Series and Parallel Combinations

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take three resistors, of two ohm, three ohm and six ohm. Join them side by side, in parallel. Will the total resistance be more than six ohm, somewhere in between, or less than two ohm? Most students say more, because we added more resistors, right? Hold on to that answer; we will settle it today. Series and parallel combinations come up often in JEE circuit questions, so let's begin.

0:32Series: same current, resistances add

Let us start with series. In series, the resistors are joined one after another, so the charge has only one path. Whatever charge goes through the first resistor must go through the next one too. So the same current I flows through each resistor. Okay? So tell me, with the same current through each resistor, what is the total resistance of the combination? Yes, exactly, they simply add! We call the resistors R one, R two, and so on, with a small one or two written below. The total voltage V is I times R one, plus I times R two, and so on, where the three dots mean and so on. The double arrow means this gives R s, the series resistance, with a small s for series. R s equals R one plus R two plus the rest. Clear? Now, let me ask you this. Two resistors are in series, one small and one large. Which one takes the bigger share of the voltage? The larger one, that is correct! The voltage divides in direct proportion to the resistances. So V one, the voltage across R one, equals R one over the sum R one plus R two, times V, the total voltage. See, its own resistance sits on top. With me? And since we keep adding resistances, R s is always more than the largest resistance in the chain. Right?

1:58Parallel: same voltage, reciprocals add

Now let us look at parallel. Here every resistor is connected between the same two points. So the same voltage V sits across each of them. Okay? But now the current has many paths, and it splits. Each branch carries V over its own resistance, and the branch currents add up to I, the total current. The total current is also V over R p, the parallel resistance, with a small p for parallel. So cancel V, and one over R p equals one over R one, plus one over R two, and so on. So in parallel, the reciprocals add. Got it? For two resistors, this gives a shortcut: R p equals R one times R two, over R one plus R two. Product over sum, easy to remember. But careful, this shortcut is only for two resistors. Clear? Now the current splits between two branches. Which branch carries the larger current, the one with the smaller resistance, or the one with the larger resistance? The smaller resistance, very good! More current takes the easier path. So the current divides inversely to the resistances. I one, the current in R one, equals R two over the sum R one plus R two, times the total current. Oh no, careful here: for the current, the other resistance sits on top, not its own. This is a common slip in the exam. Okay? And now our puzzle starts to make sense. Every new branch is one more path for the current, so the total resistance only goes down. R p is always less than the smallest resistance. Right?

3:36Two quick tools

Now, two quick tools that save time. First, take n equal resistors, and let R be the resistance of each one. In series, R s equals n times R, and in parallel, R p equals R over n. Okay? So what is the ratio of R s to R p? It is n times R, divided by R over n, which gives n squared, the small two on top. So R s is to R p as n squared is to one, and the colon means is to. This ratio comes up often. With me? Now for the second tool, take a wire of resistance R, and bend it into a circle. Pick two points on it. They divide the circle into two arcs. Let x be the fraction of the wire in one arc, so the other arc has the rest, one minus x. Clear? Now, between those two points, the two arcs are two separate paths, so they are in parallel. Their resistances are x times R, and one minus x, times R, on screen. Use product over sum, where the dot means times, and their sum is just R, the whole wire. So the resistance between the points is x, times one minus x, times R, got it?

4:50Example 1: 2, 3 and 6 ohm in series and in parallel

Alright, example one, and this is our puzzle from the start. Find the equivalent resistance of two ohm, three ohm and six ohm, first in series, and then in parallel. The Greek letter omega after each number is the symbol for ohm. Pause the video and try it first. Done? Okay, let's check. Series is easy, we just add. Two plus three plus six gives eleven ohm, more than six ohm, the largest. Right? Now for the parallel part. Some students use the shortcut here. The product is thirty six, the sum is eleven, so the answer is thirty six over eleven ohm. Is that correct? No, no, careful, that shortcut works only for two resistors! And thirty six over eleven is more than three, so it is bigger than two ohm, the smallest resistor, which is impossible in parallel. So we add the reciprocals: one half, plus one third, plus one sixth. Over a common denominator of six, that is three plus two plus one, over six, which gives exactly one. Okay? So R p is one ohm, less than two ohm, the smallest one. There is our puzzle, solved! Each resistor in parallel adds one more path for the current. Clear?

6:11Example 2: 12 ohm wire in a circle, quarter arc and diameter

Now a slightly harder one. There is a wire of resistance twelve ohm, bent into a circle. Find the resistance between two points separated by a quarter of the circumference, and then between the ends of a diameter. Pause the video and try it first. Done? Okay, let's check. The quarter arc is one fourth of the wire, so its resistance is one fourth of twelve, which is three ohm. The other arc is twelve minus three, which gives nine ohm. With me? So we have three ohm and nine ohm. Do we just add them, and get twelve ohm? No, that is exactly the trap! Both arcs start at one point and end at the other, so between these two points they are in parallel. Two resistors, so product over sum works. Three times nine is twenty seven, and three plus nine is twelve. Twenty seven over twelve gives two point two five ohm. Okay? Now let's check this with our circle tool. Here, x equals one fourth, and one minus x equals three fourths. One fourth times three fourths times twelve gives two point two five ohm again. Right? Finally, for the diameter, each half of the wire is six ohm. The two vertical lines on screen mean in parallel with. So six ohm in parallel with six ohm is six times six, over six plus six. That is thirty six over twelve, which gives three ohm. Got it?

7:44Quick revision: series adds R, parallel adds 1/R, circles

So, quick revision. Series means the same current, the resistances add, and the voltage divides in proportion to the resistances. Parallel means the same voltage, the reciprocals add, and the current divides inversely. Product over sum is only for two resistors, and for a circle, it is x, times one minus x, times R, okay? Now go and try the practice problems on this page. You can do them. All the best!