Assume that the circuit in Problem 12 is powered by a 60 Hz AC source.
Calculate the inductive reactance, X_L, as seen by the AC voltage source.
If the DC source is replaced by an AC source, the circuit would appear as follows:
L_{EQ} as seen by the AC voltage source, is shown in the simplified equivalent circuit below:
As computed in Problem 12, the combined or net inductance contributed to the circuit by the parallel and series network of inductors is L_{EQ} = 19.23 mH. Then, by applying Eq. 1.37, the inductive reactance, X_{L-EQ}, as seen by the AC voltage source V_{AC}, would be as follows:
X_L=ωL=2πfL (1.37)
X_{L-EQ}=ω.L=(2πf).L_{EQ}= 2(3.14)(60 Hz)(19.23 mH)
= 7.25 Ω