Question

What the mutual inductance of a ideal transformer?

Answer 1

Mutual Inductance

$$V _ {1} = L _ {1} \frac {d I _ {1}}{d t} \pm L _ {m} \frac {d I _ {2}}{d t}$$ $$V _ {2} = \pm L _ {m} \frac {d I _ {1}}{d t} + L _ {m} \frac {d I _ {2}}{d t}$$

- Correct sign for mutual inductance found from Lenz' law and dot convention

- Dot convention: current flowing into one dot will induce current flow out of second dot

Transformers

- A transformer is just a special case where the mutual inductance is made as large as possible by allowing both coils to share the same flux

- This is usually achieved by winding them both on a common core of high permeability material (soft iron or ferrite materials)

$$V _ {1} = j \omega L _ {1} I _ {1} + j \omega L _ {m} I _ {2}$$ $$V _ {2} = j \omega L _ {m} I _ {1} + j \omega L _ {2} I _ {2}$$

When there is no flux leakage, the mutual inductance is related to the primary and secondary inductances as

$$L _ {m} = \sqrt {L _ {1} L _ {2}}$$

For real transformers this can never be quite achieved, so we write

$$L _ {m} = k \sqrt {L _ {1} L _ {2}}$$

where $0 < k < 1$ - coefficient of coupling

Ideal Transformer

If both coils share the same flux, then Farady's law gives:

$$\frac {V _ {1}}{V _ {2}} = \frac {N _ {1}}{N _ {2}} = \frac {1}{n}$$

As the permeability of the core increases, the relationship between the primary and secondary currents approaches a limiting value set by the turns ratio:

$$\frac {I _ {1}}{I _ {2}} \Rightarrow \frac {N _ {2}}{N _ {1}} = n$$

These two relationships define an ideal transformer. This is a fictitious element (note that $\mu \rightarrow \infty$ implies infinite inductances so the impedance matrix is infinite) but a real transformer approximates this behavior.

An idea transformer has the following useful property when one winding is terminated:

$$Z _ {m} = \frac {V _ {1}}{I _ {1}} = \frac {\frac {N _ {1}}{N _ {2}} V _ {2}}{\frac {N _ {2}}{N _ {1}}} = \frac {Z _ {L}}{n ^ {2}}$$