LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

### Theory:

Length of a conductor:
The conductor's resistance ($$R$$) is directly proportional to its length ($$L$$).

Circuit with resistors

Consider the length of two conductors to be $$L_{1}$$ and $$L_{2}$$, having a resistance of $$R_{1}$$ and $$R_{2}$$, respectively. Assume that the conductors are made of the same material. Then,

$\begin{array}{l}{R}_{1}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}{L}_{1}\\ \\ {R}_{2}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}{L}_{2}\\ \\ \frac{{R}_{1}}{{R}_{2}}=\frac{{L}_{1}}{{L}_{2}}\end{array}$

Area of a cross-section of the conductor:
The conductor's resistance ($$R$$) is inversely proportional to its area of cross-section ($$A$$).

Let the cross-sectional area of two conductors be $$A_{1}$$ and $$A_{2}$$, having a resistance of $$R_{1}$$ and $$R_{2}$$, respectively.

$\begin{array}{l}{R}_{1}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\frac{1}{{A}_{1}}\\ \\ {R}_{2}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\frac{1}{{A}_{2}}\\ \\ \frac{{R}_{1}}{{R}_{2}}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\frac{{A}_{2}}{{A}_{1}}\end{array}$

Since the area of cross-section of a conductor is $A=\mathrm{\pi }{r}^{2}$, where $$r$$ is the radius of the conductor.

Here, $$r_{1}$$ and $$r_{2}$$ be the radii of two conductors. On substituting this value in the above equation, we get

$\begin{array}{l}\frac{{R}_{1}}{{R}_{2}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{A}_{2}}{{A}_{1}}\\ \\ \frac{{R}_{1}}{{R}_{2}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathrm{\pi }{r}_{2}^{2}}{\mathrm{\pi }{r}_{1}^{2}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}\end{array}$

On combining both the equations of the second and third factors, we get

$\begin{array}{l}R\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\frac{L}{A}\\ \\ R\phantom{\rule{0.147em}{0ex}}=\left(\mathit{constant}\right)\phantom{\rule{0.147em}{0ex}}\frac{L}{A}\end{array}$

$R\phantom{\rule{0.147em}{0ex}}=\mathrm{\rho }\phantom{\rule{0.147em}{0ex}}\frac{L}{A}$ ---- (eq. 1)

The constant of proportionality is given as $$\rho$$ in the above equation. It is known as an electrical resistivity or specific resistance of the material of the conductor.

The (eq. 1) can be written as,

$\mathrm{\rho }=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{RA}}{L}$

If $$L$$ $$=$$ $$1\ m$$, $$A$$ $$=$$ $$1\ m^2$$ then, the above equation becomes $$\rho$$ $$=$$ $$R$$.
The electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross-section.
Unit:
The SI unit of resistivity is ohm metre or $\mathrm{\Omega }m$.

$\begin{array}{l}\mathit{Unit}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathrm{\rho }=\phantom{\rule{0.147em}{0ex}}\frac{R×A}{L}\\ \\ =\phantom{\rule{0.147em}{0ex}}\frac{\mathit{ohm}×{\mathit{metre}}^{2}}{\mathit{metre}}\\ =\mathit{ohm}\phantom{\rule{0.147em}{0ex}}\mathit{metre}\phantom{\rule{0.147em}{0ex}}\left(\mathrm{\Omega }m\right)\end{array}$

A conductor's electrical resistivity is a measurement of its resistance to the passage of an electric current. The electrical resistivity of a material is always constant and independent of its size and shape.

Important!
Nichrome has the highest electrical resistivity, equal to $1.5×{10}^{-6}\phantom{\rule{0.147em}{0ex}}\mathrm{\Omega }m$. Hence, this conducting material is used in making heating elements.
Nichrome coil
Reference: