Galvanometer, Ammeter and Voltmeter

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two equal resistors, of one hundred ohm each, are in series across twenty volts. So each one has ten volts across it, right? Now you connect a voltmeter across one of them, and it reads about nine point five volts. So tell me, is the voltmeter broken? Keep that question in mind, because by the end of this video you will know exactly what is going on. Meter questions come up often in JEE, so let's begin.

0:33The galvanometer: deflection follows current

So, let us start with the moving-coil galvanometer, a meter that detects current. Its pointer turns through an angle theta, the Greek letter on screen, and theta is proportional to I, the current; the curly sign between them means is proportional to. Double the current, double the deflection. Okay? Two numbers describe it. Capital G is its resistance, and I g, capital I with a small g below, is the full-scale current, which sends the pointer to the end of the scale. Clear? The figure of merit, capital K, is I over theta, the current per division. Current sensitivity is the other way up, theta over I, so a small figure of merit means a sensitive galvanometer. Got it? Voltage sensitivity is theta over V, where V is the voltage across the galvanometer. By Ohm's law, that voltage is I times G, so this becomes theta over the product of I and G, with me?

1:31Ammeter: a small shunt in parallel

Now look at the left of the figure; the right half is the voltmeter, which comes next. The circle marked G is the galvanometer. We want to measure a big current, capital I, shown by the arrow, but only I g may pass through the galvanometer. So we join a small resistance, called a shunt, capital S, in parallel with it, and most of the current bypasses the coil. Okay? The galvanometer and the shunt are in parallel. So what is the same for both of them? Yes, it is the voltage across them, very good! At full scale, I g flows through the galvanometer, so its voltage is G times I g. The rest, I minus I g, flows through the shunt, so its voltage is S times that bracket. Set the two equal, as on screen. Right? Now just divide. S equals I g times G, over I minus I g. That is the shunt formula, also written under the figure. Clear? The whole ammeter is G in parallel with S; the two vertical lines mean in parallel with. The shunt is small, so this is almost exactly S, where the wavy equals sign means approximately equal. So an ammeter has a small resistance. Got it?

2:49Voltmeter: a large resistance in series

Now the right side of the figure, the voltmeter. Here a large resistance, capital R, goes in series with the galvanometer. We want full-scale deflection when the voltage across the pair is V, the range of the voltmeter. Okay? So at full scale, what current flows through G and through R? The same current, I g, through both, because they are in series, exactly! So V equals I g times the bracket, G plus R, as on screen. With me? Divide by I g, and move G across. R equals V over I g, minus G, as written under the figure. Clear? So the voltmeter's resistance, G plus R, is large. Small shunt in parallel for an ammeter, large resistance in series for a voltmeter; students often mix these up. Right?

3:41Ideal meters, and why a real voltmeter reads low

Now, an ammeter goes in series, so it must carry the very current it measures. So what resistance should an ideal ammeter have? Yes, exactly, it is zero, so it changes nothing! And a voltmeter goes in parallel, so it should draw as little current as possible. So an ideal voltmeter has infinite resistance; the sideways eight means infinity. Okay? But real meters are not ideal. Put a real voltmeter of resistance R V, capital R with a small V below, across a resistor R, as on screen. The meter is now in parallel with R, so it reads the voltage across that combination. Clear? So tell me, is that reading more than the true value, or less? It is less, very good! R in parallel with R V is smaller than R alone, so, when R shares the voltage with other resistance in series, as in our puzzle, it takes a smaller share. And the higher R V is, the smaller the error. Hold on to this for our puzzle. Got it?

4:46Topper's trick: extending a range n times

Now a trick that toppers use. You have an ammeter of resistance R A, capital R with a small A below, and you want n times its range. The meter still carries its own full-scale current, so how much must the shunt carry? The other n minus one times that current, exactly! Current divides in the inverse ratio of resistance, so the shunt is R A, over n minus one. The cross before n means times. Okay? For a voltmeter, the total resistance must become n times R V, so you add n minus one, in brackets, times R V, in series. Not n times R V, careful, that is a common slip. Clear?

5:29Example 1: shunt for a 50 ohm, 2 mA galvanometer to read 1 A

Alright, example one, an easy one. The galvanometer has fifty ohm, where the Greek letter omega means ohm, and gives full-scale deflection for two milliamperes; the small m means milli, and the capital A means ampere. Find the shunt that makes it an ammeter of range one ampere. Pause the video and try it first. Done? Okay, let's check. Same units first: two milliamperes is zero point zero zero two ampere. So G is fifty ohm, I g is zero point zero zero two ampere, and the range, capital I, is one ampere. Right? Many students put just one ampere in the bottom of the formula. But which current actually flows through the shunt? Only one minus zero point zero zero two, which is zero point nine nine eight ampere, because the galvanometer takes the rest. On top, zero point zero zero two times fifty gives zero point one. So S is zero point one over zero point nine nine eight. With me? That is about zero point one ohm, more precisely zero point one zero zero two ohm, in parallel. See how tiny the shunt is? That keeps the ammeter's resistance small. Got it?

6:49Example 2: why a 1000 ohm voltmeter reads 9.5 V, not 10 V

Now example two, our puzzle from the start. Two resistors, of one hundred ohm each, are in series across twenty volts, where capital V means volt. What does a one kilo ohm voltmeter read across one of them? The small k means kilo, so that is one thousand ohm. Pause the video and try it first. Done? Okay, let's check. Without the meter, the equal resistors share the twenty volts equally, ten volts each. So will the voltmeter read exactly ten volts? No, because the voltmeter itself joins the circuit! Its one thousand ohm is in parallel with that one hundred ohm. One hundred times one thousand, over one thousand one hundred, gives about ninety point nine ohm. With me? This is in series with the other one hundred ohm, and the twenty volts is shared in proportion to resistance. So V equals twenty times ninety point nine, over ninety point nine plus one hundred, which is one hundred and ninety point nine. Right? That gives about nine point five volts, not ten. Puzzle solved: the voltmeter was not broken, it simply changed the circuit it was measuring. With a higher voltmeter resistance, the reading comes closer to ten volts. Got it?

8:07Quick revision: shunt, series resistance, ideal meters

So, quick revision. For an ammeter, a small shunt in parallel: S equals I g times G, over I minus I g. For a voltmeter, a large resistance in series: R equals V over I g, minus G, okay? An ideal ammeter has zero resistance, an ideal voltmeter infinite, and a real voltmeter usually reads a little low. Now go and try the practice problem on this page. All the best!