Question

A piece of wire of resistance R is cut into 5 equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is Rp, then the ratio of R/Rp is *​

Answer 1

Explanation:

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Answer 2

☞︎︎︎ We know,

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$\: \: \: \: \: \: \: \mapsto \large \rm \: r∝l$

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Initial resistance of wire = R Ω

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$\large \rm \: 5 \: \: pieces \: \: of \: \: wire = \dfrac{1}{5} \: th$

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Resistance of each peice of wire

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$\large \rm \: \: \: \: \: \: \: \: = \dfrac{r}{5} \: Ω$

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☆ Now,

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We know :

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• In parallel connection, Equivalent resistance is given by -

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$\: \: \large \rm \: \dfrac{1}{r'} = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} + \dfrac{1}{r_4} + \dfrac{1}{r_5}$

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$\: \: \large \rm \: \dfrac{1}{r'} = \dfrac{1}{r/5} + \dfrac{1}{r/5} + \dfrac{1}{r/5} + \dfrac{1}{r/5} + \dfrac{1}{r/5}$

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$\: \: \large \rm \: \dfrac{1}{r'} = \dfrac{5}{r} + \dfrac{5}{r} + \dfrac{5}{r} + \dfrac{5}{r} + \dfrac{5}{r}$

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$\: \: \large \rm \: \dfrac{1}{r'} = \dfrac{25}{r}$

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$\: \: \large \rm \: {r'} = \dfrac{r}{25}$

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$\: \: \: \: \: \: \: \mapsto \: \: \large \rm finding : \: \dfrac{r}{r'}$

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$: \implies \: \large \rm \: \dfrac{r}{r'} = \dfrac{r}{r/25}$

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$: \implies \: \large \rm \: \dfrac{r}{r'} = \dfrac{25r}{r}$

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$\therefore \boxed{ \: \large \rm \: \dfrac{r}{r'} = {25} }$

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