The Potentiometer

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. To find the emf of a cell, you connect a voltmeter across it. But a real voltmeter draws a small current, so the cell loses some voltage across its internal resistance, and the reading comes out low. So how can we ever measure the true emf? Today you will meet an instrument that takes no current from the cell at balance, the potentiometer. Questions on it come up often in JEE, so let's begin.

0:30The set-up: a uniform wire and a null point

So, look at the top of the figure. The driver cell, epsilon naught, the Greek letter epsilon with a small zero below, sends a steady current through a rheostat and the wire from end A to end B, okay? The wire is uniform, so the potential falls evenly along it. Clear? Now say the whole wire has a potential difference VAB across it, capital V with a small A B below, and its length is capital L, okay? Since the potential falls evenly, tell me, what is the drop across half of the wire? Yes, exactly, half of VAB! Equal lengths get equal drops, so the drop per metre is VAB over capital L, okay? We call this k, the potential gradient. Its unit is volt per metre, written capital V, small m, to the power minus one. Right? Now the bottom of the figure. The cell we test, of emf capital E, and a galvanometer, the capital G in a circle, run from end A to J, the jockey, a sliding contact that can touch the wire anywhere. With me? Slide the jockey along the wire. At most points the galvanometer deflects, but at one special point it reads zero. So when does that happen? Yes, it happens when the drop along the wire, from A up to the jockey, exactly balances E, the emf. That point, where G reads zero, is the null point, and the length up to it, small l, is the balancing length. So E equals k times l. Clear? And here is the beauty of it. At the null point no current flows through the cell, so there is no drop across its internal resistance. We get the emf itself, not the terminal voltage, and that is why a potentiometer is preferred to a voltmeter. Got it?

2:25What decides the gradient k

Now, what decides k? The driver pushes current through three resistances in series: the wire, R wire, with the word wire below, the rheostat, R h, with a small h below, and its own internal resistance, r naught, small r with a zero below. So the current, capital I, is epsilon naught over their sum. Okay? The voltage across the wire is this current times R wire. Divide by capital L, and you get k, as on screen; the dot in the middle means times. Right? Now let me ask you this. For a precise measurement, do we want a large k, or a small k? Small k, very good! Then each volt is spread over a longer piece of wire, so the null point is found more precisely; that is greater sensitivity. To make k small, use a longer wire, or a larger resistance in series. Got it?

3:21No null point? Check two things

Now, sometimes you slide the jockey from end to end and find no null point at all. So what if VAB, the drop across the whole wire, is smaller than the emf E? Then even the full wire cannot balance the cell, and the galvanometer deflects the same way everywhere. So VAB must be greater than E; the sign on screen means greater than. Clear? And the second condition: the positive terminals of both cells must be joined to the same end, A, as in the figure. Otherwise the two voltages add instead of opposing, and again nothing balances, okay?

3:58Comparing two emfs

Next, comparing two cells. The first cell, E one, capital E with a small one below, balances at l one, and the second, E two, at l two. Same wire, same k, which cancels: E one over E two equals l one over l two. Got it? Now join the two cells in series, and let l one and l two be the new balancing lengths. When they aid each other, E one plus E two balances at l one. When they oppose, E one minus E two balances at l two. With me? So, from these two equations, how do you find E one over E two? Add them, subtract them, then divide, and k cancels again. E one over E two equals l one plus l two, over l one minus l two. This is the sum and difference method. Right?

4:52Internal resistance of a cell

Now one more use: finding the internal resistance of a cell. First, balance the cell on open circuit, at l one. No current flows, so that gives the emf: E equals k times l one. Okay? Now join a resistance, capital R, across the cell, as a shunt. The cell now gives current, so the wire balances its terminal voltage V, at l two. V equals k times l two. Clear? So tell me, which length is longer, l one or l two? It is l one, because the terminal voltage is less than the emf when the cell gives current. Combine the two balances, and the internal resistance, small r, equals R times the bracket, l one over l two, minus one. Get a negative r, and you have flipped the ratio. Got it?

5:45Example 1: emf of a cell balancing at 6.5 m on a 10 m wire

Alright, example one. We have a ten metre wire of twenty ohm, where the small m means metre and the Greek letter omega means ohm. It is joined directly to a two volt cell with negligible internal resistance. Another cell balances at six point five metres. Find its emf. Pause the video and try it first. Done? Okay, let's check. The cell is joined directly, with negligible internal resistance, so all its two volts fall across the wire. So k equals two over ten, which is zero point two volt per metre. Notice, the twenty ohm is not even needed here. Right? Now E equals k times l. Zero point two times six point five, where the cross means times, gives one point three. So the emf is one point three volts, as simple as that.

6:38Example 2: 60 ohm in series, where does 1.2 V balance?

Now a harder one. The ten metre wire has forty ohm, and it is in series with sixty ohm and a four volt ideal cell, with no internal resistance. Where does a cell of one point two volts balance? Pause the video and try it first. Done? Okay, let's check. Many students say, four volts across ten metres, so k is zero point four, and the cell balances at three metres. Is that right? No, no, wait, the sixty ohm resistor takes its share of the voltage! The total resistance is forty plus sixty, one hundred ohm, so the current is four over one hundred, which gives zero point zero four ampere; capital A means ampere. Okay? The voltage across the wire is zero point zero four times forty, which gives one point six volts. With me? So k equals one point six over ten, which is zero point one six volt per metre. Clear? Finally, l equals E over k. One point two over zero point one six gives seven point five metres, from the end A of the wire. See, the series resistance made k smaller and the balancing length longer, just as we said. Got it?

7:56Quick revision: E = kl, E1/E2 = l1/l2, r = R(l1/l2 - 1)

So, quick revision. The gradient k is VAB over capital L, and at the null point E equals k times l, with no current drawn. E one over E two equals l one over l two, and r equals R times the bracket, l one over l two, minus one. And the null point needs VAB greater than E, while a smaller k means greater sensitivity. Now go and try the practice problem on this page. All the best!