Question

a piece of wire of resustance R is cut into five equal parts. These parts are then connected in parallel. if the equivalent resistance of this combination is R', then what is the ratio R/R'?​

Answer 1

$\overline{\underline{\huge{\bf{\pink{Question}}}}}$

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel . if the equivalent resistance of this combination is R' , then what is the ratio  R / R' ?

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The equivalent resistance of 5 resistances ( R₁ , R₂ , R₃ , R₄ , R₅ ) connected in parallel will be given by

$\boxed{\tt{\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}R_2}+\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}}}$

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Let ,

initial length of wire = 5 L

initial area of cross section = A

then,

initial Resistance = R   will be

$\boxed{\purple{\sf{R = \rho \times \frac{5 \:L}{A}}}}$

Now,

∵ Wire is divided in 5 equal parts

Therefore,

new length of each wire piece = L

area of cross section of each wire piece = A

( area of cross section will remain unchanged)

so,

new resistance of each part  will be

$\sf{new\:Resistance\;of\:each\;part=\rho\times\frac{L}{A}}$

5 pieces were connected in parallel connection

so, Equivalent resistance of parallel combination = R' will be

$\sf{\frac{1}{R'}=5\times\frac{1}{ (new\:resistance\:of\:each\:part)}}\\\\\sf{\frac{1}{R'}=5\times\frac{1}{\rho\times\frac{L}{A}}}$

$\boxed{\sf{\purple{R'=\rho\times\frac{L}{5A}}}}$

▶ Now finding the ratio R / R'

$\large{\sf{\frac{R}{R'}=\frac{\rho\times\frac{5L}{A}}{\rho\times\frac{L}{5A}}}}$

$\boxed{\boxed{\large{\sf{\frac{R}{R'}=\frac{25}{1}}}}}$

Hence ,the ratio R : R' = 25 : 1