Question

find resistance between A and B​

Answer 1

Answer:

$R_{eq}=\frac{8r}{19}$(look at attachment)

Explanation:

we know that,

for parallel connection;

$\bold{\boxed{\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}}}$

for series connection;

$\bold{\boxed{R_{eq}=R_{1}+R_{2}}}$

And,

(DE) and (EC) are in series.so,

$R_{eq1}=r+r=2r$

And ($R_{eq1}$) is parallel to (DC).So,

$\frac{1}{R_{eq2}}=\frac{1}{r}+\frac{1}{R_{eq1}}$

$\frac{1}{R_{eq2}}=\frac{2}{2r}+\frac{1}{2r}$

$R_{eq2}=\frac{2r}{3}$

And,(AD) (DC) (CB) are in series.So,

$R_{eq3}=r+R_{eq2}+r$

$R_{eq3}=r+\frac{2r}{3}+r$

$R_{eq3}=2r+\frac{2r}{3}$

$R_{eq3}=\frac{8r}{3}$

And also,$R_{eq3}$ is parallel to (AE)&(EB)

So,

$\frac{1}{R_{eq4}}=\frac{1}{r}+\frac{1}{r}+\frac{1}{R_{eq3}}$

$\frac{1}{R_{eq4}}=\frac{2}{r}+\frac{3}{8r}$

$\frac{1}{R_{eq4}}=\frac{16+3}{8r}$

$R_{eq4}=\frac{8r}{19}$

HOPE THAT HELPS!!