EMF, Internal Resistance and Terminal Voltage

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a cell with an e m f of twelve volts, and connect it to a resistor. Now put a voltmeter across the two terminals of the cell. It does not read twelve volts. It reads only eleven volts! So where did that one volt go? Is the cell faulty? Hold on to your answer. By the end of this video, you will know exactly where it went. Cells and terminal voltage come up often in JEE circuit problems, so let's begin.

0:32A real cell: emf and internal resistance

So, let us start with a real cell. Every real cell comes with two numbers. First, its e m f, written capital E, the electromotive force. Second, its internal resistance, written small r, because the inside of the cell itself resists the current a little. Okay? Now connect this cell to an outside resistor, capital R, the external resistance. The current flows through R, of course. But tell me, where else must it flow to complete the loop? Through the cell itself, yes, exactly! The same current goes through R outside and through r inside, one after the other. So r is in series with R, and the total resistance is R plus r, right? So the current, capital I, equals E over R plus r. The whole of R plus r sits below the fraction line. Many students forget the small r and write just E over R, so please don't do that! Got it?

1:33Terminal voltage: what the voltmeter reads

Now, the big question. Terminal voltage, capital V, is the voltage across the two terminals of the cell, what a voltmeter there reads. When the cell drives a current I, does this voltmeter read the full e m f? No, it reads less, and here is why. The current also passes through the internal resistance, so some voltage is used up inside the cell itself. That drop is I times r, the current times the internal resistance. With me? So only what is left reaches the terminals. V equals E minus I times r. This is the most important line of today's lesson, so write it down. Clear? Now put I equals E over R plus r into it. V becomes E times R, over R plus r, and that is just I times R, the voltage across the outside resistor. Okay?

2:26Charging a cell: the sign flips

But careful, V is not always less than E, so look at a cell that is being charged. Now a bigger source pushes the current the other way through the cell, so the current I enters at its positive terminal. Right? Look at the direction carefully. The current is pushed into the positive terminal, against the e m f of the cell. So tell me, while charging, is the terminal voltage more than E, or less than E? It is more, very good! The charger must overcome the e m f, and also push the current through the internal resistance. So V equals E plus I times r. Only the sign changes. Got it? So remember the rule. When the cell drives a circuit, V is less than E, and when it is being charged, V is more than E, okay? Using minus I times r for a charging cell is a common slip in the exam, so always watch the direction of the current.

3:22Open circuit, short circuit and the V–I line

Now, two extreme cases. First, open circuit: nothing is connected, so no current flows, and I equals zero. Then I times r is zero, so V equals E, the full e m f. So the open circuit voltage is the e m f itself. Clear? Second, short circuit. Join the two terminals directly by a plain wire, so the outside resistance R is zero. What current flows now? Only r is left to limit the current, so I equals E over small r, and this is the short circuit current, the largest current this cell can drive, because R plus r is smallest when R is zero. Right? Now plot V against I, the V I line in the heading, where the dash means against. Write V equals E minus r times I, and look, that is a straight line! It meets the V axis at E, the intercept, and its slope is minus r. So the steeper the line falls, the bigger the internal resistance. With me? And two points on this line give its slope, and so r. Take two readings, I one with V one, and I two with V two, the small one and two written below. Then r equals V one minus V two, over I two minus I one. See the order: V falls when I rises, so the top and the bottom both come out positive. Okay?

4:46Example 1: 12 V cell, r = 0.5 ohm, R = 5.5 ohm, find I and V

Alright, example one, and this is our puzzle from the start. There is a cell of e m f twelve volts, with an internal resistance of zero point five ohm, driving an external resistance of five point five ohm. The capital V after a number means volt, and the Greek letter omega means ohm. Find the current and the terminal voltage. Pause the video and try it first. Done? Okay, let's check. Some students write the current as twelve over five point five amperes, right? No, no, careful, that forgets the cell's own resistance! The total resistance is five point five plus zero point five, which is six ohm. So I equals twelve over six, which gives two amperes, and the capital A means ampere. Clear? Now the terminal voltage. V equals E minus I times r, which is twelve minus two times zero point five, where the cross means times. That gives eleven volts. Got it? And there is our puzzle, solved! The missing volt was used up inside the cell itself: I times r is two times zero point five, which is one volt. Twelve minus one leaves eleven volts at the terminals. Nothing is faulty; every real cell does this. With me?

6:07Example 2: find E and r from two terminal voltage readings

Now a slightly harder one. The terminal voltage of a battery is four point five volts when the current is one ampere, and four volts when the current is two amperes. Find E, its e m f, and r, its internal resistance. Pause the video and try it first. Done? Okay, let's check. Some students say the e m f is four point five volts, the larger of the two readings, right? Oh no, careful, the terminal voltage equals the e m f only on open circuit, when no current flows. Here one ampere is flowing, so the e m f must be more than four point five volts. First find r from the two readings. r equals four point five minus four, over two minus one, which is zero point five over one, so r is zero point five ohm. Okay? Now use V equals E minus I times r for the first reading. So E equals V one plus I one times r, which is four point five plus one times zero point five. That gives five volts. With me? Let's check with the second reading too. Four plus two times zero point five also gives five volts. Both readings agree, so the e m f is five volts, and the internal resistance is zero point five ohm. See how quick that was?

7:30Quick revision: V = E - Ir, V = E + Ir, short circuit E/r

So, quick revision. When a cell drives a circuit, I equals E over R plus r, and V equals E minus I times r. When it is being charged, V equals E plus I times r. On open circuit, V equals the e m f; on short circuit, I equals E over r; and the V I line has a slope of minus r, where r is the internal resistance. Now go and try the practice problem on this page. You can do it. All the best!