Electromagnetic induction > Mutual inductance. Inductance of conductors. Transformers > Problem 11.3.18

Statement

$11.3.18.$

Show that in an ideal transformer with a short-circuited secondary winding,
the ratioI1=NN12
holds, where I1and I2are the currents, and N1and N2are
I2
the number of turns in the windings.

Solution

In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability)

the voltage induced in each winding is proportional to the number of turns

$ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$

With the secondary winding short‑circuited, the voltage $ V_2 = 0$
Since$ V_2 = N_2 \frac{d\Phi}{dt}$
this implies that the magnetic flux $\Phi $in the core must be constant

For the net flux to be zero, the total magnetomotive force must be zero

$N_1 I_1 + N_2 I_2 = 0$

From this, the ratio of magnitudes is :

$\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$

The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given.

Answer

$\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$