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The Primary & Secondary has the same number of coils/windings, but the resistors on them are different, using P=I2 × R, we get that the current in the Secondary side decreases and Voltage Increases, which is the definition of a transformer, but doesn't a step up transformer needs more windings in the secondary? (Assume an ideal transformer 100% eff.)

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  • $\begingroup$ A 1:1 transformer simply isolates the secondary load ground from the primary, unless the transformer grounds are bonded. This allows the load ground to "float." It also protects the secondary from a DC bias level, such as a 70-V audio power amplifier (primary) connected to a speaker (secondary). $\endgroup$ – Bill N Dec 20 '18 at 15:39
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$\let\Om=\Omega \def\qy#1#2{#1\,\mathrm{#2}}$ You've given inconsistent data. Apparently you think there is no potential difference (p.d.) at primary's terminals. Otherwise how could you have $\qy{20}V$ and $\qy{10}A$?

But this can't be true. Let's call $V_1$ primary's p.d., $I_1$ primary's current, $V_2$, $I_2$ secondary's quantities. You must have $V_1=V_2$, $I_1=I_2$. If $R_2$ is secondary's load, then $V_2 = R_2\,I_2$ and also $V_1 = R_2\,I_1$. It's as if in place of a transformer in primary's circuit there were the secondary's load $R_2 = \qy8\Om$.

So total resistance is $\qy2\Om + \qy8\Om = \qy{10}\Om$ and current is $I = \qy{20}V/\qy{10}\Om = \qy2A$.

The rest is up to you.

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  • $\begingroup$ Elio this is a good solution, we could also arbitrarily set say V2 = 2 x V1 and therefore I1 = 2 x I2, when we solve get I2 = 2.21 amps. But the problem does state equal windings. $\endgroup$ – PhysicsDave Dec 20 '18 at 15:56
  • $\begingroup$ @PhysicsDave ... we could also arbitrarily set ... The problem as written by OP has 6 data. Of course you may change them in infinite many ways to make them consistent. But surely you can't leave alone 20V and 10A in the primary, unless you short-circuit the transformer. $\endgroup$ – Elio Fabri Dec 20 '18 at 16:07
  • $\begingroup$ Yes, effectively the primary coil has an impedance which is effected by the resistor and current in the secondary circuit. $\endgroup$ – PhysicsDave Dec 20 '18 at 18:21
  • $\begingroup$ So basically there's no way to increase voltage with the same number of windings? $\endgroup$ – acmilan Dec 20 '18 at 19:59
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    $\begingroup$ @acmilan Absolutely no way. Faraday forbids. Equal number of windings means equal flux, so same induced emf. (Ideal transformer, of course.) $\endgroup$ – Elio Fabri Dec 20 '18 at 20:06
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It looks like you set up your own problem, and you're confused that changing the resistors to arbitrary values (or number of windings), might violate power conservation. Are you sure Pin = Pout in this case? Via kirchoff - you have to conserve power going around a closed loop, but going from one end of the transformer to the other doesn't count... Given a left side circuit and number of windings, I would think the transformer with 5 windings on each side would try to push the same voltage to the secondary as is across the primary. Then you can think of that as a voltage source in the secondary circuit, and calculate current using ohms law and the 8 ohm resistor.

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  • $\begingroup$ So what you are saying is that in this case I violated Pinput=Poutput ? $\endgroup$ – acmilan Dec 20 '18 at 8:07
  • $\begingroup$ That's true, we usually don't have any resistor on the primary side for ideal transformers for these kinds of problems. So it complicated now and we would have to calculate impedance on the primary side and this would have a reduced voltage on the primary. From there the secondary would circuit would be calculated. $\endgroup$ – PhysicsDave Dec 20 '18 at 15:12
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Yes it needs more windings, ratio needs to be x4x 2:1 in this case.

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    $\begingroup$ 4:1? Shouldn't it be 2:1 , as voltage increases by 2? $\endgroup$ – acmilan Dec 20 '18 at 7:25
  • $\begingroup$ sorry yes 2 to 1 $\endgroup$ – PhysicsDave Dec 20 '18 at 15:04
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The applicable primary voltage, in applying the power into the transformer equals the power out of the transformer, is the voltage directly across the primary winding, not the voltage source shown.

The current in the primary winding should be the difference between the primary winding voltage and the voltage source, divided by the 2 ohm resistor. If the current in the primary is 2 amperes, then voltage across the primary winding would have to be 0.

ADDENDUM

The reason you are getting secondary voltages and currents inconsistent with a 1:1 turns ratio transformer is you incorrectly calculated the primary winding current. You can’t just simply divide the voltage source by the resistor. You need to take into account the voltage across the primary winding. In order to determine the primary current (as well as the primary voltage and secondary current and voltage) you can do the following:

  1. Apply Kirchoff’s Voltage Law (KVL) to both the primary and secondary circuit. This will give you two loop equations and four unknowns (primary voltage and current, and secondary voltage and current).
  2. Since there are the same number of turns in the primary and secondary, you know the primary voltage equals the secondary voltage, and the primary current equals the secondary current. This gives you two more equations. Substitute these into the two loop equations and you can solve for all the unknowns.

Hope this helps.

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  • $\begingroup$ How do you calculate the difference between the primary winding voltage and the voltage source you just mentioned? $\endgroup$ – acmilan Dec 20 '18 at 22:25
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    $\begingroup$ You don't calculate it, it is usually given.The problem is the current you assume is in the primary cannot be so unless the primary winding voltage were zero. $\endgroup$ – Bob D Dec 20 '18 at 22:28
  • $\begingroup$ I still don't really understand why the voltage across the primary winding is 0, can you elaborate more? $\endgroup$ – acmilan Dec 21 '18 at 0:34
  • $\begingroup$ You could give an example for me to analyze it better $\endgroup$ – acmilan Dec 21 '18 at 0:56
  • $\begingroup$ Sure with a circuit diagram, but can't do it in comments format. I will make an addendum to my answer. $\endgroup$ – Bob D Dec 21 '18 at 2:07

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