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

☞︎︎︎ We know,

$\:$

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

$\:$

Initial resistance of wire = R Ω

$\:$

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

$\:$

Resistance of each peice of wire

$\:$

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

$\:$

☆ Now,

$\:$

We know :

$\:$

In parallel connection, Equivalent resistance is given by -

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

Answer 2

☞︎︎︎ We know,

$\:$

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

$\:$

Initial resistance of wire = R Ω

$\:$

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

$\:$

Resistance of each peice of wire

$\:$

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

$\:$

☆ Now,

$\:$

We know :

$\:$

In parallel connection, Equivalent resistance is given by -

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

Answer 3

☞︎︎︎ We know,

$\:$

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

$\:$

Initial resistance of wire = R Ω

$\:$

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

$\:$

Resistance of each peice of wire

$\:$

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

$\:$

☆ Now,

$\:$

We know :

$\:$

In parallel connection, Equivalent resistance is given by -

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$