Alternating Quantities

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. The supply in your home is called two hundred and twenty volts A C, okay? But the voltage at the socket keeps rising, falling and reversing, and averaged over a full cycle, it is exactly zero! So why does a heater get hot at all? And where does two hundred and twenty come from? These ideas come up often in JEE, so let's begin.

0:26Alternating voltage and current

Alright, an alternating voltage varies sinusoidally with time. Small v equals capital V nought, times sine omega t. Small v is the voltage at any instant, small t is the time, and omega, the Greek letter, is the angular frequency. V nought, the V with a small zero below, is a constant. Now tell me, what is the largest value small v can reach? It is V nought, yes, exactly! Sine omega t never goes above one. So V nought is the peak value. Next, omega equals two pi f, where f is the frequency. And capital T, the period, the time for one full cycle, equals one over f. Clear? Now the current. It is small i equals capital I nought, times sine of omega t plus phi. I nought is the peak current, and phi, the Greek letter added to omega t, is something new. Right? So phi is the phase of the current relative to the voltage: how far the current runs ahead of, or behind, the voltage. Its value depends on what is in the circuit. With me?

1:35The average value of an alternating current

Now let's average this current over time. For half the cycle the current flows one way, and for the next half, the other way, with exactly the same shape. So what is the average current over one full cycle? It is zero, that's correct; the two halves cancel completely. On screen, the angle brackets around small i mean the average of i over time. Okay? But over the positive half cycle alone, the average is not zero. It is two I nought over pi, which is zero point six three seven times I nought.

2:09Root mean square: the effective value

Got it? So now, back to our puzzle from the start. If the average current over a cycle is zero, why does a heater get hot at all? Because heating does not care about direction, yes! The rate of heating in a resistor goes as i squared, and i squared is never negative. So we square the current, take the mean, then the square root. That gives the root mean square value, capital I with the subscript r m s. On screen, the square root sign sits over the average of i squared. Okay? For a sine wave, the average of sine squared omega t over a full cycle is exactly one half. The double arrow means which gives: I r m s equals I nought over root two, that is zero point seven zero seven times I nought. Right? So a steady current equal to I r m s produces the same heat in a resistor as our alternating current. That is why the r m s value is also called the effective value. Clear? But careful, there is one condition. I nought over root two holds only for a sinusoidal current, because we used the average of sine squared. For any other shape, go back to the definition: square, take the mean, then the root. Got it?

3:25What 220 volts AC really means

Now the other half of our puzzle. Meters made for A C read r m s values, okay? So the reading tells you the heating effect directly. So two hundred and twenty volts A C means V r m s is two hundred and twenty volts; the capital V after the number is volts. Now, what is the peak voltage of this supply? The peak, V nought, is two hundred and twenty times root two, and the wavy equals sign means approximately, so about three hundred and eleven volts! So the socket swings up to three hundred and eleven volts, but heats like a steady two hundred and twenty volts. Puzzle solved, right?

4:06When the current is not a pure sine

First, a steady part plus a sine part, which comes up often. Small i equals a, plus b times sine omega t, where the letters a and b are constants. Then I r m s is the square root of a squared, plus b squared over two. We will see why in example two. Clear? And a sine plus a cosine: a times sine omega t, plus b times cos omega t. They combine into one pure sine whose peak is the root of a squared plus b squared, so the r m s value is that peak over root two. With me?

4:42Example 1: peak, rms, frequency and period of 311 sin 314t

Now, example one. An alternating voltage is small v equals three hundred and eleven times sine of three hundred and fourteen t, in volts. Find the peak value, the r m s value, the frequency and the period. Pause the video and try it first. Done? Okay, let's check. The number in front of the sine is the peak value, so V nought is three hundred and eleven volts. Easy, right? Next, V r m s is three hundred and eleven over root two, which is about two hundred and twenty volts. Look, that is our home supply again! Okay? Now the frequency. Many students see three hundred and fourteen next to t, and write three hundred and fourteen hertz. Is that right? No, no, wait, that is the trap! Three hundred and fourteen is omega, the angular frequency, in radians per second. So f is omega over two pi, that is three hundred and fourteen over two pi, about fifty hertz, H z on screen. Got it? Finally, the period, capital T, is one over f, one fiftieth of a second, with small s for seconds. That is twenty milliseconds, written m s. See how quick that was?

5:59Example 2: rms of current i = 3 + 4 sin omega t

Now a harder one. Find the r m s value of the current small i equals three plus four sine omega t, in amperes, the capital A on screen. Pause the video and try it first. Done? Okay, let's check. The largest value of i is three plus four, which is seven amperes. So the r m s value is seven over root two, about four point nine five amperes, right? Oh no, careful, this is exactly the trap! I nought over root two is only for a pure sine, and this is a steady three plus a sine. So square the current: nine, plus twenty four sine omega t, plus sixteen sine squared omega t. Okay? Now average each term. Nine is a constant, so it stays nine. Then twenty four times the average of sine omega t, plus sixteen times the average of sine squared omega t. With me? So tell me, over a full cycle, what is the average of sine omega t, and what is the average of sine squared omega t? Zero and one half, yes, exactly! So the middle term vanishes, and sixteen times one half gives eight. The mean of i squared is nine plus zero plus eight, which is seventeen. Right? So I r m s is root seventeen, about four point one two amperes, not four point nine five. Remember, always square first, then average, then take the root, and this trap can never catch you.

7:32Quick revision: mean zero, rms for heating, I0 over root 2

So, quick revision. Over a full cycle the mean is zero, and the r m s value measures the heating. For a pure sine, I r m s is I nought over root two, and A C meters read r m s values. For other shapes, square, average, then take the root. Now try the practice problems on this page. All the best!