Ohm's Law, Resistance and Resistivity

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a copper wire and pull it until it is twice as long. More length means more resistance, so the resistance simply doubles, right? But wait, are you really sure about that? Hold on to your answer; we will check it in this video. Ohm's law and resistivity come up often in JEE problems. So, let's begin.

0:27Ohm's law: V=IR

So, first, Ohm's law. Take a metallic conductor at constant temperature. Then the current I through it is proportional to the potential difference V across it. The sign on screen that looks like an open Greek alpha means is proportional to. Okay? So V equals I times R, the law in one line. Here R is the resistance, measured in ohm, and the Greek capital omega in brackets is its symbol. And never forget the condition: constant temperature. Right? For such an ohmic conductor, the graph of V against I, the V I graph on screen, is a straight line through the origin. Double the voltage, double the current. Clear? So here is a question for you. An ordinary filament bulb glows white hot. Over a wide range of voltage, is it ohmic? The answer is no. Over a wide range the filament heats up a lot, so its temperature is far from constant, and the graph is not a straight line. So filament lamps over a wide range, diodes and electrolytes are non-ohmic: V and I are not proportional. Got it?

1:37Inside the wire: the microscopic form

From the drift lesson, J, the current density, or current per unit area, equals n times e times v d. Here n is the number of free electrons per unit volume, e the size of the electron's charge, and v d, v with a small subscript d, the drift speed. And v d equals e times E times tau, over m: E is the electric field, tau, the Greek letter, the relaxation time, the mean time between collisions, and m the electron's mass. With me? So here is a small task for you. Put this v d into the formula for J, and tell me what you get. Yes, exactly: J equals n times e squared times tau, over m, all times E, where the small two up there is a power. Everything in front of E is fixed for a given metal at a given temperature, so J is proportional to E, okay? That is Ohm's law in its microscopic form: J vector equals sigma times E vector. The little arrows on top mean vectors; the current density points along the field. Sigma, the Greek letter, is the conductivity, n times e squared times tau, over m. Right? Now, resistivity. Rho, the Greek letter like a curly p, is the resistivity, simply one over sigma. So rho equals m, over n times e squared times tau. Good conductor, large sigma, small rho.

3:01Resistance and resistivity are not the same

Clear? Now, the resistance of a real wire. R equals rho times l, over capital A, where l is the length and capital A the area of cross-section. Longer wire, more resistance; thicker wire, less resistance. Makes sense, right? Now a quick question. Suppose we cut a copper wire into two equal halves. What happens to rho, the resistivity, of each half? Nothing happens to it at all. Many students say rho becomes half, but oh no, careful, that is wrong. Resistivity depends only on the material and the temperature: same copper, same temperature, same rho. Got it? But resistance also depends on the shape, the length and the area. Each half has half the length and the same area, so half the resistance. Rho belongs to the material; R belongs to the wire.

3:56Stretching a wire: the volume stays the same

Okay? Now back to our puzzle: stretching. When you pull a wire longer, it also gets thinner. So what stays the same? The volume, exactly! No metal is added or removed. So capital A times l, the volume, stays fixed. With me? Now a neat trick: in rho l over capital A, multiply the top and the bottom by l. You get rho times l squared, over capital A times l, and the bottom is the fixed volume. So R is proportional to l squared; the double arrow means it follows. Okay? In terms of area, l equals the volume over capital A, so R goes as one over capital A squared. And the area is pi r squared, with r the radius, so R goes as one over r to the power four. Right? For small changes there is a quick rule. Delta, the Greek triangle, means a small change. Delta R over R equals two times delta l over l: the fractional change in resistance is twice that in length. But only for small changes; for a big stretch, square the length ratio exactly. Clear?

5:07Example 1: resistance of a 10 m copper wire of 1 mm2 area

Alright, example one, an easy one. Find the resistance of a copper wire ten metres long, with a cross-section of one square millimetre. Rho for copper is one point seven times ten to the power minus eight ohm metre. Pause the video and try it first. Done? Okay, let's check. First, the units: one square millimetre is how many square metres? One millimetre is ten to the power minus three metre; square it, and you get ten to the power minus six square metres. Not ten to the power minus three; that is the trap here. Okay? Now use R equals rho l over capital A: one point seven times ten to the power minus eight, times ten, all over ten to the power minus six. Right? The powers of ten leave ten to the power minus one, so R is one point seven times ten to the power minus one, which is zero point one seven. So the resistance is zero point one seven ohm. Got it?

6:10Example 2: new R when stretched n times or radius halved

Example two is a little harder, and it settles our puzzle. Here it is: a wire of resistance R is stretched uniformly to n times its length; find the new resistance. And what if, instead, its radius is reduced to half by stretching? Pause the video and try it first. Done? Okay, let's check. The length becomes n times, so the new resistance is just n times R, right? No, no, wait, that is the trap! The volume is fixed, so when the length becomes n times, the area becomes capital A over n; the arrow on screen means becomes. Okay? So the new resistance, R prime, the R with a small dash, is rho times n l, divided by capital A over n. One n is on the top, and dividing by capital A over n gives another n, so the answer is n squared times R, as shown. So in our puzzle, n equals two, and the resistance becomes four times, not two times. Clear? Now part two: the radius becomes half. The area is pi r squared, so it becomes one quarter. So the resistance becomes four times R, right? Oh no, careful, we forgot something! The volume is still fixed, so one quarter of the area means four times the length. With me? So R prime is rho times four l, divided by capital A over four, which is sixteen times the old resistance. Check it with R proportional to one over r to the power four: halve the radius, and two to the power four is sixteen, the same answer. Got it?

7:50Quick revision: V = IR, J = sigma E, R = rho l/A, stretching

So, quick revision. Ohm's law: V equals I times R for an ohmic conductor at constant temperature, and inside the metal, J equals sigma times E, as vectors. Resistance is rho l over capital A, and rho depends only on the material and the temperature. And when you stretch a wire, the volume stays fixed, so R goes as l squared, or as one over r to the power four. You now know exactly why a stretched wire surprises people. Now go and try the practice problems on this page. All the best!