Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take two bulbs made for the same voltage, one rated forty watt and the other rated one hundred watt. Join them in series across that voltage. Which one glows brighter? The one hundred watt bulb, obviously, right? Hold on to that answer; our last example settles it. This is part A of our current electricity problem bank, at JEE Main level: three quick tools, then two full problems. So, let's begin.

0:33Tool 1: series and parallel

So, tool number one. Here, one ohm, two ohm and three ohm are combined all in series, then all in parallel; the capital omega on screen, the Greek letter, means ohm. Find the ratio of the two results, okay? In series, the same current flows through each, so the resistances add. R s, the R with a small s below, is one plus two plus three, which gives six ohm. Right? Now tell me, in parallel, will the result be more than one ohm, or less than one ohm? It is less, that is correct! In parallel, R p, the R with a small p below, is always less than the smallest resistance, and here the smallest is one ohm. With me? In parallel, every resistor has the same voltage, and the reciprocals add. One over R p equals one over one, plus one over two, plus one over three, which gives eleven over six. Clear? So flip it: R p is six over eleven ohm, less than one ohm, as we said. The ratio, R s over R p, is six times eleven over six, which gives eleven. Got it?

1:44Tool 2: a wire drawn longer

Now, tool number two. Here, a wire is drawn to double its length. By what factor does its resistance change? Twice the length, so twice the resistance, right? No, no, careful, that is the trap! We start from R equals rho l over capital A, where l is 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. When a wire is drawn out, the amount of metal stays the same, so the volume, l times capital A, stays fixed. Okay? So if the length doubles, the area must halve. The arrow on screen means becomes, and the double arrow means which gives. So l becomes two times l, which gives, capital A becomes capital A over two. Right? Now R prime, the R with a small dash, is the new resistance. R prime equals rho times two times l, over capital A over two. That is four times rho l over capital A, which is four times R, four times the old resistance. Clear? So at constant volume, R is proportional to l squared; the sign like an open alpha means is proportional to. The resistance becomes four times, not two times. With me?

3:04Tool 3: galvanometer to voltmeter

Now, tool number three. Here, a one hundred ohm galvanometer gives full scale deflection at one milliampere, written m A, and it is to read ten volts, the V after the ten. Find the series resistance. Look at the figure, where the circle G is the galvanometer. On the left, a small shunt S in parallel turns it into an ammeter; on the right, a large resistance R in series turns it into a voltmeter, and each has its formula below. Today we need the voltmeter. Okay? So G, the galvanometer's resistance, is one hundred ohm. And I g, the I with a small g below, is its full scale current, which is one milliampere, that is, ten to the power minus three ampere, the capital A on screen. Right? At full scale, the current I g flows through G and R, one after the other. So I g times, G plus R, equals V, the ten volts we want to read. With me? So the series resistance is ten over zero point zero zero one, which gives ten thousand ohm, right? No, no, wait, ten thousand ohm is the resistance of the whole meter! The galvanometer's own one hundred ohm is part of it. So R equals V over I g, minus G, exactly the voltmeter formula in the figure: ten over zero point zero zero one, minus one hundred. Clear? Ten over zero point zero zero one gives ten thousand, and ten thousand minus one hundred gives nine thousand nine hundred ohm, in series. It is large, as a voltmeter's series resistance should be. Got it?

4:42Example 1: EMF and internal resistance from two readings

Alright, example one. Here, a cell reads one point eight volts when it gives one ampere, and one point six volts when it gives two amperes. Find E, its e m f, and small r, its internal resistance. Pause the video and try it first. Done? Okay, let's check. The reading is V, the terminal voltage. While the cell drives a current I, the voltage I times r is lost inside it, so V equals E minus I times r, okay? Many students say the e m f is simply the bigger reading, one point eight volts. Is that right? No, it is not, and that is the trap! Even at one ampere, some voltage is lost inside the cell. So write V equals E minus I times r for each reading. We get one point eight equals E minus one times r, and one point six equals E minus two times r, with me? Now subtract the second equation from the first. The e m f cancels, and we are left with zero point two equals r, so the internal resistance is zero point two ohm. Clear? Put it back into the first equation: E equals one point eight plus one times zero point two, which gives two volts. So the e m f is two volts, more than both readings, and the internal resistance is zero point two ohm.

6:08Example 2: 40 W and 100 W bulbs in series, power ratio

Got it? And now, our puzzle from the start, in the bank's words. Here, a forty watt bulb and a one hundred watt bulb, with the same rated voltage, are in series on that voltage; the W after each number stands for watt. Find the ratio of their powers. Pause the video and try it first. Done? Okay, let's check. The rating, P nought at V nought, fixes the resistance; the small zero below marks the rated value. So R equals V nought squared over P nought. Both bulbs have the same V nought, so R is proportional to one over P nought. Okay? So tell me, which bulb has the bigger resistance, the forty watt one or the one hundred watt one? The forty watt bulb, yes, exactly! R forty, the R with a small forty below, over R one hundred, equals one hundred over forty, which gives five over two. The lower rating means the bigger resistance. Right? Now, in series, the same current I flows through both bulbs. So the convenient form is P equals I squared R, and with the same current, the power is proportional to R, the resistance. With me? So the powers are in the same ratio as the resistances. P forty to P one hundred, with the colon on screen meaning to, equals R forty to R one hundred, which is five to two. Clear? So the forty watt bulb glows brighter, and our puzzle is solved! In series, the lower rated bulb gets the bigger share of the power.

7:46Quick revision: resistors, voltmeter, V = E - Ir, bulbs

Got it? So, quick revision. In series the resistances add, in parallel the reciprocals add, and a drawn wire keeps its volume, so R is proportional to l squared. For a voltmeter, the series R equals V over I g, minus G, the galvanometer's resistance. For a cell driving a current, V equals E minus I times r, okay? And for bulbs of the same rated voltage in series, the lower rating glows brighter. Now try the practice problem on this page. All the best!