Question

If n equal resistances are first connected in series and then connected in parallel, the ratio of the maximum to the minimum resistance is ​

Answer 1

Equivalent resistance in series (maximum) :-

$: \implies \tt R_{eq(s)} = R_{1} + R_{2}.... + R_{n}$

$: \implies \tt R_{eq(s)} = nR \: \: \: \: \: \: ....(1)$

Equivalent resistance in parallel (minimum) :-

$: \implies \tt \dfrac{1}{R_{eq(p)} }= \dfrac{1}{R_{1} }+ \dfrac{1}{R_{2}}.... + \dfrac{1}{R_{n}}$

$: \implies \tt \dfrac{1}{R_{eq(p)}} = \dfrac{n}{R}$

$: \implies \tt R_{eq(p)} = \dfrac{R}{n}\: \: \: \: \: \: ....(2)$

Eq(1)/Eq(2)

$: \implies \tt \dfrac{R_{eq(s)}}{R_{eq(p)}}= \dfrac{nR}{ \dfrac{R}{n} }$

$: \implies \tt \dfrac{R_{eq(s)}}{R_{eq(p)}}= {n}^{2}$