Chapter 17, Problem 1PP

Inductive Circuits

Fill in all the missing values. Refer to the following formulas:

$\begin{array}{l}{\text{X}}_{\text{L}}=2\text{πfL}\\ \text{L=}\frac{{\text{X}}_{\text{L}}}{2\text{πf}}\\ \text{f =}\frac{{\text{X}}_{\text{L}}}{2\text{πL}}\end{array}$

Inductance (H)	Frequency (Hz)	Inductive Reactance ( $\Omega$)
1.2	60
0.085		213.628
	1000	4712.389
0.65	600
3.6		678.584
	25	411.459
0.5	60
0.85		6408.849
	20	201.062
0.45	400
4.8		2412.743
	1000	40.841

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

$\begin{array}{l}{X}_L=2\pi fL\mathrm{...}\text{ (1)}\\ L=\frac{X_L}{\text{2}\pi f}\mathrm{...}\text{ (2)}\\ f=\frac{X_L}{\text{2}\pi L}\mathrm{...}\text{ (3)}\\ \end{array}$

$\begin{array}{l}\left(1\right)X_L=2\pi fL\\ =2\pi \times 60\times 1.2\\ =452.39\Omega \\ \\ \left(2\right)f=\frac{X_L}{\text{2}\pi L}\\ =\frac{213.628}{\text{2}\pi \times 0.085}\\ =400Hz\\ \\ \left(3\right)L=\frac{X_L}{\text{2}\pi f}\\ =\frac{4712.389}{\text{2}\pi \times 1000}\\ =0.75H\end{array}$

$\begin{array}{l}\left(4\right)X_L=2\pi fL\\ =2\pi \times 600\times 0.65\\ =2450.44\Omega \\ \\ \left(5\right)f=\frac{X_L}{\text{2}\pi L}\\ =\frac{678.584}{\text{2}\pi \times 3.6}\\ =30Hz\\ \\ \left(6\right)L=\frac{X_L}{\text{2}\pi f}\\ =\frac{411.459}{\text{2}\pi \times 25}\\ =2.61H\end{array}$

$\begin{array}{l}\left(7\right)X_L=2\pi fL\\ =2\pi \times 60\times 0.5\\ =188.50\Omega \\ \\ \left(8\right)f=\frac{X_L}{\text{2}\pi L}\\ =\frac{6408.849}{\text{2}\pi \times 0.85}\\ =1200Hz\\ \\ \left(9\right)L=\frac{X_L}{\text{2}\pi f}\\ =\frac{201.062}{\text{2}\pi \times 20}\\ =1.6H\end{array}$

$\begin{array}{l}\left(10\right)X_L=2\pi fL\\ =2\pi \times 400\times 0.45\\ =1130\Omega \\ \\ \left(11\right)f=\frac{X_L}{\text{2}\pi L}\\ =\frac{2412.743}{\text{2}\pi \times 4.8}\\ =80Hz\\ \\ \left(12\right)L=\frac{X_L}{\text{2}\pi f}\\ =\frac{40.481}{\text{2}\pi \times 1000}\\ =6.44mH\end{array}$