Question

Calculate the equivalent resistance Req of the network shown in Fig. 3.87 if R1 = 2R2 = 3R3 = 4R4 etc. and R11 = 3

Answer 1

The equivalent resistance of the network shown is: $R_{eq}=16.463\Ohm$

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 $R_{1}=11R_{11}$ This means that:

$R_{1}=11(3\Ohm)$

Therefore:

$R_{1}=33\Ohm$

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:

$R_{n}=\frac{R_{1}}{n}$

so:

$R_{2}=\frac{R_{1}}{2}=\frac{33}{2}\Ohm$

$R_{3}=\frac{R_{1}}{3}=\frac{33}{3}\Ohm=11 \Ohm$

$R_{4}=\frac{R_{1}}{4}=\frac{33}{4}\Ohm$

$R_{5}=\frac{R_{1}}{5}=\frac{33}{5}\Ohm$

$R_{6}=\frac{R_{1}}{6}=\frac{33}{6}\Ohm=\frac{11}{2}\Ohm$

$R_{7}=\frac{R_{1}}{7}=\frac{33}{7}\Ohm$

$R_{8}=\frac{R_{1}}{8}=\frac{33}{8}\Ohm$

$R_{9}=\frac{R_{1}}{9}=\frac{33}{9}\Ohm=\frac{11}{3}\Ohm$

$R_{10}=\frac{R_{1}}{10}=\frac{33}{10}\Ohm$

$R_{11}=3\Ohm$

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 $R_{10}$ and $R_{11}$ are parallel, so we can calculate their equivalent resistance.

$R_{1011}=\frac{R_{10}R_{11}}{R_{10}+R_{11}}$

Which yields:

$R_{1011}=\frac{(\frac{33}{10}\Ohm)(3\Ohm)}{\frac{33}{10}\Ohm+3\Ohm}$

$R_{1011}=\frac{11}{7}\Ohm$

Now, $R_{8}$, $R_{9}$ and $R_{1011}$ are connected in series, so we can calculate their equivalent resistance like this:

$R_{891011}=R_{8}+R_{9}+R_{1011}$

$R_{891011}=\frac{33}{8}+\frac{11}{3}+\frac{11}{7}$

$R_{891011}=\frac{1573}{168}\Ohm$

Now,  we can see that $R_{7}$ and $R_{891011}$ are parallel, so we can calculate their equivalent resistance.

$R_{7891011}=\frac{R_{7}R_{891011}}{R_{7}+R_{891011}}$

Which yields:

$R_{7891011}=\frac{(\frac{33}{7}\Ohm)(\frac{1573}{168})}{\frac{33}{7}\Ohm+\frac{1573}{168}\Ohm}$

$R_{7891011}=\frac{4719}{1505}\Ohm$

Now, $R_{5}$, $R_{6}$ and $R_{7891011}$ are connected in series, so we can calculate their equivalent resistance like this:

$R_{567891011}=R_{5}+R_{6}+R_{7891011}$

$R_{567891011}=\frac{33}{5}\Ohm+\frac{11}{2}\Ohm+\frac{4719}{1505}\Ohm$

$R_{567891011}=15.236\Ohm$

Next,  we can see that $R_{4}$ and $R_{567891011}$ are parallel, so we can calculate their equivalent resistance.

$R_{4567891011}=\frac{R_{4}R_{567891011}}{R_{4}+R_{567891011}}$

Which yields:

$R_{4567891011}=\frac{(\frac{33}{4}\Ohm)(15.236\Ohm)}{\frac{33}{4}\Ohm+15.236\Ohm}$

$R_{4567891011}=5.352\Ohm$

Now, $R_{2}$, $R_{3}$ and $R_{4567891011}$ are connected in series, so we can calculate their equivalent resistance like this:

$R_{234567891011}=R_{2}+R_{3}+R_{4567891011}$

$R_{234567891011}=\frac{33}{2}\Ohm+11\Ohm+5.352\Ohm$

$R_{234567891011}=32.852\Ohm$

Finally, we can see that $R_{1}$ and $R_{234567891011}$ are parallel, so we can calculate their equivalent resistance.

$R_{eq}=\frac{R_{1}R_{234567891011}}{R_{1}+R_{234567891011}}$

Which yields:

$R_{eq}=\frac{(33\Ohm)(32.852\Ohm)}{33\Ohm+32.852\Ohm}$

$R_{eq}=16.463\Ohm$

See attached picture for images on how to reduce the circuit.

You can find further information in the following link.

brainly.com/question/21538325?referrer=searchResults