The equivalent resistance of the network shown is: [tex]R_{eq}=16.463\Ohm[/tex]
So, first of all, we can start by determining what each of the resistances is equal to. In this case we can start by saying that [tex]R_{1}=11R_{11}[/tex] This means that:
[tex]R_{1}=11(3\Ohm)[/tex]
Therefore:
[tex]R_{1}=33\Ohm[/tex]
We can now use this value to find the value of the other resistances. The given condition for the resistances can be represented with the following formula:
[tex]R_{n}=\frac{R_{1}}{n}[/tex]
so:
[tex]R_{2}=\frac{R_{1}}{2}=\frac{33}{2}\Ohm[/tex]
[tex]R_{3}=\frac{R_{1}}{3}=\frac{33}{3}\Ohm=11 \Ohm[/tex]
[tex]R_{4}=\frac{R_{1}}{4}=\frac{33}{4}\Ohm[/tex]
[tex]R_{5}=\frac{R_{1}}{5}=\frac{33}{5}\Ohm[/tex]
[tex]R_{6}=\frac{R_{1}}{6}=\frac{33}{6}\Ohm=\frac{11}{2}\Ohm[/tex]
[tex]R_{7}=\frac{R_{1}}{7}=\frac{33}{7}\Ohm[/tex]
[tex]R_{8}=\frac{R_{1}}{8}=\frac{33}{8}\Ohm[/tex]
[tex]R_{9}=\frac{R_{1}}{9}=\frac{33}{9}\Ohm=\frac{11}{3}\Ohm[/tex]
[tex]R_{10}=\frac{R_{1}}{10}=\frac{33}{10}\Ohm[/tex]
[tex]R_{11}=3\Ohm[/tex]
So now that we know what each resistance is equal to, we can go ahead and analyze the circuit.
In this case we can see that [tex]R_{10}[/tex] and [tex]R_{11}[/tex] are parallel, so we can calculate their equivalent resistance.
[tex]R_{1011}=\frac{R_{10}R_{11}}{R_{10}+R_{11}}[/tex]
Which yields:
[tex]R_{1011}=\frac{(\frac{33}{10}\Ohm)(3\Ohm)}{\frac{33}{10}\Ohm+3\Ohm}[/tex]
[tex]R_{1011}=\frac{11}{7}\Ohm[/tex]
Now, [tex]R_{8}[/tex], [tex]R_{9}[/tex] and [tex]R_{1011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:
[tex]R_{891011}=R_{8}+R_{9}+R_{1011}[/tex]
[tex]R_{891011}=\frac{33}{8}+\frac{11}{3}+\frac{11}{7}[/tex]
[tex]R_{891011}=\frac{1573}{168}\Ohm[/tex]
Now, we can see that [tex]R_{7}[/tex] and [tex]R_{891011}[/tex] are parallel, so we can calculate their equivalent resistance.
[tex]R_{7891011}=\frac{R_{7}R_{891011}}{R_{7}+R_{891011}}[/tex]
Which yields:
[tex]R_{7891011}=\frac{(\frac{33}{7}\Ohm)(\frac{1573}{168})}{\frac{33}{7}\Ohm+\frac{1573}{168}\Ohm}[/tex]
[tex]R_{7891011}=\frac{4719}{1505}\Ohm[/tex]
Now, [tex]R_{5}[/tex], [tex]R_{6}[/tex] and [tex]R_{7891011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:
[tex]R_{567891011}=R_{5}+R_{6}+R_{7891011}[/tex]
[tex]R_{567891011}=\frac{33}{5}\Ohm+\frac{11}{2}\Ohm+\frac{4719}{1505}\Ohm[/tex]
[tex]R_{567891011}=15.236\Ohm[/tex]
Next, we can see that [tex]R_{4}[/tex] and [tex]R_{567891011}[/tex] are parallel, so we can calculate their equivalent resistance.
[tex]R_{4567891011}=\frac{R_{4}R_{567891011}}{R_{4}+R_{567891011}}[/tex]
Which yields:
[tex]R_{4567891011}=\frac{(\frac{33}{4}\Ohm)(15.236\Ohm)}{\frac{33}{4}\Ohm+15.236\Ohm}[/tex]
[tex]R_{4567891011}=5.352\Ohm[/tex]
Now, [tex]R_{2}[/tex], [tex]R_{3}[/tex] and [tex]R_{4567891011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:
[tex]R_{234567891011}=R_{2}+R_{3}+R_{4567891011}[/tex]
[tex]R_{234567891011}=\frac{33}{2}\Ohm+11\Ohm+5.352\Ohm[/tex]
[tex]R_{234567891011}=32.852\Ohm[/tex]
Finally, we can see that [tex]R_{1}[/tex] and [tex]R_{234567891011}[/tex] are parallel, so we can calculate their equivalent resistance.
[tex]R_{eq}=\frac{R_{1}R_{234567891011}}{R_{1}+R_{234567891011}}[/tex]
Which yields:
[tex]R_{eq}=\frac{(33\Ohm)(32.852\Ohm)}{33\Ohm+32.852\Ohm}[/tex]
[tex]R_{eq}=16.463\Ohm[/tex]
See attached picture for images on how to reduce the circuit.
You can find further information in the following link.
https://brainly.com/question/21538325?referrer=searchResults
