Symmetry, Balanced Bridges and Special Networks

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a ladder of resistors, two ohm in each series link and one ohm in each rung, going on forever. Infinitely many resistors, so infinite resistance, right? No, no, wait, the real answer is less than three ohm! Such networks come up often in JEE circuit questions, and today we learn the tools that crack them. So, let's begin.

0:28Equal potentials: remove or join

Here is our first tool, and the most useful one. Suppose two points in a network are at the same potential. Then there is no potential difference between them, okay? So tell me, if a resistor sits between two such points, how much current flows through it? Yes, exactly, it is zero! The current I equals V over R, the voltage across it over its resistance R, and here V is zero. So we may simply remove that resistor. Clear? And the opposite move works too: two such points may be joined by a wire, and nothing changes. Remove or join, whichever is simpler. Got it?

1:09The balanced Wheatstone bridge

The most famous case is the Wheatstone bridge on your screen. Four resistors, P, Q, R and S, form a diamond with corners A, B, C and D, as you see. The battery E is connected across A and C, and the middle arm B D has G, a galvanometer, a meter that detects current. With me? Now look at the two paths. The top path shares the battery voltage between P and Q, and the bottom path shares it between R and S, right? When both paths share it in the same ratio, that is, P over Q equals R over S, the points B and D sit at the same potential, and we call the bridge balanced, okay? You can see this in the note beside the figure too. So V B, that is V with a small B below, the potential at B, equals V D, the potential at D, right? And I G, that is I with a small G below, the current through the galvanometer, is zero. Clear? So here is my question for you. In a balanced bridge, the arm B D carries no current at all. Does the resistance sitting in that arm change anything? That's right, it changes nothing at all! Arm B D carries no current, so we may remove it, or short it with a plain wire, and no current in the network changes. Remember this one, it will save you in the first example. Got it?

2:35Infinite ladder: the ladder hides inside itself

Now the infinite ladder in general. Each section has a series link R one, that is R with a small one below, and a rung R two, with a small two below. Okay? Here is the clever question. If we cut off the first section, one link and one rung, what is left? Yes, exactly, the same infinite ladder! Forever minus one section is still forever. So if the letter x stands for the resistance of the whole ladder, what is left also has resistance x. Got it? Now put the first section back. The rest of the ladder, x, is in parallel with the rung R two, and that pair is in series with the link R one. So x equals R one, plus R two times x, over R two plus x, product over sum. With me?

3:23Results to remember: the cube and star–delta

Next, results to learn by heart. Twelve equal resistors R sit on the edges of a cube. Across a body diagonal, the opposite corners, the resistance is five sixths of R; across a face diagonal, three fourths of R; and across one edge, seven twelfths of R, okay? Now the star and the delta. Three resistors in a triangle between points a, b and c form a delta, and three meeting at one centre form a star. The arrow means we convert the delta into a star. Right? Here R a b, with a small a b below, is the delta resistor between points a and b. The star arm R a equals the product of the two delta resistors at point a, R a b times R c a, over the sum of all three. Cyclically means the same rule for b and c. Clear? So, a quick question for you. If all three delta resistors are equal to R, what is each arm of the star? Yes, it is R times R, over three R, which is R over three! So an equal delta of R becomes an equal star of R over three. Got it?

4:35Example 1: balanced bridge, 5 ohm middle arm, find R AC

Alright, example one. This bridge has P equal to two ohm, Q four, R three, S six, and five ohm in the middle arm; the Greek letter omega is the symbol for ohm. Find the equivalent resistance between A and C, okay? Pause the video and try it first. Done? Okay, let's check. Many students first ask where the five ohm goes, in series or in parallel. But what should you check first? Yes, check whether the bridge is balanced! P over Q is two over four, which is one half, and R over S is three over six, also one half. Equal ratios, so the bridge is balanced. Right? So the five ohm in arm B D carries no current at all; it is a decoy, and that is the trap here. Just remove it. Okay? Now, P and Q in series give two plus four, which is six ohm, and R and S give three plus six, which is nine ohm. Each path sits in its own round brackets, and the two vertical lines mean in parallel with. So, six times nine, over six plus nine. With me? That is fifty four over fifteen, which gives three point six ohm. So RAC, with a small A C below, the resistance between A and C, is three point six ohm. Clear? Now a quick check with our first tool. B and D are at the same potential, so we may join them instead. Two and three in parallel give one point two, four and six give two point four, and together that makes three point six ohm again. Got it?

6:17Example 2: infinite ladder, 2 ohm links, 1 ohm rungs

Now a harder one, and it is our puzzle from the start. Find the resistance of an infinite ladder with two ohm in each series link and one ohm in each rung. Pause the video and try it first. Done? Okay, let's check. Here R one is two and R two is one, so x equals two, plus one times x, over one plus x, where the dot means times. That is two, plus x over one plus x. Okay? Now multiply both sides by one plus x. We get x times the bracket one plus x, equals two times the bracket one plus x, plus x. Collect the terms; the double arrow means this gives x squared, with a small two on top, minus two x, minus two, equals zero. Right? Now the quadratic formula; the sign with a plus over a minus means plus or minus. x equals two, plus or minus the square root of four plus eight, all over two. Root twelve is two root three, so x equals one plus or minus root three. With me? So we get two roots, one plus root three, and one minus root three. Which one do we keep? We keep the positive one, of course! One minus root three is negative, and a resistance can never be negative; that is the trap. So x is one plus root three ohm, about two point seven three ohm; the wavy equals sign means approximately. Infinitely many resistors, still less than three ohm. Our puzzle is solved!

7:52Quick revision: balanced bridge, ladder, cube, star-delta

So, quick revision. Equal potentials: remove the resistor between them, or join them by a wire. Balanced bridge: P over Q equals R over S, and the middle arm drops out. Infinite ladder: cut off one section, the same ladder is left, and that gives the equation for x. Cube: five sixths, three fourths and seven twelfths of R, and an equal delta of R becomes a star of R over three. You did really well today. Now try the practice problems on this page. All the best!