Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here are two different questions from this chapter, both about a coil and a resistor connected to a battery. The first one asks: when does the current reach half its final value? The second one asks: when does the resistor heat up at exactly the rate at which the coil stores energy? Two completely different questions, and yet, in terms of the time constant, they have the same answer. So, is that just a coincidence? Hold on to your guess. Today, we will first see how to use this unit's problem bank, and then solve both questions from it. Circuits like these come up often in JEE, so let's begin.

0:45What is in this chapter

So, first, let us see what this chapter is. It is the problem bank for the whole unit on electromagnetic induction and A C: mixed solved problems, in three parts. Okay? Part A has fourteen solved problems, at JEE Main level. Part B has twelve longer solved problems, at JEE Advanced level. With me? And Part B combines induction with mechanics and with circuits, so one problem can need ideas from several chapters at once. Right? And Part C has ten concept checks, on the ideas most often tested in statement form. Got it?

1:22Part A: a 30-minute test

Now, when should you attempt Part A of this chapter? On the very first day of the unit, just to have a look? No, not on the first day. The course says to attempt it only after Chapters one to six, okay? That is because its problems are mixed: Faraday's law, motional emf, L R and L C circuits, A C, LCR circuits and transformers, all together. In the exam, no question tells you which chapter it comes from. Right? So finish every chapter first, and then attempt Part A as a real thirty minute test: solve each problem yourself, and only then check the solution. Clear? Now, quick maths. Fourteen problems, and thirty minutes on the clock. So how much time can you give each problem? Thirty over fourteen gives about two point one, and the wavy equals sign means approximately equal. So you have a little over two minutes for each problem. Keep moving, and do not get stuck on one.

2:22Part C: two concept checks

With me? Now, a taste of Part C, with two checks from the course. The first one: does a changing magnetic flux always produce a current? The answer is, not always. It always produces an emf; that is Faraday's law. But a current flows only if the circuit is closed. Okay? Now the second check. Take an ideal inductor, a coil with no resistance, in a D C circuit. How does it behave just after the switch is closed, and how after a long time? At first, it acts as an open circuit, because the current through it cannot jump. And in the steady state, the current has stopped changing, so the ideal inductor is just a plain wire. And the current starting from zero is exactly where our examples begin. Got it?

3:13Example 1: time for half the final current when tau is 2 ms

Now, Example one, and it is from Part A, okay? An L R circuit has a time constant of two milliseconds; the m s on screen means milliseconds. How long after switching on does the current reach half its final value? This is the first question from our puzzle. Pause the video, and try it first. Done? Okay, let's check. The left graph shows the current, small i, growing with time, small t; the right one shows decay, which we do not need today. i equals i nought, the i with a small zero below, times one minus e to the power minus t over tau, where e is the base of natural logarithms. And i nought, the final current, is curly E over R, the battery's emf over the resistance. With me? Tau, the Greek letter, is the time constant: L over R, the self-inductance over the resistance. Here it is two milliseconds. Right? Now, many students say: half the current, so half the time constant, which is one millisecond. Is that right? No, no, careful, that is the trap! The current does not grow in a straight line. Look at the curve: it rises fast at first and then keeps slowing down, and at one tau it is already at zero point six three times i nought, the dashed line. So set i equal to i nought over two, as on screen: i nought times one minus e to the power minus t over tau, equals i nought over two. Clear? Cancel i nought on both sides, and we get e to the power minus t over tau equals one half. Now take the natural log of both sides. The double arrow on screen means "which gives", and we get t equals tau times l n two, where l n stands for the natural log. Okay? The natural log of two is zero point six nine three. So t equals two times zero point six nine three, which is about one point three nine milliseconds. So the answer is one point three nine milliseconds, and not one millisecond. Got it?

5:14Example 2: heating rate equals inductor energy storing rate

Now Example two, from Part B, the harder part. Again, a coil and a resistor are connected to a battery. At what time is the rate of heating in the resistor equal to the rate at which energy is being stored in the inductor? Yes, this is the second question from our puzzle. Pause the video, and try it first. Done? Okay, let's check. The resistor heats at the rate i squared R, right? The inductor stores energy one half L i squared, so it stores energy at the rate d over d t of that, which is L times i times d i over d t. Okay? Now set the two rates equal: i squared R equals L times i times d i over d t. Once the current has started, i is not zero; the equals sign with a slash through it means not equal. So cancel one i from both sides: i R equals L times d i over d t. Clear? But now we have two unknowns, i and d i over d t. What else do we know about this circuit that links them? The battery, yes, exactly! Going around the circuit, its emf is shared between the resistor and the coil: curly E equals i R plus L times d i over d t. So L times d i over d t equals curly E minus i R, with me? Now put this back into our equation: i R equals curly E minus i R, okay? So two times i R equals curly E, and i equals curly E over two R, as on screen. But curly E over R is i nought, the final current. So the current is exactly i nought over two, half its final value! Right? And when is the current half its final value? We found that in Example one: at t equals tau times l n two. So that is the answer. Now, our puzzle from the start: it was no coincidence at all. Both questions ask for the moment the current is half its final value. Beautiful, isn't it?

7:16Quick revision: Part A test, both LR answers at tau ln 2

So, quick revision. Part A is a thirty minute test after Chapters one to six; Part B is at JEE Advanced level; and Part C has the concept checks. In an L R circuit, the current reaches half its final value at tau times l n two, not at tau over two. And the resistor heats as fast as the coil stores energy just when the current is half its final value, so again at tau times l n two. Now set a thirty minute clock, open Part A, and begin. All the best!