Unit of Specific Resistance Explained

Resistivity or specific resistance of a material is a measure of the resistance, which it offers to the flow of current through it. It is an intrinsic property of a material. Specific resistance depends on the composition, temperature, pressure of the material. The reciprocal of resistivity is defined as specific conductance which amounts to the ability to conduct electricity. The value of resistivity is very low for conductors and very high for insulators. The resistivity of a material is a scalar quantity.  Like any other physical quantity, the description of specific resistance requires a number (magnitude), associated with a unit. 

Definition of Resistivity

The resistance R of a conductor depends on its length L, cross-section A and its composition. For a fixed cross-section, the resistance is proportional to the length of the conductor. Whereas the resistance is inversely proportional to the cross-section for a fixed length. These two dependencies can be written down combinedly as,

R \[\propto\] \[\frac{L}{A}\]

R =  \[\frac{\rho L}{A}\]

Here, is a proportionality constant which is known as the specific resistance. The constant has different values for different materials. Specific resistance depends on the physical properties of the material e.g. density and composition. For unit length and unit cross-sectional area i.e.  L=1 and A=1,

\[\rho\] = R

This condition can be used to define resistivity.

(Image will be uploaded soon)

Resistivity

Definition:  The resistance of a homogeneous chunk of a material of unit length and unit cross-section is defined as the resistivity or specific resistance of the material. Quantitatively,

\[\rho\] =  \[\frac{RA}{L}\]

The conductivity of material is defined as the inverse of resistivity,

\[\sigma\] =  \[\frac{1}{\rho}\]

Unit of Resistivity

From the specific resistance formula,

Unit of \[\rho\] = \[\frac{Unit \; of\; R\; \times \; Unit \; of \; A}{Unit \;of \;L}\]

One useful unit is obtained when resistance R is expressed in Ohm (Ω) and distances are expressed in centimeters (cm). In this convention, the unit of resistivity is Ohmcm (Ω . cm) .

If the distances are expressed in meter (m), the SI unit of specific resistance is given by Ohmm (Ω . m) . The SI unit of conductivity is Ohm-1 m-1 or siemens m-1 (S . m-1).

Metals are good conductors of electricity whereas insulators can carry little to zero current. The specific resistance of metals is very low ( \[\sim 10^{-8}\]Ω . m )but for typical insulators, the values are very large ( \[\sim 10^{16}\]Ω . m ). The values of resistivity at 200C  for some standard materials are listed below.

The Specific Resistance of Various Materials

Resistivity of Copper

The specific resistance of copper is 1.68 x 10-8 Ω. m (200C) i.e. the resistance between two opposite surfaces of a copper cube of side 1 m is 1.68 x 10-8 Ω at temperature 200C. The conductivity of copper is5.96 x 107 Sm-1 . Due to its very low resistivity and high conductivity, the resistance of copper to the flow of current is negligible. In electrical circuits, copper wires are widely used to conduct electricity.

Solved Examples

l. Find the length and cross-sectional area of a copper wire made from a chunk of copper having a mass of 10 g, if the resistance of the copper wire is 2 Ω. The density and resistivity of copper wire are 9 g/cm3 and  1.8 x 10-6 Ω . cm respectively.

Solution: The density of copper is d = 9 g/cm3  and the mass of the chunk is m = 10 g such that the volume of the copper wire is 

V = \[\frac{m}{d}\]

V = \[\frac{10}{9}\] cm3

If the length of the wire is L and the cross-section is A, the volume is given by V=LA. Therefore,

LA = \[\frac{10}{9}\] cm3         (1)

Substituting R = 2Ω and \[\rho\] = 1.8 x 10-6 Ω . cm in the formula of specific resistance,

 1.8 x 10-6 =  2 x \[\frac{A}{L}\]

\[\frac{L}{A}\] = \[\frac{10}{9}\] x 106 cm-1       (2)

Multiplying equations (1) and (2),

LA x   \[\frac{L}{A}\] = (\[\frac{10}{9}\])2 x 106 cm2

L = 11.1 m

Dividing equation (1) by (2),

LA x \[\frac{A}{L}\] = 10-6 cm4

A = 0.1 mm2

The length and cross-sectional area of the copper wire are 11.1 m and 0.1 mm2  respectively.

ll. A 5 mm diameter wire is produced from a chunk of metal. Another wire of diameter 1 cm is produced from an identical chunk. What is the ratio of the resistance of the two wires?

Solution: Resistance of a wire of length L  and cross-section A  is R = \[\frac{\rho L}{A}, where \rho\] is the resistivity of the material. The mass and density of the chunk are m and  D respectively. If the diameter of a wire of volume V is d,

A = \[\frac{\pi d^{2}}{4}\]

V =  \[\frac{m}{D}\]

L =  \[\frac{V}{A}\]

Therefore, the resistance of a wire of diameter d is,

R = \[\frac{16 \rho m}{\pi D a^{4}}\]

According to the problem, mass and density of the two wires of diameters  \[d_{1}\] = 5 mm=0.5cm and \[d_{2}\] = 1 cm are same such that the ratio of resistance is,

\[\frac{R_{1}}{R_{2}}\] = \[(\frac{d_{2}}{d_{1}})^{4}\]

\[\frac{R_{1}}{R_{2}}\]  =  \[(\frac{1 \; cm }{0.5\; cm})^{4}\]

\[\frac{R_{1}}{R_{2}}\] = 16

The ratio of the resistance of the wires is 16:1. 

Did You Know?

• Specific resistance depends on the ambient temperature. It increases with increasing temperature for metals. For glass, however, at very high temperatures, the resistivity decreases considerably.

• Superconductors have zero resistivity in the superconducting state (at very low temperature).

• The resistivity of semiconductors decreases with increasing temperature.

More about Specific Resistance

Students often consider Physics to be a tough subject but it isn't like that. It is theoretical, conceptual and also has a formula base. 

It is based on many other subjects too. Mostly, it would need good Mathematical skills. Apart from that, you just need the right approach to ace Physics. 

You must have understood the unit of specific resistance and also have got clarity on the same. 

Physics is not a cup of everybody's tea but that doesn't make any sense if you just make a perception about it and whenever you study it just dive into it with the right mindset.

You must use the below-mentioned strategies for mastering your preparation of the Physics subject. 

• Think conceptually

Physics needs thinking of the concepts with a bigger view. Like other subjects, it doesn't call for a simple review of the chapter. It calls for critical conceptual thinking. Your concepts need to be clear from the foundation level, you can't miss the starting chapters and jump on to the chapters in the last. Break the problem into tit bits and start bringing them together by applying concepts and that's how your understanding makes the problem 10 times easier. 

• Keep up with studying

In physics, you need to keep up with your reading, you can't miss any class as the concepts are interlinked and you may find yourself as a misfit in the class if you miss any of the classes. As in any other subject like History, if you miss some of the classes, you may skip a war but get back on track from the next topic. But if you do the same with Physics, you may lose all the courses. So be consistent and take your time to understand the given subject and related topics.

• Catch up on maths

Having good math skills can make your hold stronger on physics. If your math skills are not up to the mark, you should try enhancing them. It is a great benefit to have Maths as your strength because then, solving the problems of Physics will become easier. Physics is always roaming around maths and you should be strong at that, especially the chapters of trigonometry, algebra and calculus. 

• Uninterrupted studying

Students shall make sure that whenever they sit to study, their study place should be free of distractions. It is very important to do uninterrupted studying because being distracted now and then will make you lose track. You might often forget things that you've studied till then. Hence, this step will help you get the best out of your studies.

• Review and understand 

Another important tip that you shall be following is reviewing the study material. You might also consider reviewing the old question papers or mock tests to gain a better understanding of the exam pattern and know about how the questions are asked. Also, checking out for homework questions and textbook examples can be beneficial too. 

• Prepare for practical part also 

Students often tend to focus only on the theory part but you shouldn't forget that the practical part also contains 30% of your total marks. And ultimately, when you prepare for it you can also improve your theoretical knowledge because, in the end, everything is interconnected. 

Subject-related additional tips: 

• Understand the logic behind the derivations. 

• Practice the numerical section. 

• For theory, refer to the class notes. 

• Select the reference book wisely. 

• Focus on the presentation. 

• Familiarise yourself with the marking scheme that is followed. 

• Other important Tips: 

• Prepare a good study action plan. 

• Make chapter-wise notes. 

• Understand how questions are asked. 

• Know about the weightage of each topic. 

• Do not skip topics. 

• The most on self-evaluation. 

To conclude, there may be times when doing something seems to be very difficult but in those times, you have to remind yourself that to get somewhere, you have to leave the place that you are currently stuck at. This means that to get closer to your dreams, you have to keep moving ahead, despite anything and everything.