Ideal transformer principle and emf equation

An ideal transformer is one with zero winding resistance, no leakage flux (all flux links both windings), no hysteresis or eddy-current loss, and a core of infinite permeability, so it needs no magnetising current. Its efficiency is 100% and its voltage regulation is zero.

Operating principle

A transformer works on mutual induction (Faraday's law).

           phi (alternating)
     +---------------------+
 o---|]                   [|---o
 V1  |]  N1     core  N2  [|   V2 -> load
 o---|]                   [|---o
     +---------------------+
   primary             secondary

1. An ac voltage $V_1$ is applied to the primary winding of $N_1$ turns.
2. The current sets up an alternating flux φ in the laminated iron core.
3. The flux links the secondary winding of $N_2$ turns and induces in it an emf $E_2$ of the same frequency. The same flux induces a back emf $E_1$ in the primary that balances $V_1$.
4. When a load is connected, $I_2$ flows; the primary automatically draws a matching current so that $N_1I_1 = N_2I_2$ and $V_1I_1 = V_2I_2$.
5. Since emf per turn is the same in both windings, $V_1/V_2 = N_1/N_2$: the voltage is stepped up or down without changing frequency.

EMF equation

Let the flux be sinusoidal: $\phi = \phi_m\sin\omega t$, $\omega = 2\pi f$.

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\ &= N_1\omega\phi_m\sin(\omega t - 90^\circ) \end{aligned}$$

So $e_1$ lags φ by 90°. Its maximum value is

$$E_{1m} = N_1\omega\phi_m = 2\pi f N_1\phi_m$$

RMS value:

$$\begin{aligned} E_1 &= \frac{E_{1m}}{\sqrt2} = \frac{2\pi}{\sqrt2} f N_1 \phi_m \\ E_1 &= 4.44\, f N_1 \phi_m = 4.44\, f N_1 B_m A \end{aligned}$$

Similarly for the secondary:

$$E_2 = 4.44\, f N_2 \phi_m$$

Transformation ratio

$$\frac{E_2}{E_1} = \frac{N_2}{N_1} = K, \qquad \frac{E_1}{N_1} = \frac{E_2}{N_2} = 4.44 f\phi_m \ (\text{emf per turn})$$

For an ideal transformer $V_1 = E_1$, $V_2 = E_2$ and $I_1/I_2 = N_2/N_1 = K$.

Example: a 50 Hz transformer with $N_1 = 500$ and $\phi_m = 9$ mWb gives $E_1 = 4.44\times50\times500\times0.009 = 999$ V.