0
$\begingroup$

let’s suppose we have an ideal transformer with 500 windings on the primary side and 5 windings on the secondary side. The primary voltage is 230 V and the primary current 16 A. So the equations tell us that the secondary voltage should be 2.3 V and the secondary current 1600 A.

My question is: Are the 1600 A on the secondary site a maximum upper bound for the secondary current? So when we for example have a nail in the secondary current which has a very low resistance it should reach the 1600 A. But if we have something with higher resistance it should reach a current less than 1600 A. But the current cannot be higher than 1600 A even with a resistance almost zero, right? Because that would contradict conservation energy. Am I Right?

$\endgroup$

3 Answers 3

2
$\begingroup$

You cannot prescribe both the voltage and the current of primary the coil. If the transformer, assuming an ideal one, is connected to a voltage source, then the primary voltage is given and the current in both the primary and secondary coils are determined by the terminating load resistance.

Specifically, $N_1=500, N_2=5$, $V_1=V_s=230V$ and therefore $V_2=V_1\tfrac{N_2}{N_1}=2.3V$. Now if you have on this a resistance $R_2$, then the current through that load and the secondary coil will be $I_2=\frac{V_2}{R}$ while the current in the primary coil will be (flux conservation) $I_1=I_2 \tfrac{N_2}{N_1}=\frac{V_1}{R_2}\tfrac{N_2^2}{N_1^2}$ showing how the primary current depends on the load attached to the secondary.

$\endgroup$
3
  • $\begingroup$ So primary current will decrease as well? $\endgroup$
    – Blue2001
    Oct 26, 2023 at 14:42
  • $\begingroup$ But could we reach a higher value than 1600 A ? $\endgroup$
    – Blue2001
    Oct 26, 2023 at 14:43
  • $\begingroup$ yes, on paper in the secondary, you can. I say "on paper" because "on paper" we have assumed an "ideal" transformer. I leave it you to check that $V_1I_1=V_2I_2$, so energy is conserved. But when the transformer is not an "ideal" one, which is always, if you short the secondary, you may blow the primary or the secondary or both, and if there is a ferromagnetic core then it may saturate well before the current reaches such high levels, etc. $\endgroup$
    – hyportnex
    Oct 26, 2023 at 14:55
1
$\begingroup$

Forget the transformer for the moment. Suppose you have a circuit with where the voltage is $230$ V and the current is $16$ A. You can figure out the resistance. You can change the resistance. But if you do, you will also change the current.

Now look at the transformer. You have a primary voltage and current. From that you figure out the secondary voltage and current. From that you figure out the resistance in the secondary circuit.

Yes you can change the resistance. But if you do, you change the current in both sides of the transformer.

$\endgroup$
1
  • $\begingroup$ Thanks for all the comments! You’ve solved my confusion! :) $\endgroup$
    – Blue2001
    Oct 26, 2023 at 22:00
1
$\begingroup$

"But the current cannot be higher than 1600 A even with a resistance almost zero, right? Because that would contradict conservation energy."

Not because it would contradict energy conservation, but because if $I_{sec} > 1600$ A, then $I_{pri} > 16$ A, and this would 'blow' (that is melt) the fuse in the plug if the transformer primary is (via a flex and plug) plugged into a domestic 'mains' socket in the UK. [In fact the fuse should have blown at 13 A!]

I wondered if you had the misconception that the primary current was somehow fixed at 16 A, rather than depending on the secondary current. I thought that an example might help ...

Suppose that a load of 0.23 $\Omega$ is connected across the 2.3 V secondary. The load current will be 10 A (and the power into the load will be 23 J/s). Since we are assuming no energy loss to heat/internal energy in the transformer, the primary current must be 0.10 A (so 23 J/s into the transformer and 23 J/s out; $V_pI_p = V_sI_s$). Try calculating the primary current for even lower load resistances.

$\endgroup$
1
  • $\begingroup$ Thanks for all the comments! You’ve solved my confusion! :) $\endgroup$
    – Blue2001
    Oct 26, 2023 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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