LEARNATHON

III

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

Learn more### 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

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

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

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

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 $1.5\times {10}^{-6}\phantom{\rule{0.147em}{0ex}}\mathrm{\Omega}m$. Hence, this conducting material is used in making heating elements.*Nichrome coil*

Reference:

https://upload.wikimedia.org/wikipedia/commons/5/5a/2_resistors_in_series.png