0
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ Looks like it's $1/(f\cdot10^{-3})$, not $10^{-3}/f$. $\endgroup$
    – Kyle Kanos
    Commented Jun 8, 2017 at 21:47
  • $\begingroup$ Also, use a ruler $\endgroup$
    – innisfree
    Commented Jun 8, 2017 at 22:37
  • $\begingroup$ No actually, in my book it is 10^-3/f, as written in the figure. @KyleKanos $\endgroup$
    – Asmaa
    Commented Jun 9, 2017 at 1:15
  • $\begingroup$ @innisfree, and then what? $\endgroup$
    – Asmaa
    Commented Jun 9, 2017 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
    Commented Jun 9, 2017 at 2:16

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.