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 $R\text{'}$, then the ratio $\frac{R}{R\text{'}}$ is:

• A. $\frac{1}{25}$
• B. $\frac{1}{5}$
• C. $5$
• D. $25$

Answer 1

The correct option is D

$25$

Step 1: Given data

Resistance of wire $=R$

Resistance of combination $=R\text{'}$

Step 2: Formula used

Resistance,$R=\rho \frac{l}{A}$

$l$is the length of the wire

$A$is the area of cross-section
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When connected in parallel, the resistance is given by $\frac{1}{R\text{'}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}$, where $R_1,R_2,R_3,R_4$ and $R_5$ are resistances of the five parts.

Since $R$ is directly proportional to the length of the wire. Therefore the resistance of every part after cutting the wire into $5$ parts is $R_1=R_2=R_3=R_4=R_5=\frac{R}{5}$.

Step 3: Calculating the ratio,

All the resistances are connected in parallel, let $R\text{'}$ be the equivalent resistance.

Resistance of all $5$ resistors $R_1=R_2=R_3=R_4=R_5=\frac{R}{5}$

Substituting the values in the formula,

$R_1,R_2,R_3,R_4$ and $R_5$ are resistances of the five parts.

$\begin{array}{rcl}\frac{1}{R\text{'}}& =& \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\\ & =& \frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}\\ & =& \frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}\\ & =& \frac{5+5+5+5+5}{R}\\ & =& \frac{25}{R}\end{array}$

$R$ is the resistance of a wire.

Solving further to find the ratio,

$$\begin{gathered} \frac{1}{R\text{'}}=\frac{25}{R} \\ \Rightarrow \frac{R}{R\text{'}}=25 \end{gathered}$$

So, If the equivalent resistance of this combination is $R\text{'}$, then the ratio $\frac{R}{R\text{'}}$ is $25$.

Hence, option D is correct.