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In the figure below, my book shows how to plot the graphical representation between the capacitive reactance ($X_C$) on the vertical axis, and ($1/f$) on the horizontal axis. It gives the values in the first table in the figure ($f$ is the frequency of the current in the circuit).

$$ \text{Table 1} \\ \begin{array}{l|ccccc} \hline X_C & 1000 & 500 & 250 & 200 & 125 \\ \hline f & 10 & 20 & 40 & 50 & 80 \\ \hline \end{array} $$

$$ \text{Table 2} \\ \begin{array}{l|ccccc} \hline X_C~(\Omega) & 1000 & 500 & 250 & 200 & 125 \\ \hline \frac{1}{f} \cdot 10^{-3}~(\mathrm{Hz^{-1}}) & 100 & 50 & 25 & 20 & 12.5 \\ \hline \end{array} $$

The book starts solving the problem by making the second table and the values in it. I don't understand how we got the values in table two regarding ($1/f$). Let's take for example the first value of frequency in table $1$, i.e. $10\,\rm Hz$, in the second table it will become $100$, but when we do the maths, it will go like this:

$$\frac{1}{10} = 0.1 \quad \Rightarrow \quad 0.1 \cdot 10^{-3} = 1 \cdot 10^{-4}$$

and that's not $100 \cdot 10^{-3}$ as the second table shows.

How it converted the values in Table 1 to the values in Table 2?

Note: I am just a beginner, so excuse me if the question is a silly one, and I would appreciate a simple explanation.

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    $\begingroup$ Looks like it's $1/(f\cdot10^{-3})$, not $10^{-3}/f$. $\endgroup$ – Kyle Kanos Jun 8 '17 at 21:47
  • $\begingroup$ Also, use a ruler $\endgroup$ – innisfree Jun 8 '17 at 22:37
  • $\begingroup$ No actually, in my book it is 10^-3/f, as written in the figure. @KyleKanos $\endgroup$ – Asmaa Jun 9 '17 at 1:15
  • $\begingroup$ @innisfree, and then what? $\endgroup$ – Asmaa Jun 9 '17 at 1:16
  • $\begingroup$ Urm use it draw straight lines... @KyleKanos is saying that the numbers in the tables are consistent if there's a typo in the first col, second row of table 2 $\endgroup$ – innisfree Jun 9 '17 at 2:16

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