3
$\begingroup$

I just learned about the basics of transformers. I am aware that according to the law of energy conservation, power input should be equal to power output. However, when I think about this with ohm's law, in the secondary coil, power remains the same but voltage is increased implies resistance is somehow increased. But what causes this "increase in resistance"? I looked through some of the previous posts on this topic but I think all of them don't explicitly explain this increase in resistance. Maybe there is some more advanced stuff that is not covered in our high-school syllabus, but I want to know more. Thanks in advance!

$\endgroup$
6
  • $\begingroup$ Forget the output circuit - how the transformer may couple to that is irrelevant. Just looking at the transformer, one coil puts power/energy into a magnetic field, the other coil takes it out. By changing the ratio of windings, the relative voltage changes, so the current into a matched load must change to keep the power constant. If the load is not matched, all bets are off on actually getting the input power into the output load. $\endgroup$
    – Jon Custer
    Aug 4 at 15:59
  • $\begingroup$ @Jon Yes, I understood what you meant but the point of my problem is how does the load magically adjusted itself so that the power matches. $\endgroup$ Aug 4 at 16:06
  • 1
    $\begingroup$ Why does you floor lamp only take 1 amp off a 15 amp circuit? The transformer can provide the full power at the step-up(down) voltage, but that doesn't mean some random load will take it all. $\endgroup$
    – Jon Custer
    Aug 4 at 16:08
  • $\begingroup$ Ok, but I want to know will the $R$ in the secondary coil change when the turn ratio of the transformer changes? If yes, why? $\endgroup$ Aug 4 at 16:14
  • $\begingroup$ The resistance of the secondary coil is whatever the resistance of that wire is - it does not know anything about the turn ratio. If the resistance of the transformer itself has a major influence on the operation of the transformer than something is wrong. If you put an infinite resistance load on the output, you would measure full voltage and no current. If you put a (near)-zero resistance load on the output you would get lots of current (limited perhaps by the transformer wiring) and collapse the voltage output. If you put a matched load on you can get optimal power transfer. $\endgroup$
    – Jon Custer
    Aug 4 at 16:18
2
$\begingroup$

You have $P_1=P_2$, or $V_1^2/R_1=V_2^2/R_2$, where $R_1$ is the apparent resistance of the primary circuit. If you change $R_2$ you are changing both, the power and the apparent resistance $R_1$ ("apparent" because is not linked to any actual resistance), but not the voltages.

$\endgroup$
2
  • $\begingroup$ Why changing $R_2$ will affect $R_1$? Aren't the primary and secondary circuits in a transformer separate, where the only linkage between them is the magnetic flux? $\endgroup$ Aug 4 at 17:02
  • $\begingroup$ Yes, as I said, it an apparent resistance. But you can forget about it, the basic answer is that the power changes in both sides, that is, as in any electrical circuit, if you fix the input V then the power changes with R $\endgroup$ Aug 4 at 17:04
3
$\begingroup$

but I want to know will the 𝑅 in the secondary coil change when the turn ratio of the transformer changes? If yes, why?

The $R$ of the secondary coil will not change because the $R$ for the primary and secondary coils is zero for an ideal transformer in which power in equals power out. If they had any resistance, some energy would dissipated as heat and therefore not be available to a load connected to the secondary.

but then according to $P=\frac{V^2}{R}$, when the power is step-up(or down), the power will change. What is wrong with this argument?

If the secondary voltage is greater than the primary voltage, then the primary current has to be greater than the secondary current since

$$\frac{V_{p}}{V_{s}}=\frac{I_{s}}{I_{p}}$$

In other words, the power will change for both the primary and the secondary

Hope this helps.

$\endgroup$
7
  • $\begingroup$ but then according to $P=\frac{V^2}{R}$, when the power is step-up(or down), the power will change. What is wrong with this argument? $\endgroup$ Aug 4 at 16:28
  • $\begingroup$ Did you mean $$P=\frac{V^2}{R}$$? $\endgroup$
    – Bob D
    Aug 4 at 16:30
  • $\begingroup$ And when you say $R$ do you mean the resistance connected to the secondary winding? $\endgroup$
    – Bob D
    Aug 4 at 16:34
  • $\begingroup$ Yes, that's right $\endgroup$ Aug 4 at 16:35
  • $\begingroup$ Are you saying the output power of the secondary is greater than the input power to the primary? I'm not sure what you are arguing. $\endgroup$
    – Bob D
    Aug 4 at 16:49

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .