Chapter 20, Problem 12TP

To determine

Ratio of the resistivities of the given table.

Answer to Problem 12TP

The required ratio of the resistivity will be ${\text{ρ}}_{\text{1}}{\text{:ρ}}_{\text{2}}{\text{:ρ}}_{\text{3}}=1:2:0.5$.

Explanation of Solution

Given:

Wire	Length	Diameter
Wire $\text{1}$	$\text{2m}$	$\text{1cm}$
Wire $2$	$\text{1m}$	$\text{0}\text{.5 cm}$
Wire $3$	$\text{1m}$	$\text{1cm}$

Formula used:

Resistance in the wire

  $\text{R=}\text{ρ}\frac{\text{l}}{\text{A}}$

Where,

  $\text{R}$ = resistance of the conductor.

  $\text{l}$ = length of the conductor.

  $\text{A}$ = cross sectional area of the conductor.

Calculation:

We know that,

  $\text{1cm = }\frac{\text{1}}{\text{100}}\text{m}$

Resistance in the wire

  $\text{R=}\text{ρ}\frac{\text{l}}{\text{A}}$

Where,

  $\text{R}$ = resistance of the conductor.

  $\text{l}$ = length of the conductor.

  $\text{A}$ = cross sectional area of the conductor.

Now, if the resistance of two or more conductor is same then the required ratio of the resistivity is as follows:

  $\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}\text{=}\frac{{\text{L}}_{\text{1}}{\text{ρ}}_{\text{1}}{\text{A}}_{\text{2}}}{{\text{L}}_{\text{2}}{\text{ρ}}_{\text{2}}{\text{A}}_1}$

Now, putting ${\text{R}}_{\text{1}}$ and ${\text{R}}_{\text{2}}$ as $\text{R}$.

Therefore, the above equation becomes,

  $\begin{array}{l}\frac{\text{R}}{\text{R}}\text{=}\frac{{\text{L}}_{\text{1}}{\text{ρ}}_{\text{1}}{\text{A}}_{\text{1}}}{{\text{L}}_{\text{2}}{\text{ρ}}_{\text{2}}{\text{A}}_{\text{2}}}\\ \text{1=}\frac{{\text{L}}_{\text{1}}{\text{ρ}}_{\text{1}}{\text{A}}_{\text{1}}}{{\text{L}}_{\text{2}}{\text{ρ}}_{\text{2}}{\text{A}}_{\text{2}}}\\ {\text{L}}_{\text{1}}{\text{ρ}}_{\text{1}}{\text{A}}_{\text{1}}={\text{L}}_{\text{2}}{\text{ρ}}_{\text{2}}{\text{A}}_{\text{2}}\end{array}$

Substitute, ${\text{A}}_{\text{1}\text{}}\text{=}\text{}\text{π}\frac{{\text{D}}_1{}^{\text{2}}}{\text{4}}$ and ${\text{A}}_{2\text{}}\text{=}\text{}\text{π}\frac{{\text{D}}_2{}^{\text{2}}}{\text{4}}$

  $\begin{array}{l}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\frac{\text{π}\frac{{\text{D}}_{\text{1}}{}^{\text{2}}}{\text{4}}{\text{L}}_{\text{2}}}{\text{π}\frac{{\text{D}}_{\text{2}}{}^{\text{2}}}{\text{4}}{\text{L}}_{\text{1}}}\\ \frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\frac{{\text{D}}_{\text{1}}{\text{L}}_{\text{2}}}{{\text{D}}_{\text{2}}{\text{L}}_{\text{1}}}\end{array}$

Put, $\text{2m}$ for ${\text{L}}_{\text{1}}$, $\text{1m}$ for ${\text{L}}_{\text{2}}$, and $\text{1cm}$ for ${\text{D}}_{\text{1}}$

  $\text{0}\text{.5}\text{cm}$ for ${\text{D}}_{\text{2}}$ in the above equation,

  $\begin{array}{l}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\frac{\text{1m}{\left(\text{1cm}\left(\frac{\text{1m}}{\text{100cm}}\right)\right)}^{\text{2}}}{\text{2m}{\left(\text{0}\text{.5cm}\left(\frac{\text{1m}}{\text{100cm}}\right)\right)}^{\text{2}}}\\ \text{=}\text{2}\end{array}$

Now, Again put, $\text{2m}$ for ${\text{L}}_{\text{1}}$, $\text{1m}$ for ${\text{L}}_{\text{2}}$, and $\text{1cm}$ for ${\text{D}}_{\text{1}}$

  $\text{1}\text{cm}$

  ${\text{D}}_{\text{2}}$ in the equation,

  $\begin{array}{l}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\frac{\text{1m}{\left(\text{1cm}\left(\frac{\text{1m}}{\text{100cm}}\right)\right)}^{\text{2}}}{\text{2m}{\left(\text{1cm}\left(\frac{\text{1m}}{\text{100cm}}\right)\right)}^{\text{2}}}\\ \text{=}\text{0}\text{.5}\end{array}$

The ratio of the resistivity will be ${\text{ρ}}_{\text{1}}{\text{:ρ}}_{\text{2}}{\text{:ρ}}_{\text{3}}=1:2:0.5$.

Conclusion:

Thus, by using the equation of the resistance and rearranging, required ratio is obtained.