Electromagnetic induction > Mutual inductance. Inductance of conductors. Transformers > Problem 11.3.24

Statement

$11.3.24.$

Two coils are wound on the iron core. The magnetic flux generated by each
coil does not leave the core and is divided equally in its branches. When coil
1 is connected to an AC circuit with a voltage of 40 V, the voltage across coil 2
is 10 V. What is the voltage at the open terminals of coil 1 if coil 2 is connected
to an AC circuit with a voltage of 10 V?

Solution

When an alternating voltage is applied to a coil, the total flux generated is split into two identical paths, so that the other coil, placed in one of the branches, is only linked by half of the total flux.

In the first case, with$ V_1 = 40\ \text{V} $on coil 1, a voltage$ V_2 = 10\ \text{V} $appears on coil 2. The voltage ratio is

$\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$

from which it follows that $N_2 = N_1/2$ (coil 2 has half the number of turns as coil 1).

When the connection is reversed and $V_2' = 10\ \text{V}$ is applied to coil 2, the generated flux divides just as before. Coil 1 is in one branch and receives half of the total flux. The voltage induced in it is

$V_1' = N_1 \frac{d}{dt}!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\, V_2'$

Substituting $N_2 = N_1/2 $ gives

$V_1' = V_2' = 10\ \text{V}$

Answer

$\boxed{V_1' = V_2' = 10\ \text{V}}$