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Resistivity

Electricity I - Concepts

Resistance Depends on Length

Resistance Depends on Area of Cross Section

Resistivity

Class - 9th Aus Edu Subjects

Concept Explanation

Resistivity

Resistivity of a conductor is defined as the resistance of a conductor of unit length and unit area of cross-section. Its SI unit is ohm-metre (Q-m). The resistivity of a material does not depend on its length or thickness but depends on the nature of the substance and temperature. It is a characteristic property of the material of the conductor and varies only, if its temperature changes

$large dpi{100} large rho =frac{RA}{L}$

$large therefore$ The SI unit of resistivity

$large rho =frac{ohmtimes m^{2}}{m}=ohmtimes meter(Omega m)$

Resistance Depends on Temperature:

Thus, the resistance of a pure metallic wire is directly proportional to its absolute temperature. The resistivity of alloys also increases with rise in temperature, but this increase in much smaller compared to pure metals. On decreasing temperature of some metals, the resistance becomes zero at a certain temperature.

For example : Resistance of mercury becomes zero at 4.2 K temperature.

There are certain alloys such as manganin, constantan, nichrome, etc.., whose resistivity is little affected by temperature, that is, their temperature coefficient of resistance is negligible. On account of their high resistivity and negligible temperature coefficient of resistance, these alloys are used to make wires for standard resistance, resistance boxes, etc.

Illustration: A wire of given material having length l and area of cross-section A has a resistance of $10;Omega$ . What would be the resistance of another wire of the same material having length $l/4$ and area of cross-section 2.5 A?

Solution: For first wire, length = l, area of cross-section = A

and resistance, $R_{1}=10;Omega ,$

i.e., $large R_{1}=frac{rho l}{A}=10;Omega$

$large rho =frac{10A}{l}$ ....(i)

For second wire, length = $large l/4$ , area of cross-section = 2.5 A

and resistance

, $large R_2=rho frac{l/4}{2.5A}$

Putting the value from Equation I

$large R_2=frac{10A}{l};X;frac{l}{4times2.5A}= 1;Omega$

So, the resistance of that wire is $large 1Omega$ .

Sample Questions

(More Questions for each concept available in Login)

Question : 1

The resistivity does not change if
A. The material is changed		B. The temperature is changed
C. The shape of the resistor is changed		D. Both matenal and temperature are changed
Right Option : C
View Explanation

Question : 2

The resistance of a wire of length 100 cm and of uniform area of cross section $0.030:cm^2$ , is found to be $2.0Omega$ .Calculate the specific resistance of wire.
A. 0.0006 $Omega:cm$	B. 0.0005 $:Omega:cm$	C. 0.0004 $:Omega:cm$	D. 0.0003 $:Omega:cm$
Right Option : A
View Explanation

Question : 3

An electrician wishes to cut a copper wire( $rho;=;$ $1.8:X:10^-^8$ $Omega:m$ that has no more than $11:Omega$ of resistance .The wire has a radius of 0.725 mm. What is the length of wire ?
A. 380 m	B. 960 m	C. $10.5:X: 10^2 m$	D. 2.6 cm
Right Option : C
View Explanation