Equivalent resistance between A and B is (up to two decimal places)

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Solution

$\in= (i_{1} + i - 2)$ Req -(1)
$\in= (i_{1} + i_{5} + i - 2) - (2)$
$i_{1} + i_{3} = i_{2} - (3) & i_{3} + i_{4} = i_{5} - (4)$
$i_{1} + i_{5} = i_{2} + i_{n} - (5), i_{1} = i_{3} + i_{5} - (6), i_{5} + i_{n} = i_{2} + i_{3} - (7)$
Solving equls we get $2 i_{1} = i_{2} + 4 i_{3} - (a)$
and $6 i_{1} = 5 i_{2}, i_{5} = \frac{4}{5} i_{1} & i_{3} = \frac{i_{1}}{5}$
$\in= i_{1} + \frac{4}{5} i_{1} + \frac{6}{5} i_{1} \Rightarrow\in= \frac{15}{5} i_{1} - (x)$
By $(1) =\in= (i_{1} + \frac{6}{5} i_{1})$ Req $\Rightarrow\in= \frac{11}{5} i_{1}$ Req -(y)
By $(x) = (y) \frac{15}{5} i_{1} = \frac{11}{5} i_{1}$ Req = Req $= \frac{15}{11}$

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