The mutual inductance **M** of two coupled inductances **L**_{1} and **L**_{2} is equal to the mutually induced voltage in one inductance divided by the rate of change of current in the other inductance:
**M = E**_{2m} / (di_{1}/dt)
**M = E**_{1m} / (di_{2}/dt)
If the self induced voltages of the inductances **L**_{1} and **L**_{2} are respectively **E**_{1s} and **E**_{2s} for the same rates of change of the current that produced the mutually induced voltages **E**_{1m} and **E**_{2m}, then:
**M = (E**_{2m} / E_{1s})L_{1}
**M = (E**_{1m} / E_{2s})L_{2}
Combining these two equations:
**M = (E**_{1m}E_{2m} / E_{1s}E_{2s})^{½} (L_{1}L_{2})^{½} = k_{M}(L_{1}L_{2})^{½}
where **k**_{M} is the mutual coupling coefficient of the two inductances **L**_{1} and **L**_{2}.
If the coupling between the two inductances **L**_{1} and **L**_{2} is perfect, then the mutual inductance **M** is:
**M = (L**_{1}L_{2})^{½} |