Mixed Problems: Set 1 (Current, Resistivity and Networks)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a five ohm wire and stretch it to three times its length. Three times the length, so three times the resistance, fifteen ohm, right? Hold on to that answer. Today is our first set of mixed problems, on current, resistivity and networks. Problems like these come up often in JEE Main. So, let's begin.

0:27Idea 1: stretching a wire

Idea number one is our puzzle. Here, a five ohm wire is stretched to three times its length; the capital omega on screen, the Greek letter, means ohm. Find its new resistance. We start from R equals rho l over capital A, where R is the resistance, l the length, and capital A the area of cross-section. Rho, the Greek letter, is the resistivity; it depends only on the material and the temperature, so stretching leaves it unchanged. Okay? So the length becomes three times. But when you pull a wire longer, what else changes? Yes, exactly, the area! The wire gets longer and thinner, but the amount of metal stays the same. So the volume, l times capital A, stays fixed. Right? So capital A equals the volume over l, and the double arrow on screen means which gives. Put that in, and R equals rho l squared, over the volume. Rho and the volume are fixed, so R is proportional to l squared; the sign like an open alpha means is proportional to. With me? Now, the arrow from l to three l means the length becomes three times l, and R prime, the R with a small dash, is the new resistance. Three squared is nine, and nine times five gives forty five ohm. So our puzzle's answer is forty five ohm, not fifteen. Clear?

1:52Idea 2: drift speed and the area trap

Now, idea number two: drift speed. Here, a wire of cross-section five square millimetres, written m m squared, carries five amperes; the A after the five stands for ampere. And n, the number of free electrons per unit volume, is five times ten to the power twenty eight per cubic metre, written m to the power minus three. Find the drift speed. We use I equals n e capital A v d, where I is the current, e the electron's charge, capital A the area, and v d, the v with a small d below it, the drift speed. So v d equals I over n e capital A, okay? Before we put in numbers, the area must be in square metres. One millimetre is ten to the power minus three metres, so five square millimetres is five times ten to the power minus three square metres, right? No, no, careful, that is the trap! Square millimetres means we square the ten to the power minus three as well, and that gives ten to the power minus six. So capital A is five times ten to the power minus six square metres, not ten to the power minus three. Got it? Now put everything in, with e, the electron's charge, equal to one point six times ten to the power minus nineteen. Then v d equals five, over, five times ten to the power twenty eight, times one point six times ten to the power minus nineteen, times five times ten to the power minus six. With me? In the bottom, five times one point six times five gives forty, and the powers of ten give ten to the power three. So the bottom is four times ten to the power four. Five over that gives one point two five times ten to the power minus four metres per second, written m s to the power minus one. Clear?

3:36Idea 3: series and parallel

Now, idea number three: networks. Here, three ohm and six ohm in parallel are in series with nine ohm. Find the equivalent resistance. First, the rules. In series, the same current flows through each, and R s, the R with a small s below, equals R sub one plus R sub two. Okay? In parallel, each has the same voltage across it, and the reciprocals add: one over R p, the R with a small p below, equals one over R sub one, plus one over R sub two. Right? For just two resistors in parallel, this gives R p equals R sub one times R sub two, over, R sub one plus R sub two. That is product over sum. With me? So tell me, three ohm and six ohm in parallel, what do you get? Yes, exactly! Product eighteen, over sum nine, gives two ohm. The two vertical bars on screen mean in parallel with. Then the nine ohm is in series, so R equals two plus nine, which gives eleven ohm. Got it?

4:36Example 1: 16 ohm wire bent into a circle, quarter arc

Alright, let us try example one. Here, a wire of sixteen ohm is bent into a circle. Find the resistance between two points a quarter of the circumference apart. Pause the video and try it first. Done? Okay, let's check. The points are a quarter of the way round, so the resistance is a quarter of sixteen, which is four ohm, right? No, no, wait, the current can go both ways round the ring! The two points cut the ring into two arcs, one quarter and three quarters of it. Same wire, same rho, same area, so the resistance is proportional to the length. Okay? So the short arc is a quarter of sixteen, which is four ohm, and the long arc is three quarters of sixteen, which is twelve ohm. Right? Both arcs join the same two points, so they are in parallel. Product over sum: four times twelve, over, four plus twelve, is forty eight over sixteen, which gives three ohm. Got it? Now a topper's shortcut from the course. If the arcs are the fractions x and one minus x of the ring, the resistance is x times, one minus x, times R, the resistance of the whole ring. One quarter times three quarters times sixteen gives three ohm again, very good!

5:54Example 2: infinite ladder, 3 ohm links and 6 ohm rungs

Now a harder one. An infinite ladder has three ohm series links and six ohm rungs. Find its input resistance. Each section is a three ohm link along the line and a six ohm rung across, repeating forever. Pause the video and try it first. Done? Okay, let's check. We cannot add infinitely many resistors one by one. So let the input resistance be x, our unknown. Okay? Now the key question. Remove the first link and the first rung. What is the resistance of the ladder that is left? Yes, exactly, it is still x! Infinity minus one section is the same infinite ladder. So the whole ladder is the first three ohm link, in series with the six ohm rung in parallel with x, the rest. Clear? Product over sum again: six in parallel with x is six times x, over, six plus x. So x equals three, plus six times x over, six plus x. With me? Multiply both sides by six plus x and simplify. We get x squared minus three x minus eighteen equals zero. This factorises as, x minus six, times, x plus three, equals zero. Right? So x is six or minus three. Resistance cannot be negative, so the input resistance is six ohm. Remember the trick: removing one section of an infinite network leaves the same network, and that is the whole idea, okay?

7:28Quick revision: stretched wire, drift speed, ring, ladder

So, quick revision. First, a stretched wire keeps its volume, so R is proportional to l squared. Second, v d equals I over n e capital A, with the area in square metres: one square millimetre is ten to the power minus six square metres. Third, two in parallel give product over sum, and a ring gives two arcs in parallel. Fourth, removing one section of an infinite ladder leaves the same x, unchanged. Now try the practice problem on this page. All the best!