Question 16.1: The inductor shown in Figure 16–8 has an inductance of 0.8 H......

The first step is to determine the amount of inductive reactance of the inductor:

X_{L} = 2πfL

X_{L} =  2 × 3.1416 × 60 Hz × 0.8

X_{L} = 301.594 Ω

Because inductive reactance is the current-limiting property of this circuit, it can be substituted for the value of R in an Ohm’s law formula:

I = {\frac{\operatorname{E}}{X_{L}}}

I = {\frac{120\,\mathrm{V}}{301.594\,\Omega}}

I = 0.398 A

If the amount of inductive reactance is known, the inductance of the coil can be determined using the formula

L = {\frac{X_{L}}{2\pi f }}