# Conservation of energy and ohms law in a step up transformer

In an ideal step up transformer with a constant resistance attached to the secondary coil, how is energy conserved and ohms law followed at the same time?

$$\frac{V_p}{V_s}=\frac{N_p}{N_s}=\frac {I_s}{I_p}$$

For constant $V_p$ and $N_p$ ,

$$V_s \propto N_s$$

If total resistance to secondary coil is constant(including wires) , $V_s \propto I_s$ according to ohm's law.

But from first relation, $V_s \propto \frac {1} {I_s}$

What's happening here?

• These ideas can be generalized to any linear impedance operator and one gets the idea of the impedance "reflected" through the transformer. This allows power engineers to transform the load of power customers from a low impedance, high current consuming one to an equivalent power, high impedance, low current consuming one, thus lowering transmission losses. See my answers here and here for more info. I'm sure Alfred would know this too, he's concentrating on your direct problem in his answer. – WetSavannaAnimal Oct 5 '13 at 4:07
• – Always Confused Jun 29 '16 at 19:17

But from first relation, $V_s \propto \frac {1} {I_s}$

Actually, no. We have:

$N_s = N_p \dfrac{I_p}{I_s}$

So, $V_s \propto \frac {1} {I_s}$ is not true since $I_p$ is not constant.

The secondary voltage and current are proportional by Ohm's Law:

(1) $V_s = R_L I_s$

The power is conserved:

(2) $V_p I_p = V_s I_s = R_L I^2_s$

The secondary current is related to the primary current:

(3) $I_s = \dfrac{N_p}{N_s}I_p$

Thus:

$V_p I_p = R_L (N_p/N_s)^2 I^2_p$

$\rightarrow \dfrac{V_p}{I_p} = (N_p/N_s)^2R_L$

• Further to Alfred's answer (I am sure he knows this already): the ideas can be generalized to any linear impedance operator and one comes up with the idea of a "transformed" impedance "reflected" by the transformer (terminology comes from transmission lines theory, where one thinks of travelling waves , so one speaks of a "Transformer's reflected impedance"). See my answers here and here for more info. I'm sure Alfred would know this too, he's concentrating on your direct problem in his answer. – WetSavannaAnimal Oct 5 '13 at 4:12