

  • \$L_S\$ is the secondary leakage inductance

  • \$R_S\$ is the secondary copper loss

If there is no loading on the secondary, \$R_S\$ and \$L_S\$ won't affect the output voltage. If the magnetization inductance and core losses (\$L_M\$ and \$R_C\$) are both very high impedances, then some leakage components on the primary (\$L_P\$ and \$R_P\$) can be tolerated without much detriment to the equation in question (providing there is still no load on the secondary).

Is transformer 1 an ideal transformer, or transformer 2

T1 might be an ideal transformer, T2 is not an ideal transformer

so in transformer 2, conceptually, less magnetic field lines go through the secondary coil than the primary coil, therefore the equation would not hold.

Correct.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.