Air Core Transformer

As the name implies, the air-core transformer does not have a ferromagnetic core to link the primary and secondary coils. Rather, the coils are placed sufficiently close to have a mutual inductance that will establish the desired transformer action. In Fig. 1, current direction and polarities have been defined for the air-core transformer. Note the presence of a mutual inductance term M, which will be positive in this case, as determined by the dot convention. From past analysis in this chapter, we now know that for the primary circuit.
We found in Chapter 11 that for the pure inductor, with no mutual inductance present, the mathematical relationship resulted in the following useful form of the voltage across an inductor: Similarly, it can be shown, for a mutual inductance, that will result in Equation (1) can then be written (using phasor notation) as and or For the secondary circuit, and or Substituting into Eq. (6) results in Solving for $\mathbf{I}_{s}, we have and, substituting into Eq. (4), we obtain Thus, the input impedance is or, defining $\mathbf{Z}_{p}=R_{p}+j X_{L_{p}}$, $\mathbf{Z}_{s}=R_{s}+j X_{L_{s}}$ and $X_{m} \angle 90^{\circ}=+j \omega M$ we have and The term $(\omega M)^{2}(\mathbf{Z}_{s}+\mathbf{Z}_{L}) $ is called the coupled impedance, and it is independent of the sign of $M$ since it is squared in the equation. Consider also that since $(\omega M)^{2}$is a constant with $ 0^{\circ} $ phase angle, if the load $ \mathbf{Z}_{L} $ is resistive, the resulting coupled impedance term will appear capacitive due to division of $ \left(\mathbf{Z}_{s}+R_{L}\right) $ into $ (\omega M)^{2} $. This resulting capacitive reactance will oppose the series primary inductance $L_{p}$, causing a reduction in $ \mathbf{Z}_{i} $. Including the effect of the mutual term, the input impedance to the network will appear as shown in Fig. 2.