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Transformer Math Check

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ITT-Joe

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I've got a question regarding transformer math. I would like someone else to confirm that I have the concept correct as I'm a tad bit confused and the many books and online content I've covered does not explicitly state what I'm asking.

(By the way, yes, this is my first post. I just found this forum and after having a look around, I decided to register. Think of this as my introduction, sort of.)

Anyway, the way I understand math related to transformers is this;

Voltage Primary (Vp) over Voltage Secondary (Vs) is equal (=) to Number of Turns Primary (Np) over Number of Turns Secondary (Ns).
(Vp over Vs = Np over Ns)

So if I have a transformer with an unknown Vp, an Vs of 25 volts, an Np of 20, an Ns of 100. The Vp (or voltage primary) would be 5 volts, correct??
(Vp = ??, Vs = 25v, Np = 20, Ns = 100)

Hopefully you understand this. If not, I'll draw up a quick diagram to perhaps make it more clear, as typing fractions is not possible. Thank you in advance for the assistance in clearing up my confusion.
 
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Yes, in a transformer the primary and secondary voltages and currents are proportional to the turn’s ratio.

Np/Ns=Vp/Vs=Ip/Is

However, that’s for an “ideal transformer” and real world transformers are not “ideal”. You have copper loss, core loss, coupling coefficients, leakage inductance… etc. to take into consideration.
 
Yes, in a transformer the primary and secondary voltages and currents are proportional to the turn’s ratio.

Np/Ns=Vp/Vs=Ip/Is

However, that’s for an “ideal transformer” and real world transformers are not “ideal”. You have copper loss, core loss, coupling coefficients, leakage inductance… etc. to take into consideration.

CORRECTION: The ratio is

Np/Ns=Vp/Vs=Is/Ip
 
You have it.

Understanding transformers is no different than understanding gear reduction or multiplication, even so much as the effect each step-up or down has on reversing current flow is analogous to the gears turning in opposite directions.
 
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