Maximum Power, Fuses and Transmission

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Power stations send electricity to our cities through long wires, at thousands of volts. But high voltage is dangerous, right? So why not use a lower, safer voltage? There must be a strong reason, and today we will find it. On the way, we will get the most power out of a battery, and see why a fuse wire melts. Questions on maximum power come up often in JEE, so let's begin.

0:32How much power reaches the load?

So, take a battery with an e m f, capital E, and an internal resistance, small r, the resistance inside the battery. Across it we connect a load of resistance capital R, outside the battery. With me? The current I is E over the sum, capital R plus small r, and the load gets a power I squared times capital R, right? So P equals E squared times capital R, over capital R plus small r, the whole squared. Now tell me, to take the most power from this battery, should the load be tiny or huge? It is neither of them! If capital R is almost zero, almost no power is used in it. If capital R is huge, the current is almost zero, so again almost no power. The peak is in between, exactly at capital R equals small r. Then the maximum power, P with a small max below, is E squared over four times small r, okay? But the same current also flows through small r, inside the battery. So at maximum power, what fraction is wasted inside? It is exactly half, yes! Same current, and capital R equals small r, so I squared small r inside equals I squared capital R outside. The efficiency is only fifty percent; that is the percent sign on screen. So maximum power does not mean maximum efficiency. Clear?

1:57Same power from two different loads

Now, one more result. Take any power P less than P max; the pointed sign means less than. Two different loads take this power, R one and R two; the small one and two below just name them. Got it? Let me ask you this: is there a simple link between these two loads? Yes, there is a link! R one times R two equals small r squared. The power formula gives a quadratic in capital R, and the product of its two roots is small r squared. We will check this in example two.

2:32Fuse wire: when does it melt?

Okay? Now, let us look at the fuse wire. Be careful, small r is now the radius of the wire, not an internal resistance. The resistance of the wire is rho times l, over pi r squared, where rho, the Greek letter, is the resistivity, and l is the length. The heat produced each second is I squared times that. With me? Each second, the wire loses heat through its surface, in proportion to the surface area, two pi r l; the stretched alpha sign means proportional to. Right? In the steady state, heat produced equals heat lost. Equate them, and the length cancels. We get I squared over r squared on the left, and just r on the right, so I squared is proportional to r cubed. The double arrow means this gives: I is proportional to r to the power three over two. Okay? Now, a longer fuse wire of the same material and radius has more resistance, so it melts at a smaller current, right? No, no, careful, that is the trap! The longer wire produces more heat, but it also has more surface to lose heat, both in proportion to the length. So the fusing current is independent of the length.

3:46Why power lines use high voltage

Got it? Now, back to our puzzle. The power station sends a power P at a voltage V, so the line carries a current I equal to P over V, okay? Here capital R is the resistance of the line itself. So the power lost as heat in the line is I squared R, that is, P squared R, over V squared. Right? Now, some students write the loss as V squared over R, and get a bigger loss at a higher voltage! What is wrong with that? Look, V is the sending voltage, not the voltage across the line; the drop across the line is only I times R, okay? So V squared over R, with this V, is wrong here. Clear? So, the higher the voltage, the smaller the current, and the smaller the loss. That is why power lines use high voltage, and that solves our puzzle.

4:37Example 1: 100 kW over a 2 ohm line at 10 kV and at 1 kV

Alright, example one. One hundred kilowatt is sent through a line of two ohm; k W is kilowatt, and the Greek letter omega is the symbol for ohm. Find the power lost if it is sent at ten kilovolt, written k V, and at one kilovolt. Pause the video and try it first. Done? Okay, let's check. The current is the power over the voltage: ten to the power five watt, over ten to the power four volt. So the current is ten ampere; the letter A stands for ampere. Right? The loss is I squared R, ten squared times two, which gives two hundred watt, where W stands for watt. That is tiny, out of one hundred kilowatt. Clear? Now at one kilovolt, the voltage is ten times smaller, so the loss is ten times bigger, right? No, wait, it is much worse than that! Ten to the power five over ten to the power three gives a current of one hundred ampere, ten times more. With me? And the loss goes as the current squared. One hundred squared times two gives twenty thousand watt, or twenty kilowatt, a hundred times more. That is a huge waste, all lost as heat! Got it?

5:50Example 2: 12 V, 2 ohm battery gives 16 W, find the loads

Now a harder one. Here, a twelve volt battery, where this V stands for volt, of internal resistance two ohm delivers sixteen watt to a load. Find the possible loads, and the maximum power it can deliver. Pause the video and try it first. Done? Okay, let's check. Put E equal to twelve and small r equal to two in our power formula. So one hundred and forty four times capital R, over capital R plus two, the whole squared, equals sixteen. Right? Cross multiply, and one hundred and forty four times capital R equals sixteen times, capital R plus two, the whole squared. Divide by sixteen and expand to get capital R squared, minus five capital R, plus four, equals zero. Clear? Before you solve it, tell me, how many loads can take sixteen watt here? There are two of them, as we learnt! The quadratic factorises as capital R minus one, times capital R minus four, so the load is one ohm or four ohm. The question asks for the possible loads, so give both, okay? Now check our rule. R one times R two is one times four, which is four, and that is two squared, exactly small r squared. With me? Finally, the maximum power comes at capital R equals small r, that is, two ohm. P max is E squared over four times small r, that is, one hundred and forty four over eight, which gives eighteen watt. And since sixteen is less than eighteen, two loads were possible. Got it?

7:27Quick revision: P max at R = r, fuse current, line loss

So, quick revision. Maximum power, E squared over four times small r, comes at capital R equals small r, with only fifty percent efficiency; and two loads with the same power multiply to small r squared. The fusing current goes as the radius to the power three over two, whatever the length. And the line loss is P squared R, over V squared, so high voltage means small loss. Now go and try the practice problems on this page. All the best!