Question 16.3: An inductor with negligible resistance is connected to a 480......

The first step in solving this problem is to determine the amount of inductance of the coil. Because the resistance of the wire used to make the inductor is negligible, the current is limited by inductive reactance. The inductive reactance can be found by substituting X_{L} for R in an Ohm’s law formula:

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

X_{L}={\frac{\mathrm{480\ V}}{24\ A}}

X_{L} = 20 Ω

Now that the inductive reactance is known, the inductance of the coil can be found using the formula

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

Note: When using a frequency of 60 hertz, 2 × π× 60 = 376.992. To simplify calculations, this value is generally rounded to 377. Because 60 hertz is the major frequency used throughout the United States, 377 should be memorized because it is used in many calculations:

{\mathsf{L}}={\frac{\mathsf{20\ \Omega}}{{337\ {\mathsf{Hz}}}}}

L = 0.053 H

Because the inductance of the coil is determined by its physical construction, it does not change when connected to a different frequency. Now that the inductance of the coil is known, the inductive reactance at 400 hertz can be calculated:

X_{L} = 2πfL

X_{L} = 2 × 3.1416 × 400 Hz × 0.053

X_{L}  = 133.204 Ω

The amount of current flow can now be found by substituting the value of inductive reactance for resistance in an Ohm’s law formula:

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

{\mathsf{I}}={\frac{480\ V}{133.204\ \Omega}}

I = 3.603 A