The Galvanometer: Half-Deflection Method

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. You want the resistance of a galvanometer, and all you have is a cell, a high resistance box, a shunt and two keys. So how can the galvanometer measure its own resistance? That is exactly what we will find out today. And one more puzzle for later: if a shunt of one hundred ohms halves the deflection, is the galvanometer exactly one hundred ohms? This experiment belongs to experimental skills in JEE, and questions on it can come up. So, let's begin.

0:35The circuit and the two readings

Look at the circuit on the left of the figure. On the far left is the cell, the long thin line and the short thick line, and capital E is its e m f, the electromotive force. At the top is capital R, a high resistance from a resistance box. And the circle marked capital G is our galvanometer, where capital G also stands for its resistance, the quantity we want. Okay? In the middle is the box marked capital S, the shunt, which can be joined in parallel with the galvanometer. It is joined through the key K two, the K with a small two below it, just under the shunt. And the key at the bottom left, K one, with a small one below it, is the main key. The notes on the right sum up the method, and we will build them line by line. Right? Step one: close K one, and keep K two open. The current flows through capital R and the galvanometer, and we get a deflection theta, this Greek letter. We choose capital R so that theta is large, and an even number of divisions. Clear? Now let me ask you this: why do we want the deflection to be large, and why an even number of divisions? Very good if you thought about reading the scale. The reason is that a large deflection is read with a small percentage error, right? And an even number of divisions has an exact half on the scale, which we need in step two. Got it? In this first reading, the galvanometer current is capital I with a small g below it. It equals capital E, over capital R plus capital G, and it also equals small k times theta. Here small k is the figure of merit, the current for a deflection of one division. With me? Step two: close K two as well, and adjust the shunt, capital S, until the deflection drops to theta over two, exactly half the first one. Okay?

2:30Why G is nearly S, and the exact G = RS/(R - S)

Now, why does this work? Capital R is several thousand ohms, and that keeps the galvanometer current, I g, within the range of the galvanometer. Right? It also makes capital R much greater than capital G, and the double angle sign on screen means much greater than. So when we connect the shunt, the main current hardly changes. Clear? Now the galvanometer shows half the deflection, so it carries half the current. Where does the other half go, and what does that tell you about capital G and capital S? Yes, exactly, the other half goes through the shunt. Two resistances in parallel that carry equal currents must be equal, so capital G is approximately capital S, the wavy equals sign. Got it? But that is only approximate. The exact result is capital G equals capital R times capital S, over capital R minus capital S, the fraction on screen. And as the bracket says, it comes close to capital S only when capital R is much greater than capital S, okay?

3:34Figure of merit and sensitivity

Now the same first reading gives us one more number. The figure of merit, small k, is the current for a deflection of one division. So small k equals capital I over theta, which is capital E, over capital R plus capital G in brackets, times theta. With me? From it, the current sensitivity is one over small k, and the voltage sensitivity is one over the product of small k and capital G, okay? If the scale has n divisions, the full scale current, I g, is n times small k. Right? And with I g known, we can convert the galvanometer. For an ammeter of range capital I, use a shunt of I g times capital G, over capital I minus I g. For a voltmeter of range capital V, use a series resistance of capital V over I g, minus capital G, clear?

4:25Deflection cut to theta/n: G = (n - 1)RS/(R - (n - 1)S)

Now a small variation. Sometimes the shunt cuts the deflection not to half, but to theta over n, okay? Then capital G equals n minus one in brackets, times capital R times capital S, over capital R minus n minus one in brackets, times capital S, and yes, it looks heavy. But we can test it very quickly, right? So let me ask you, what happens if you put n equal to two? Yes, exactly, n minus one becomes one, and we get back capital R times capital S, over capital R minus capital S, our half deflection result. So this one formula covers both cases. Got it?

5:08Example 1: G with R = 5000 ohm and a 100 ohm shunt

Now the examples, the easier one first. In a half deflection experiment, capital R is five thousand ohms, and the shunt needed is one hundred ohms. The omega sign on screen means ohms. Find the resistance of the galvanometer. Pause the video and try it first. Done? Okay, let's check. And remember our second puzzle from the start. The shunt is one hundred ohms, and since capital G is about capital S, is the galvanometer simply one hundred ohms? No, not exactly, so let us use the exact formula. Capital G equals five thousand times one hundred, over five thousand minus one hundred, where the cross is the times sign. Right? Working it out, that is five hundred thousand over four thousand nine hundred, which gives about one hundred and two ohms. Clear? So that settles our puzzle. The galvanometer is close to one hundred ohms, but not exactly, because capital G equals capital S only approximately, when capital R is much greater than capital S, got it?

6:16Example 2: R = 6000 ohm, 60 ohm shunt cuts deflection to 1/3

Now the harder one. With capital R equal to six thousand ohms, a shunt of sixty ohms reduces the deflection to one third. Find capital G, the resistance of the galvanometer. Pause the video and try it first. Done? Okay, let's check. Can we use our half deflection formula, capital R times capital S, over capital R minus capital S, here? No, we cannot, so be careful, because the deflection is cut to one third, not half. So this is theta over n with n equal to three, and n minus one is two. Okay? So capital G equals two times capital R times capital S, over capital R minus two times capital S, from our formula. Putting in the numbers, that is two times six thousand times sixty, over six thousand minus one hundred and twenty. Right? Working it out, that is seven hundred and twenty thousand over five thousand eight hundred and eighty, which gives about one hundred and twenty two ohms. If you got this on your own, very well done!

7:24Quick revision: G = RS/(R - S), figure of merit, theta/n

So, quick revision. First, for half deflection, capital G equals capital R times capital S, over capital R minus capital S, which is nearly capital S when capital R is much greater. Second, the figure of merit, small k, is capital E, over capital R plus capital G, times theta, and the current sensitivity is one over small k. Third, if the deflection is cut to theta over n, put n minus one in the formula. Now go and try the practice problems on this page, and all the best!