Ideal transformer no-load and loaded, ratios

An ideal transformer has no resistance, no leakage flux, no core loss, and a core of infinite permeability.

(i) No-load operation

   I1 ~ 0     N1 : N2     I2 = 0
 o---->--+   ||     ||   +---o
         )   ||     ||   (        
  V1     )   ||core ||   (  V2 (open)
 o-------+   ||     ||   +---o

The secondary is open, so $I_2 = 0$. A sinusoidal voltage $V_1$ on the primary draws a very small magnetising current $I_\mu$, which creates the alternating flux $\phi = \phi_m\sin\omega t$ in the core. This flux links both windings and induces:

$$e_1 = -N_1\frac{d\phi}{dt},\qquad e_2 = -N_2\frac{d\phi}{dt}$$

In an ideal transformer $V_1 = E_1$ and $V_2 = E_2$. Dividing,

$$\frac{V_1}{V_2} = \frac{E_1}{E_2} = \frac{N_1}{N_2} = a$$

where $a$ is the turns ratio.

(ii) Loaded operation

When a load is connected, the secondary current $I_2$ flows and its mmf $N_2I_2$ opposes the flux. The flux would fall, so the primary draws an extra current $I_1$ to restore it, until the two mmfs balance (neglecting $I_\mu$):

$$N_1I_1 = N_2I_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1} = \frac{1}{a}$$

Since there are no losses, the power in equals the power out:

$$V_1I_1 = V_2I_2 \Rightarrow \frac{V_1}{V_2} = \frac{I_2}{I_1}$$

Results

$$\boxed{\frac{V_1}{V_2} = \frac{E_1}{E_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}}$$

The voltage ratio equals the turns ratio, and the current ratio is its inverse. In phasor terms, $V_1$ leads the flux by 90° (in the sign convention $V_1 = -E_1$), and on load $I_1$ and $V_1$ have the same phase angle $\phi$ as $I_2$ and $V_2$.