Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two bulbs, one of forty watt and one of one hundred watt, both made for the same voltage. Now join them in series, and put that same voltage across the pair. Which bulb takes more power, and so glows brighter? The hundred watt bulb, surely, right? Keep your guess. Today, we will first see how to use this unit's problem bank, and then we will settle this puzzle with a problem from it. Questions like this come up often in JEE, so let's begin.

0:36What is in this chapter

So, first, let us see what this chapter is. It is the problem bank for the whole unit on current electricity: mixed solved problems, in three parts. Okay? Part A has ten solved problems, at JEE Main level. Part B has eight problems at JEE Advanced level, and each one combines several ideas. With me? And Part C has six concept checks, on the statements that are often tested as true or false. And each part ends with one practice problem on the page, for you to solve. Got it?

1:10Part 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 seven, okay? That is because its problems are mixed: drift speed, resistance, cells, meters, the potentiometer, power, and R C circuits, where a resistor and a capacitor work together. In the exam, no question tells you which chapter it comes from. Right? So finish every chapter first, and then attempt it as a real thirty minute test: solve each problem yourself first, and only then check the solution. Clear? Now, quick maths. Ten problems, and thirty minutes on the clock. So how much time can you give each problem? Thirty over ten gives three, so you have three minutes for each problem. Keep that number in your head while you solve. With me?

2:08Part C: two statement checks

Now, a taste of Part C, with two checks from the course. The first one: why is a potentiometer preferred to a voltmeter for measuring the e m f, the electromotive force, of a cell? The answer is this: at balance, the potentiometer draws no current from the cell. So it measures the e m f itself, and not a terminal voltage, which is what you read once current flows. Okay? Now the second check. Does the balance of a Wheatstone bridge depend on the e m f of the battery? Put in a bigger battery, and does the balance change? The answer is no. The balance depends only on the ratio of the arm resistances, and not on the battery at all. Statements like this come up often, so read them carefully. Got it?

2:57Example 1: which glows brighter, 40 W or 100 W in series

Now, Example one, from Part A, and it is our puzzle from the start. Forty watt and one hundred watt bulbs, made for the same rated voltage, are in series on that voltage; the capital W stands for watt. Find the ratio of their powers. Pause the video, and try it first. Done? Okay, let's check. Many students say the hundred watt bulb is stronger, so it must take more power here too. Is that right? No, no, careful, that is exactly the trap, because those watts are rated values, at the full rated voltage. So first, find the resistances. At the rated voltage, capital V, the power is V squared over R, so R equals V squared over P rated, the P with the small word rated below it. The double arrow means implies. The voltage is the same for both bulbs, so R is proportional to one over P rated; the sign like a stretched Greek alpha means proportional to. Okay? So, R forty, the R with a small forty below it, over R hundred is one hundred over forty, which is five over two. The forty watt bulb has the larger resistance, more than twice as much. Right? Now, the two bulbs are in series. So, what is the same for both of them? The current, yes, exactly! With the same current, capital I, the power is I squared R, so P is proportional to R, okay? When the current is common, I squared R is the convenient form. Clear? So the powers are in the same ratio as the resistances: P forty to P hundred is five to two, where the colon means ratio. The forty watt bulb takes more power, so it glows brighter. Our puzzle is solved: if you picked the hundred watt bulb, you fell for the trap. Got it?

4:52Example 2: steady charge on a capacitor between two branches

Now Example two, and this one is from Part B, a hard one. Here, a twelve volt ideal battery feeds two branches; the capital V after the twelve means volt. The first branch has two ohm, then four ohm, with the junction B between them. The second has four ohm, then two ohm, with the junction D, okay? The Greek letter omega means ohm. Then a one microfarad capacitor joins B and D; the Greek letter mu with a capital F means microfarad. Find its steady charge. Pause the video, and try it first. Done? Okay, let's check. First, a question for you. Once everything settles, in the steady state, does any current flow through the capacitor? No, it does not. In the steady state, a capacitor carries no current. So we leave it out while finding the currents, and each branch is two plus four, which is six ohm. Clear? The battery is ideal, with no internal resistance, so the full twelve volts is across each branch. The current in each branch is twelve over six, which is two ampere; the capital A means ampere. Right? Now, many students say: both branches have the same two resistors, two ohm and four ohm, so the junctions B and D are at the same potential, and the charge is zero. Is that right? No, no, wait, that is the trap, because the order of the resistors matters! Start from the positive end of the battery. To reach B, the current crosses the two ohm first: a drop of two times two, which is four volts, and the cross on screen means times. To reach D, it crosses the four ohm first: a drop of two times four, which is eight volts. With me? So the junction D is below B by eight minus four, that is four volts. We write this as V B minus V D, the potential at B minus the potential at D, with the point named by the small letter below each V, and it equals four volts, not zero. Okay? Now the charge. Capital Q, the charge, equals capital C, the capacitance, times, in brackets, V B minus V D, the potential difference we just found: one microfarad times four volts, which gives four microcoulomb. Here, the mu with a capital C means microcoulomb. So the answer is four microcoulomb, not zero. Got it? And look, Part B's practice problem on the page continues this very problem. The capacitor is replaced by a two ohm resistor, and you have to find the current in it. Try it yourself, and check your answer on the page.

7:39Quick revision: test plan, P = I squared R, capacitor rule

So, quick revision. Part A is a thirty minute test after Chapters one to seven; Part B is at JEE Advanced level; and Part C has the concept checks. For bulbs of the same rated voltage in series, the current is the same, so the power is I squared R, and the lower-rated bulb takes more power. And in the steady state, a capacitor carries no current, so find its voltage from the drops along each branch. Now set a thirty minute clock, open Part A, and begin. All the best!