The Metre Bridge and Resistivity

Transcript

0:00Introduction

Hello everyone, let us start with a small puzzle. Two students measure the same unknown resistance on a metre bridge. The first chooses a known resistance that gives the null point at twenty centimetres, and the second gets it at fifty. Both read the scale to a millimetre. So whose answer would you trust more, or are they equally good? Keep your guess, and we will settle it with real numbers. The metre bridge is part of experimental skills in JEE, and it comes up often. So, let's begin.

0:33The metre bridge: a Wheatstone bridge

Look at the figure, where the orange line from capital A to capital B is a uniform wire, one metre long, the small m on screen. That is one hundred centimetres, the c m on screen, okay? The metre bridge is really a Wheatstone bridge. The two parts of this wire, on either side of the point capital J, form two of its arms. Right? The two boxes above the wire are the gaps. We put a known resistance, capital R, in the left gap, and the unknown, capital S, in the right gap. The cell is the pair of short lines at the very top. With me? The circle marked capital G is the galvanometer, and the arrow from it is the jockey, touching the wire at capital J, okay? Where the galvanometer shows no deflection, we have the null point, at a distance l from the end capital A, in centimetres.

1:25Balance: from lengths to resistance

Clear? Now, the wire is uniform, so the resistance of each part is proportional to its length. At the null point the bridge is balanced. So what is capital R over capital S equal to? Yes, exactly, it is the ratio of the two lengths of wire. Capital R over capital S equals l, over one hundred minus l, the two lengths marked by the arrows below the wire, okay? The known resistance pairs with the left part, and the unknown with the right part. Turn it around, and capital S equals capital R times one hundred minus l, over l. Got it? And once we know capital S, we get the resistivity of the wire. Rho, this Greek letter, equals pi times d squared times capital S, over four times capital L, okay? Capital L is the length of the wire in the gap, and small d is its diameter, so pi times d squared over four is its area of cross section. Right?

2:25Why the null point should be near the middle

Now, how do we choose capital R? We choose it so that the null point lies near the middle, between forty and sixty centimetres. But why the middle? Let's see. The little triangle is capital delta, and delta l is the error in reading the null point. The same delta l affects both lengths, l and one hundred minus l. So delta S over S, the relative error in capital S, is delta l over l, plus delta l over one hundred minus l, okay? Add the two fractions, and we get one hundred times delta l, over l times one hundred minus l. To make this error small, the product at the bottom must be as large as possible. So where on the wire is that product the greatest? Yes, exactly, it is greatest when l equals one hundred minus l, so at fifty centimetres. That is the best place for the null point. Clear? Now back to our puzzle. Both students read to a millimetre, so delta l is zero point one centimetre. At twenty centimetres, the error is one hundred times zero point one, over twenty times eighty. That is about zero point six three per cent, and the wavy sign means approximately. Okay? At fifty centimetres it is one hundred times zero point one, over fifty times fifty, which gives zero point four zero per cent. So the null point in the middle gives the smaller error, and our puzzle is settled. With me?

3:55Interchange the gaps, and end corrections

Next, a good habit in the lab. We interchange the two gaps, capital R and capital S, and take the mean of the two results. Okay? We do this because it reduces the effect of the resistance at the ends of the wire, and of any non-uniformity in the wire. Right? Alpha and beta, the two Greek letters, are the end corrections at the left and right ends, in centimetres of wire. We add them to the two lengths, so capital R over capital S equals l plus alpha, over one hundred minus l plus beta. Clear?

4:29Care at the bench

Now a few points of care. The connecting strips are made of thick copper, so their resistance is negligible. Got it? The jockey is tapped gently on the wire, and never slid along it. Now, why such a strict rule? What goes wrong with sliding? Sliding scrapes the wire and makes its cross section uneven. Then the resistance is no longer proportional to the length, and our whole formula fails. With me? And one last fact. The cell and the galvanometer may be interchanged, and the balance is not disturbed. Okay?

5:03Example 1: find S with 10 ohm and null point at 40 cm

Now the examples, the easier one first. With ten ohms in the left gap, the null point is at forty point zero centimetres, and the omega sign on screen stands for ohm. Find the resistance in the right gap. Pause the video and try it first. Done? Okay, let's check. Here l is forty point zero centimetres, so the right part of the wire, one hundred minus l, is sixty point zero centimetres. Right? So capital S equals ten times sixty point zero, over forty point zero, which gives fifteen ohms. Okay? Now a quick check. Here l is less than fifty centimetres, so capital S is greater than capital R, the greater than sign on screen, and fifteen is indeed more than ten. This check catches a flipped fraction at once.

5:56Example 2: resistance and resistivity of a 1.20 m wire

Clear? Now the harder one. With five ohms in the left gap, the null point is at sixty two point five centimetres. The unknown wire is one point two zero metres long, and zero point five zero millimetres in diameter, the m m on screen. Find its resistance and its resistivity. Pause the video and try it first. Done? Okay, let's check. One hundred minus sixty two point five is thirty seven point five. So capital S equals five times thirty seven point five, over sixty two point five, which gives three point zero ohms. As expected, it is less than five ohms, because the null point is to the right of the middle. With me? Now the resistivity, and here is a trap. Many students put zero point five zero straight into the formula for the diameter. So is that right, or should something change first? That would be wrong, so be careful, because the diameter is in millimetres. Change it to metres first, zero point five zero times ten to the power minus three metre. And capital L is one point two zero metres. Okay? Putting everything in, rho equals pi, times zero point five zero times ten to the power minus three, all squared, times three point zero, over four times one point two zero. Right? That works out to about four point nine times ten to the power minus seven ohm metre, the unit of resistivity. Notice that the diameter is squared, so its unit really matters. If you got this on your own, very well done!

7:40Quick revision: R/S = l/(100 - l), null point near middle

So, quick revision. First, at balance, capital R over capital S equals l over one hundred minus l, and from capital S the formula on screen gives rho, the resistivity. Second, keep the null point near the middle, because the error is least at fifty centimetres. Third, interchange the gaps and take the mean, and tap the jockey, never slide it. Now go and try the practice problems on this page, and all the best!