Combinations of Cells

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You have twelve identical cells, each of one point five volts with an internal resistance of zero point five ohm, and you want the largest possible current through a resistor of one point five ohm. How should you join them? All twelve in series gives the biggest e m f, so that must give the biggest current, right? Hmm, it is not that simple, and that is the point of today's lesson. Combinations of cells come up often in JEE circuit questions, so let's begin.

0:34Cells in series

So, let us begin with n identical cells. Each one has an e m f, capital E, and an internal resistance, small r, the resistance inside the cell itself. Okay? Now join them in series, the positive terminal of one to the negative terminal of the next. What is the total e m f, and what is the total internal resistance? Both simply add, yes, exactly! The e m f becomes n times E, and the internal resistance becomes n times small r. Clear? Now a classic mistake. Suppose k of these cells are put in the wrong way round. Those k cells no longer help, so the e m f is n minus k, times E, right? No, no, careful, that is exactly the trap! Each reversed cell does not just stop helping; it pushes against the others. So each one costs you two times E, and the e m f is n minus two k, in brackets, times capital E, with me? And now look at the internal resistance. The current still passes through every cell, reversed or not, so it stays n times small r. Got it?

1:47Cells in parallel

Now join m identical cells in parallel, all the positive terminals together, and all the negative terminals together. Is the e m f now m times E? No, it stays just E, the e m f of one cell, because the cells sit side by side between the same two points. Right? But their internal resistances are now in parallel, like m equal resistors. So the internal resistance becomes small r over m. Clear? So which is better, series or parallel? Suppose capital R, the external resistance, is much larger than small r. Which grouping gives more current? It is series, very good! When capital R is huge, the internal resistance hardly matters, so the larger e m f wins. The double greater than sign means much greater than. And when capital R is much less than small r, the double less than sign, parallel wins, because cutting the internal resistance matters most.

2:48Mixed grouping: m rows of n cells

Okay? Now we mix the two. Make m rows, with n cells in series in each row. Each row has an e m f of n times E, and a resistance of n times small r. Got it? Now join these m rows in parallel. Each row is like one big cell, so what is the e m f of the whole battery? Still n times E, exactly, just like identical cells in parallel! But the resistance is divided by m, so it becomes n times small r, over m. Right? Now connect capital R, the external resistance. The current, capital I, equals n times E, over capital R plus n times small r over m. That small fraction in the bottom is the internal resistance of the battery. With me? And here is the topper's trick. With the total number of cells fixed, the current is the largest when capital R equals n times small r over m. In words, the external resistance equals the internal resistance of the battery. Clear?

3:52Different cells in parallel

Now the last case, cells with different e m f values in parallel, like terminals together. Their internal resistances are in parallel, so one over r e q equals the sum of one over r i. The Greek capital sigma means add over all the cells, the small i below names each cell, and the small e q means equivalent. Okay? Now, what about the e m f of this combination? Many students simply add the e m f values, as in series. Is that correct? No, that works only in series. E e q equals the sum of E i over r i, divided by the sum of one over r i. So it is an average of the e m f values, each weighted by one over its own internal resistance, not their sum. Got it? And a cell turned the other way simply enters the top with a minus sign, as minus E i. Right?

4:47Example 1: four 1.5 V cells in series across 4 ohm, find I

Alright, example one, an easy one to warm up. Four cells, each of one point five volts and zero point five ohm, are joined in series and drive four ohm. The capital V after a number means volt, and the Greek letter omega means ohm. Find the current. Pause the video and try it first. Done? Okay, let's check. In series, the e m f is four times one point five, which gives six volts, where the cross means times. The internal resistance is four times zero point five, which gives two ohm. Clear? Now some students divide six volts by four ohm, and get one point five amperes, with the capital A for ampere, right? Oh no, careful, that forgets the internal resistance! So capital I equals six, over four plus two, which is six over six, and that gives one ampere. Okay?

5:45Example 2: best grouping of 12 cells for maximum current

Now example two, and this is our puzzle from the start. Twelve cells, each of one point five volts and zero point five ohm, must send the largest current through one point five ohm. How should they be grouped, and what is the current? Pause the video and try it first. Done? Okay, let's check. Take m rows, with n cells in each row. There are twelve cells in all, so n times m equals twelve. Right? Now, which condition gives the largest current? External resistance equals internal resistance, yes! So n times zero point five, over m, equals one point five. That gives n over m equals three, so n equals three times m, where the double arrow means this gives. With me? Put that into n times m equals twelve. Three times m squared, with the small raised two, equals twelve, so m squared is four, and m is two. Then n is six, so we need two rows of six cells. There is our puzzle, solved, and it is not all twelve in series! Got it? Now let's find the current. Each row has six cells in series, so the e m f is six times one point five, which gives nine volts. The internal resistance is six times zero point five, over two, which gives one point five ohm, equal to the external resistance, as planned. Clear? So capital I equals nine, over one point five plus one point five. Nine over three gives three amperes, the largest current possible here. See how quick that was?

7:26Quick revision: series, parallel, mixed and unequal cells

So, quick revision. In series, the e m f is n times E and the resistance n times small r, and k reversed cells leave n minus two k, times capital E, okay? In parallel, the e m f stays E and the resistance is small r over m; for rows, the current is largest when the external resistance equals the internal resistance. For different cells in parallel, use the weighted average of the e m f values. Now go and try the practice problem on this page. You can do it. All the best!