Joule Heating and Electric Power

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a sixty watt bulb and a forty watt bulb, both made for two hundred and twenty volts. Which one glows brighter? The sixty watt one, obviously, right? Yes, if you join them side by side, in parallel. But join them one after the other, in series, and the forty watt bulb glows brighter! How can the smaller bulb win? Let us find out. Questions on bulbs and heaters come up often in JEE, so let's begin.

0:35Current heats a resistor: power and heat

So, a current flowing through a resistor makes it hot. This is Joule heating, the heating effect of current. If V is the voltage across it and I is the current through it, the power delivered, P, equals V times I, okay? Now, a question for you. Let R be the resistance of the resistor. Can you write this power using only the current and R? Yes, exactly, we use Ohm's law for an ohmic resistor at constant temperature, V equals I times R, and P becomes I squared R; the small two above I means squared. Or put I equals V over R, and you get V squared over R as well. Three forms, one and the same power. Right? And the heat produced in a time t is H, which equals I squared times R times t. Power times time, nothing more. Clear?

1:30A rating fixes the resistance

Now, every bulb and heater carries a label, its rating: a power P nought at a voltage V nought. The small zero below each letter marks the rated value. Okay? Which of our three power formulas gives the resistance straight from the rating? Yes, it is V squared over R! Rearrange it, and R equals V nought squared over P nought. So the rating fixes the resistance of the bulb. With me? Now, we run the same bulb on half its rated voltage. Half the voltage, so half the power, right? No, no, careful, that is the trap! We take the resistance as constant, so the power goes as V squared. Half the voltage gives only one quarter of the power. In general, P equals P nought times V squared over V nought squared. Got it?

2:24Series or parallel: which bulb glows brighter?

Now, back to our puzzle. Two bulbs are joined in series. What is the same for both of them? Yes, exactly, it is the current! In series, the same current flows through both bulbs. So use I squared R, and the power is proportional to R; the sign on screen that looks like a stretched alpha means proportional to. Okay? Now look at the resistance. For bulbs rated for the same voltage, R equals V nought squared over P nought, so the lower-rated bulb has the larger resistance. So in series, the lower-rated bulb glows brighter. Right? Now join them in parallel. Here the voltage is the same for both, so use V squared over R; the power is proportional to one over R, okay? So in parallel the smaller resistance wins, and, again for the same rated voltage, the higher-rated bulb glows brighter. So the rule is: when the current is common, think I squared R; when the voltage is common, think V squared over R, every time. Clear?

3:28Total power, and the unit on your bill

Now the total power. Take two bulbs rated P one and P two at the same voltage, and use them on that voltage; the small one and two below just name the bulbs. With me? Now, in series, is the total power simply P one plus P two? No, it is not, and this catches many students! In series the total resistance goes up, so the power goes down. The total is P one times P two, over P one plus P two, which is less than either bulb alone. Okay? In parallel, each bulb works at its rating, so the total is simply P one plus P two. Right? And finally, the electricity bill. One kilowatt hour, written k W h, is one kilowatt used for one hour. That is three point six times ten to the power six joules, the J on screen, and on your bill it is called one unit. Got it?

4:20Example 1: 1000 W heater, resistance, current and kWh

Alright, example one. We have a heater rated one thousand watt, two hundred and twenty volt; the W and the V on screen stand for watt and volt. Find its resistance, its current, and the energy it uses in two hours. Pause the video and try it first. Done? Okay, let's check. Resistance first. R equals V nought squared over P nought, that is two hundred and twenty squared, over one thousand. That gives forty eight thousand four hundred over one thousand, so R is forty eight point four ohm; the Greek letter omega is the symbol for ohm. Clear? Now the current. We know the power and the voltage. Which formula gives the current fastest? Perfect, it is P equals V times I! So I equals P over V, that is, one thousand over two hundred and twenty, which gives about four point five five ampere. The letter A stands for ampere, and the wavy equals sign means approximately. Okay? Now the energy in two hours. Here is a topper's shortcut: skip joules and seconds completely. One thousand watt is one kilowatt, so the energy E is one kilowatt times two hours, which gives two kilowatt hours, or two units on the bill; the small h is for hour. Right? And for joules, multiply by three point six times ten to the power six. Two times that gives seven point two times ten to the power six joules. With me?

5:53Example 2: 60 W and 40 W bulbs in series on 220 V

Now our puzzle from the start, JEE style. Sixty watt and forty watt bulbs, both rated for two hundred and twenty volts, are joined in series across two hundred and twenty volts. Find the power of each. Which one glows brighter? Pause the video and try it first. Done? Okay, let's check. First, there is a trap here. Will the sixty watt bulb really give sixty watts here? No, it will not, because in series each bulb gets only part of the two hundred and twenty volts. So we start with the resistances. R sixty, that is R with a small sixty below, is two hundred and twenty squared over sixty, about eight hundred and six point seven ohm. And R forty is two hundred and twenty squared over forty, which gives one thousand two hundred and ten ohm. Okay? In series, the resistances add, to two thousand and sixteen point seven ohm. So the current is two hundred and twenty over two thousand and sixteen point seven, which gives about zero point one zero nine ampere, the same in both bulbs. Right? Now, for each bulb, we use I squared R with this same current. P sixty, the power in the sixty watt bulb, works out to nine point six watt. And P forty, the power in the forty watt bulb, works out to fourteen point four watt. Clear? So the forty watt bulb glows brighter, exactly as our rule said: same current, larger resistance, more power. And that solves our puzzle from the start. Well done if you got it! Now a quick check that toppers use. The total is nine point six plus fourteen point four, which gives twenty four watt. Our series formula gives sixty times forty, over sixty plus forty, that is two thousand four hundred over one hundred, which is twenty four watt too.

7:45Quick revision: power formulas, ratings, bulb brightness

Got it? So, quick revision. Power is V times I, or I squared R, or V squared over R, and the heat is I squared times R times t. The rating fixes the resistance, R equals V nought squared over P nought, and at constant resistance, the power goes as V squared. And for bulbs rated for the same voltage, in series the lower-rated bulb glows brighter, while in parallel the higher-rated one does. Now go and try the practice problems on this page. All the best!