# An Ideal Transformer

In a transformer assumed to be transformer, power in the primary is equal to power in the secondary. So in a sense, the power in the secondary is 'fixed'. Output voltage in the secondary is also fixed by the turns ratio and the voltage in the primary. Then, The current tapped from the secondary coil should also be fixed $I=\frac{P}{V}$.

Doesn't the resistance in the circuit connected to the secondary coil also affect the current tapped from the secondary coil? So if a load resistance in the secondary coil circuit is varied, would the current still be fixed the same $I=\frac{P}{V}$ or would it change accordingly with $I=\frac{V}{R}$

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Hello William & welcome to Physics.SE. Please use the resources of this site for mentioning Formulas..! It's called LaTeX. You'll learn on your way anyways... It's not at all a problem. Your question would be revised for better clearance..! :) –  Waffle's Crazy Peanut Oct 3 '12 at 14:01

The primary and secondary powers are equal but not fixed.

In an ideal transformer with turns ratio $n$ (= # secondary turns / # primary turns) fed by an ideal ac voltage source $V_p$, the secondary current $I_s$ is determined by the transformed primary voltage (that is, the secondary voltage $V_s=n V_p$) and load resistance $R$ ($I_s=V_s/R = n V_p/R$), and the primary current is just the transformed secondary current ($I_p= nI_s$).

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@CrazyBuddy, I believe that his $n$ is a turns ratio rather than an efficiency. –  Alfred Centauri Oct 3 '12 at 14:49
@AlfredCentauri: OMG, I didn't notice that..! Thanks Alfred. Please don't mind it. I was a bit confused of those powers & efficiencies..! :) –  Waffle's Crazy Peanut Oct 3 '12 at 14:53
Yes n is the turns ratio. I will edit to make that clearer. –  Art Brown Oct 3 '12 at 15:46

If the load changes on the secondary, the primary voltage and/or current will change too.

Don't forget that the source connected to the primary "sees" the load connected to the secondary transformed through the square of the turns ratio.

$\dfrac{V_p}{I_p} = \dfrac{N^2_p}{N^2_s} \dfrac{V_s}{I_s}$

Assuming a voltage source driving the primary, if the load connected to the secondary is changed, the secondary current will indeed change but the primary current changes too and in the correct proportion.

The source doesn't actually "see" the (ideal) transformer, rather, from the source's perspective, there is an impedance connected equal to:

$Z_p = \dfrac{N^2_p}{N^2_s} Z_s$

So, for example, if an ideal voltage source is connected to the primary of an ideal transformer with a turns ratio of $2:1$ and the load impedance is doubled, the current in the secondary will be halved while the current in the primary will be quartered since the impedance the source "sees" is quadrupled.

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