Question

In the given network, find the equivalent resistance between A and B.

Answer 1

The equivalent circuit of $IMAGE01$ is given in $IMAGE02$

From left side,

$R_1$ and $R_2$ are in series

$\therefore R_{eq1}=R_1+R_2=5+5=10\Omega$

Now, $R_{eq1}$ and $R_3$ are in parallel

$\therefore R_{eq2}=R_{eq1}\parallel R_3=\frac{R_{eq1}\times R_3}{R_{eq1}+R_3}=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$

Again,

$R_{eq2}$ and $R_4$ are in series

$\therefore R_{eq3}=R_{eq2}+R_4=5+5=10\Omega$

Now, $R_{eq3}$ and $R_5$ are in parallel

$\therefore R_{eq4}=R_{eq3}\parallel R_5=\frac{R_{eq3}\times R_5}{R_{eq3}+R_5}=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$

and $R_{eq4}$ and $R_6$are in series

$\therefore R_{eq5}=R_{eq4}+R_6=5+5=10\Omega$

and $R_{eq5}$ and $R_7$ are in parallel

$\therefore R_{eq6}=R_{eq5}\parallel R_7=\frac{R_{eq5}\times R_7}{R_{eq5}+R_7}=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$

and $R_{eq6}$ and $R_8$ are in series

$\therefore R_{eq7}=R_{eq6}+R_8=5+5=10\Omega$

and finally $R_{eq7}$ and $R_9$ are in parallel

$\therefore R_{eq}=R_{eq7}\parallel R_9=\frac{R_{eq7}\times R_9}{R_{eq7}+R_9}=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$

Therefore, the equivalent resistance between $A$ and $B$ is $5\Omega$