WEBVTT 00:00:00.035 --> 00:00:02.543 In a direct current circuit, a resistor's 00:00:02.543 --> 00:00:06.897 current is the voltage divided by the resistance, and that is the end of it. 00:00:07.094 --> 00:00:10.331 Alternating current has a second question hiding inside it, 00:00:10.414 --> 00:00:13.515 and the whole of this lecture comes out of that question. 00:00:14.911 --> 00:00:19.161 Here is a sinusoidal source, and the load it is driving is a coil. 00:00:19.265 --> 00:00:22.377 A big motor winding, say, or a transformer. 00:00:22.574 --> 00:00:26.487 Time runs across, and both volts and amps are read up the side. 00:00:27.284 --> 00:00:31.093 The yellow curve is the voltage the source puts across that coil. 00:00:31.289 --> 00:00:37.560 It is a sine wave, of amplitude V m, going round at angular frequency omega. 00:00:38.356 --> 00:00:41.178 And here is the current the coil draws. 00:00:41.375 --> 00:00:46.078 Same shape, same frequency, and that is the first thing worth noticing: 00:00:46.274 --> 00:00:50.638 a sinusoidal source in a linear circuit gives you sinusoidal everything, 00:00:50.721 --> 00:00:52.741 at one single frequency. 00:00:53.283 --> 00:00:55.002 But look where it sits. 00:00:55.198 --> 00:00:56.938 The voltage reaches its peak here, 00:00:57.021 --> 00:01:00.911 and the current does not reach its own peak until later, over here. 00:01:01.824 --> 00:01:05.319 That gap is a quarter of a cycle. Ninety degrees. 00:01:05.423 --> 00:01:06.896 The current is running late, 00:01:06.979 --> 00:01:11.310 and in the language of the trade we say it lags the voltage by an angle phi. 00:01:11.735 --> 00:01:15.277 For a pure coil that lag is exactly ninety degrees. 00:01:15.473 --> 00:01:19.897 For a real motor, which has resistance in its windings as well, it will be somewhere 00:01:19.897 --> 00:01:24.008 between zero and ninety, and finding it will be our job shortly. 00:01:24.804 --> 00:01:26.953 So where does the lag come from? 00:01:27.150 --> 00:01:31.191 From one relation, and it is the defining property of an inductor. 00:01:31.503 --> 00:01:36.194 The voltage across a coil is L times the rate of change of the current through it. 00:01:36.991 --> 00:01:38.081 Read that carefully, 00:01:38.164 --> 00:01:41.320 because it is not a statement about how much current is flowing. 00:01:41.403 --> 00:01:44.828 It is a statement about how fast that current is changing. 00:01:45.625 --> 00:01:47.786 Now put the two together on the picture. 00:01:47.889 --> 00:01:51.466 Here is the instant the current is climbing at its very steepest. 00:01:51.662 --> 00:01:53.321 Look at the red tangent: 00:01:53.404 --> 00:01:56.261 that is the biggest slope anywhere on the green curve. 00:01:56.802 --> 00:02:00.878 And at that same instant, the yellow voltage is at its maximum. 00:02:01.074 --> 00:02:04.930 Steepest current, largest voltage, at the same moment. 00:02:05.982 --> 00:02:09.094 Now take the instant the current is at its own peak. 00:02:09.291 --> 00:02:12.914 Its slope there is zero: the red tangent is flat. 00:02:13.110 --> 00:02:16.524 And the voltage at that instant is passing through zero. 00:02:17.321 --> 00:02:19.931 So the voltage tracks the slope of the current, 00:02:20.014 --> 00:02:24.253 and the slope of a sine is a cosine, which is a quarter cycle ahead of it. 00:02:24.566 --> 00:02:26.561 The lag is not a quirk of the coil. 00:02:26.644 --> 00:02:28.223 It is calculus. 00:02:29.365 --> 00:02:30.561 Write it out. 00:02:30.874 --> 00:02:35.240 Take the current as I m sine of omega t minus ninety degrees. 00:02:35.344 --> 00:02:39.616 Differentiate it, multiply by L, and what comes back is a sine 00:02:39.616 --> 00:02:44.098 of omega t with no shift at all, which is the voltage we started with. 00:02:44.895 --> 00:02:47.958 And hold on to that factor out front, omega L. 00:02:48.041 --> 00:02:52.454 It has the units of ohms, and it is going to become half of the answer. 00:02:53.967 --> 00:02:55.522 Before we go on, let me tell you 00:02:55.522 --> 00:02:59.575 where this ends, so that every step has somewhere to be walking towards. 00:03:00.372 --> 00:03:04.795 First, we replace those two sinusoids by two arrows on a turning wheel. 00:03:04.899 --> 00:03:07.034 That picture is called a phasor diagram, 00:03:07.117 --> 00:03:10.299 and it is the single most useful thing in this subject. 00:03:11.095 --> 00:03:15.228 Second, the angle between those arrows turns Ohm's law into something that works 00:03:15.228 --> 00:03:17.748 for coils and capacitors as well. 00:03:17.945 --> 00:03:19.617 That is impedance. 00:03:20.530 --> 00:03:25.395 Third, we multiply voltage by current and find that the power splits in two. 00:03:25.592 --> 00:03:27.030 One part does work. 00:03:27.113 --> 00:03:29.922 The other part sloshes back and forth twice every cycle 00:03:29.922 --> 00:03:31.827 and delivers nothing at all. 00:03:32.531 --> 00:03:35.641 And fourth, we put one capacitor across the load, 00:03:35.724 --> 00:03:38.114 watch the current arrow swing back into line, 00:03:38.197 --> 00:03:42.865 and deliver exactly the same useful power with noticeably less current in the wires. 00:03:43.177 --> 00:03:47.625 That last one is worth real money to a factory, and we will see exactly why. 00:03:48.728 --> 00:03:52.083 Two sinusoids at the same frequency carry only three pieces 00:03:52.083 --> 00:03:53.802 of information between them: 00:03:53.999 --> 00:03:58.922 one amplitude, a second amplitude, and the angle by which one trails the other. 00:03:59.235 --> 00:04:03.229 Everything else is the same frequency ticking along underneath. 00:04:03.844 --> 00:04:07.618 So here is a picture that keeps exactly those three things. 00:04:08.414 --> 00:04:12.119 A yellow arrow, whose length is the amplitude of the voltage. 00:04:12.315 --> 00:04:16.856 And a green arrow, shorter, whose length is the amplitude of the current. 00:04:18.256 --> 00:04:22.622 The green one is set behind the yellow one by the phase angle phi, 00:04:22.725 --> 00:04:27.684 and that angle is fixed. It is built into the picture and it will not change. 00:04:28.294 --> 00:04:31.558 Drop each arrow tip onto the vertical axis. 00:04:31.661 --> 00:04:35.420 Those two dashed projections are the heights we are going to plot, 00:04:35.504 --> 00:04:37.560 on the time axes over here. 00:04:38.356 --> 00:04:43.535 Now spin both arrows, counterclockwise, at the angular frequency omega. 00:04:43.732 --> 00:04:48.028 They turn together, rigidly, like two spokes of one wheel. 00:04:48.825 --> 00:04:52.285 Watch what the projections trace out as the wheel turns. 00:04:52.482 --> 00:04:57.209 The yellow height rises, falls, goes negative, comes back. 00:04:57.405 --> 00:05:01.632 That is a sine wave, and here it is being drawn instant by instant. 00:05:02.684 --> 00:05:04.716 Keep going for a second turn. 00:05:05.331 --> 00:05:09.673 The green projection does the same thing, but it arrives at each landmark a quarter 00:05:09.673 --> 00:05:13.993 of a cycle later, because its arrow is a quarter of a turn behind. 00:05:14.790 --> 00:05:18.354 So the two waveforms we started this lecture with are the shadows 00:05:18.354 --> 00:05:20.874 of two arrows on one turning wheel. 00:05:20.978 --> 00:05:26.087 That is the whole content of the phasor idea, and everything else follows from it. 00:05:26.628 --> 00:05:29.125 And now the move that makes it useful. 00:05:29.321 --> 00:05:32.291 Both arrows turn at the same rate, forever. 00:05:32.375 --> 00:05:35.592 So the turning tells us nothing we did not already know. 00:05:35.695 --> 00:05:39.988 What carries all the information is the two lengths and the angle between them, 00:05:40.072 --> 00:05:42.105 and those never change. 00:05:43.316 --> 00:05:48.216 So stop the wheel. Photograph it at one instant and throw the rotation away. 00:05:48.412 --> 00:05:50.120 What is left is this. 00:05:50.661 --> 00:05:53.340 Line the voltage arrow up along the reference direction 00:05:53.424 --> 00:05:55.793 and call its angle zero. 00:05:55.990 --> 00:05:57.554 We are free to do that, 00:05:57.638 --> 00:06:01.192 because only the difference between the two angles ever mattered. 00:06:01.988 --> 00:06:05.623 The current arrow then sits below it, at minus phi. 00:06:05.819 --> 00:06:08.746 For an inductive load it is always below: 00:06:08.943 --> 00:06:10.093 lagging. 00:06:10.796 --> 00:06:12.678 Two arrows, frozen. 00:06:12.875 --> 00:06:18.514 That is a phasor diagram, and each arrow is a phasor: a length and an angle, 00:06:18.598 --> 00:06:21.002 standing in for a whole sinusoid. 00:06:22.217 --> 00:06:25.746 One more piece of bookkeeping before we use it. From here on, 00:06:25.746 --> 00:06:31.598 when I write V or I as a length, I mean the root mean square value, not the peak. 00:06:32.050 --> 00:06:37.321 Two forty volts from a wall socket is an r m s number, and using r m s 00:06:37.321 --> 00:06:41.479 throughout keeps the power formulas free of stray factors of one half. 00:06:42.275 --> 00:06:47.199 So we have replaced two moving curves by two still arrows, and lost nothing. 00:06:47.512 --> 00:06:51.889 Now let us put those arrows to work and get an Ohm's law that a coil obeys. 00:06:52.946 --> 00:06:54.978 A real motor winding is a resistance 00:06:54.978 --> 00:06:59.262 and an inductance in series, and one current flows through both of them. 00:06:59.575 --> 00:07:03.558 So let us draw that current first, and hang everything else off it. 00:07:04.471 --> 00:07:08.825 Across the resistance, voltage and current stay in step. 00:07:09.277 --> 00:07:12.548 So the resistor's voltage phasor lies flat along the current, 00:07:12.632 --> 00:07:15.222 and its length is R times I. 00:07:16.019 --> 00:07:20.745 Across the coil it is different, and we did the calculus for this in the opening. 00:07:21.197 --> 00:07:25.632 The coil's voltage leads its current by ninety degrees, so its phasor stands 00:07:25.632 --> 00:07:29.835 square to the current, and its length is omega L times I. 00:07:30.632 --> 00:07:35.656 That quantity omega L is called the inductive reactance, X sub L, 00:07:35.740 --> 00:07:38.667 and like resistance it is measured in ohms. 00:07:38.863 --> 00:07:43.810 The j in front of it is the algebra's way of saying: turn it ninety degrees. 00:07:44.351 --> 00:07:46.159 The voltage law still holds, 00:07:46.243 --> 00:07:49.982 provided we add the phasors as vectors rather than as numbers. 00:07:50.179 --> 00:07:50.945 So the source 00:07:50.945 --> 00:07:55.985 voltage is the resistor's voltage and the coil's voltage joined head to tail. 00:07:57.037 --> 00:08:01.264 And there is the whole geometry of an inductive load in one triangle. 00:08:01.461 --> 00:08:03.733 The source voltage is the hypotenuse, 00:08:03.817 --> 00:08:06.891 and it sits ahead of the current by an angle phi. 00:08:06.975 --> 00:08:09.786 That angle is what we are about to name. 00:08:11.182 --> 00:08:13.795 Factor the current out of both terms. 00:08:13.991 --> 00:08:18.041 What is left multiplying it is R plus j X sub L, 00:08:18.125 --> 00:08:21.597 and that combination is the impedance, bold Z. 00:08:22.138 --> 00:08:27.491 It is a complex number, and its two parts are exactly the two sides of the triangle. 00:08:27.688 --> 00:08:32.251 Its magnitude is the square root of R squared plus X squared. 00:08:32.861 --> 00:08:35.007 And its angle is the arctangent of X 00:08:35.091 --> 00:08:39.735 over R, which is the same phi we have been carrying since the first waveform. 00:08:40.187 --> 00:08:44.286 The reactance sets the phase; the resistance holds it back. 00:08:45.199 --> 00:08:49.681 So Ohm's law survives into alternating current unchanged in form. 00:08:49.785 --> 00:08:53.315 Voltage phasor equals impedance times current phasor. 00:08:53.511 --> 00:08:56.504 The only thing that has grown is that all three are complex, 00:08:56.588 --> 00:08:58.818 and the angles carry the timing. 00:09:00.215 --> 00:09:01.686 Let us put numbers on it, 00:09:01.770 --> 00:09:05.161 because these numbers will follow us to the end of the lecture. 00:09:05.358 --> 00:09:07.967 A motor on a two hundred and forty volt supply, 00:09:08.051 --> 00:09:12.789 with eight ohms of winding resistance and six ohms of reactance. 00:09:13.841 --> 00:09:17.228 The triangle is now measured in ohms rather than volts, 00:09:17.312 --> 00:09:21.992 which is the same picture divided through by the current. Eight along, six up. 00:09:22.533 --> 00:09:27.697 So the magnitude of the impedance is the square root of sixty four plus thirty six, 00:09:27.781 --> 00:09:30.088 which is the square root of a hundred: 00:09:30.172 --> 00:09:31.590 ten ohms. 00:09:32.386 --> 00:09:36.137 Now the current is two hundred and forty volts over ten ohms. 00:09:36.334 --> 00:09:37.890 Twenty four amps. 00:09:38.087 --> 00:09:41.687 That is the number a clamp meter on the supply lead would read. 00:09:42.483 --> 00:09:45.871 And the phase angle is the arctangent of six over eight, 00:09:45.955 --> 00:09:48.882 which is thirty six point eight seven degrees. 00:09:49.078 --> 00:09:51.668 The current lags the voltage by that much. 00:09:52.279 --> 00:09:53.440 Impedance ten ohms 00:09:53.440 --> 00:09:58.650 at thirty seven degrees, current twenty four amps lagging by thirty seven degrees. 00:09:58.734 --> 00:10:00.198 Hold on to both. 00:10:00.394 --> 00:10:03.155 In the next section we multiply the voltage by the current, 00:10:03.239 --> 00:10:05.968 and that angle is about to cost somebody money. 00:10:07.056 --> 00:10:11.202 Power at any instant is voltage times current at that instant. 00:10:11.514 --> 00:10:17.250 No phasors, no averages: just the product of the two numbers, moment by moment. 00:10:18.047 --> 00:10:22.447 Here are our two waveforms again, the voltage in yellow and the current lagging 00:10:22.447 --> 00:10:24.828 it by thirty seven degrees in green. 00:10:25.532 --> 00:10:27.817 And here in magenta is their product, 00:10:27.900 --> 00:10:31.082 the instantaneous power flowing into the motor. 00:10:31.879 --> 00:10:33.435 Two things about it. 00:10:33.887 --> 00:10:38.113 First, it oscillates at twice the supply frequency, because it is a product 00:10:38.113 --> 00:10:42.526 of two sinusoids: two humps for every one cycle of the voltage. 00:10:43.322 --> 00:10:46.919 Second, and this is the interesting one, it goes negative. 00:10:47.002 --> 00:10:49.834 Here, and here again half a cycle later. 00:10:49.917 --> 00:10:54.329 Negative power means energy flowing out of the motor and back into the supply. 00:10:54.812 --> 00:10:56.310 The coil's magnetic field 00:10:56.310 --> 00:11:01.024 is storing energy on one part of the cycle and handing it straight back on the next. 00:11:01.221 --> 00:11:03.332 Nothing is consumed by that exchange. 00:11:03.415 --> 00:11:04.914 It just sloshes. 00:11:05.966 --> 00:11:08.750 Let us separate the sloshing from the working. 00:11:08.833 --> 00:11:11.943 Expand the product of the two sines with the usual identity, 00:11:12.026 --> 00:11:14.558 and it falls into exactly two terms. 00:11:15.099 --> 00:11:16.316 Look at the first term. 00:11:16.399 --> 00:11:22.040 It is V I cosine phi, multiplied by one minus cosine of two omega t. 00:11:22.123 --> 00:11:25.929 That bracket never goes negative: it swings between zero and two, 00:11:26.012 --> 00:11:27.801 and its average is one. 00:11:28.226 --> 00:11:31.626 So that term has an average value of V I cosine phi, 00:11:31.709 --> 00:11:33.834 and it is always into the load. 00:11:33.938 --> 00:11:37.770 That is energy actually leaving the supply and not coming back. 00:11:38.567 --> 00:11:43.444 The second term is V I sine phi times sine of two omega t. 00:11:43.640 --> 00:11:46.599 That is a pure sinusoid at double frequency, 00:11:46.682 --> 00:11:48.935 sitting symmetrically about zero. 00:11:49.132 --> 00:11:52.407 Its average over a cycle is exactly nothing. 00:11:54.221 --> 00:11:56.765 Here are those two terms drawn separately. 00:11:56.961 --> 00:11:59.191 The blue one is the working term. 00:11:59.388 --> 00:12:03.507 Notice that it never dips below the axis: it is one-directional, 00:12:03.590 --> 00:12:05.611 always from supply to load. 00:12:06.408 --> 00:12:11.761 Its average is this dashed level, and that average is what we call the real power P. 00:12:11.865 --> 00:12:16.080 V times I times cosine phi, measured in watts. 00:12:16.277 --> 00:12:19.992 That is the number that turns the shaft and heats the windings. 00:12:20.905 --> 00:12:22.729 And here is the red one. 00:12:22.925 --> 00:12:27.222 Equal time above the axis and below it, twice every cycle. 00:12:27.674 --> 00:12:31.018 Energy out, energy back, out, back. 00:12:31.443 --> 00:12:33.487 Its average is zero. 00:12:33.684 --> 00:12:37.388 Over any whole number of cycles it delivers not one joule. 00:12:37.840 --> 00:12:39.952 But its amplitude is not zero, 00:12:40.035 --> 00:12:46.816 and that amplitude has a name: reactive power Q, V I sine phi, measured in var. 00:12:47.612 --> 00:12:50.780 And the plain product of the voltage and current magnitudes, 00:12:50.863 --> 00:12:55.622 with no angle in it at all, is the apparent power S, in volt amps. 00:12:55.705 --> 00:12:59.908 That is what a voltmeter and an ammeter multiplied together would tell you. 00:13:00.821 --> 00:13:05.266 The ratio of the real power to the apparent power is the power factor, 00:13:05.349 --> 00:13:08.401 and by the algebra it is simply cosine phi. 00:13:08.484 --> 00:13:12.443 A load whose current is in step has a power factor of one. 00:13:12.547 --> 00:13:16.911 Our motor, lagging thirty seven degrees, has cosine of thirty seven, 00:13:16.994 --> 00:13:18.829 which is zero point eight. 00:13:20.225 --> 00:13:24.220 Now put our numbers in, and draw the three powers as a triangle. 00:13:24.416 --> 00:13:29.221 It is the impedance triangle from the last section, scaled up by the current squared, 00:13:29.304 --> 00:13:32.219 so it has the same shape and the same angle. 00:13:33.016 --> 00:13:38.099 Two forty volts times twenty four amps is five point seven six kilovolt amps 00:13:38.182 --> 00:13:39.750 of apparent power. 00:13:39.947 --> 00:13:42.316 That is the whole hypotenuse. 00:13:43.020 --> 00:13:44.864 Multiply by the power factor, 00:13:44.947 --> 00:13:49.708 zero point eight, and the real power is four point six one kilowatts. 00:13:49.812 --> 00:13:53.806 That is the useful part, and it is the horizontal side. 00:13:54.603 --> 00:13:56.970 The vertical side is the reactive power, 00:13:57.053 --> 00:14:02.232 three point four six kilovar, and the power factor is zero point eight lagging. 00:14:03.028 --> 00:14:05.757 So here is the puzzle the customer raises. 00:14:05.954 --> 00:14:10.471 The reactive power delivers no energy at all. Its average is zero. 00:14:10.667 --> 00:14:13.025 Why should anybody be charged for it? 00:14:13.508 --> 00:14:15.549 And the answer is not in the energy. 00:14:15.632 --> 00:14:16.896 It is in the current. 00:14:16.979 --> 00:14:20.193 The utility does not send a phase angle down its cables. 00:14:20.276 --> 00:14:22.297 It sends twenty four amps. 00:14:23.001 --> 00:14:28.154 Losses in a transmission line go as the current squared times the line resistance. 00:14:28.237 --> 00:14:30.117 Not as the useful power. 00:14:30.200 --> 00:14:32.116 As the current, squared. 00:14:32.913 --> 00:14:37.951 A customer taking the same four point six kilowatts at unity power factor would draw 00:14:37.951 --> 00:14:40.170 only nineteen point two amps. 00:14:40.622 --> 00:14:41.990 Compare the squares: 00:14:42.073 --> 00:14:46.845 twenty four squared over nineteen point two squared is one point five six. 00:14:47.642 --> 00:14:52.088 Fifty six percent more loss in the cables, and fifty six percent more conductor 00:14:52.088 --> 00:14:56.002 and transformer capacity needed to serve the same useful load. 00:14:56.798 --> 00:14:59.794 That is why the tariff has a power factor clause. 00:14:59.991 --> 00:15:03.600 The wires and the transformers are sized by the hypotenuse, 00:15:03.683 --> 00:15:08.013 and the customer with the poor power factor is asking for a bigger hypotenuse 00:15:08.013 --> 00:15:10.278 to get the same horizontal side. 00:15:11.191 --> 00:15:13.421 Which raises the obvious question. 00:15:13.524 --> 00:15:17.832 If the trouble is that the current arrow is tilted, can we tilt it back? 00:15:18.973 --> 00:15:21.714 Here is the motor's phasor picture as we left it. 00:15:21.910 --> 00:15:23.292 Voltage along the reference, 00:15:23.292 --> 00:15:27.391 current lagging by thirty seven degrees, twenty four amps long. 00:15:28.188 --> 00:15:30.566 Now think about what a capacitor does. 00:15:30.649 --> 00:15:34.921 Its current is C times the rate of change of its voltage, which is the mirror image 00:15:34.921 --> 00:15:36.582 of the coil relation. 00:15:37.007 --> 00:15:39.920 So where the coil's current lagged by ninety degrees, 00:15:40.003 --> 00:15:42.230 the capacitor's current leads by ninety. 00:15:42.313 --> 00:15:45.402 It is reactive current pointing the other way. 00:15:45.943 --> 00:15:47.950 Put one in parallel with the motor. 00:15:48.033 --> 00:15:52.013 Both see the same supply voltage, because that is what parallel means, 00:15:52.096 --> 00:15:55.499 and the supply now has to provide the sum of the two currents. 00:15:56.110 --> 00:15:58.082 The motor's current is unchanged. 00:15:58.165 --> 00:16:00.380 It has no idea the capacitor is there: 00:16:00.463 --> 00:16:05.561 same voltage across it, same impedance, same twenty four amps at the same angle. 00:16:06.013 --> 00:16:09.590 That is the crucial point and it is worth pausing on. 00:16:10.387 --> 00:16:14.171 What changes is what the supply cable carries, which is the motor's current 00:16:14.171 --> 00:16:18.457 plus the capacitor's current, added as phasors, head to tail. 00:16:19.369 --> 00:16:23.968 So watch what happens to that magenta line current as we make the capacitor bigger. 00:16:25.364 --> 00:16:29.184 The motor's current has a vertical component: twenty four times the sine 00:16:29.184 --> 00:16:33.968 of thirty seven, which is fourteen point four amps of lagging reactive current. 00:16:34.764 --> 00:16:38.898 The capacitor's current points straight up, so it cancels part of that vertical 00:16:38.898 --> 00:16:44.146 component. The line current shortens, and its angle to the voltage closes up. 00:16:44.850 --> 00:16:46.847 A bigger one takes out more. 00:16:47.044 --> 00:16:49.782 The magenta line current is getting shorter and flatter, 00:16:49.865 --> 00:16:52.884 while the green motor current has not moved at all. 00:16:53.681 --> 00:16:56.721 And at fourteen point four amps of capacitor current, 00:16:56.804 --> 00:16:59.858 the two vertical components cancel exactly. 00:17:00.055 --> 00:17:02.924 The line current lies flat along the voltage. 00:17:03.407 --> 00:17:07.062 What is left in the supply cable is only the horizontal component: 00:17:07.145 --> 00:17:11.778 twenty four times cosine thirty seven, which is nineteen point two amps. 00:17:12.575 --> 00:17:16.407 The phase angle is now zero and the power factor is one. 00:17:16.511 --> 00:17:19.159 The current is back in step with the voltage. 00:17:19.584 --> 00:17:24.412 And to size the capacitor, we ask for the reactive power it must supply: 00:17:24.495 --> 00:17:27.372 two forty volts times fourteen point four amps, 00:17:27.455 --> 00:17:29.880 which is three point four six kilovar. 00:17:29.963 --> 00:17:33.226 Exactly the reactive power the motor was drawing from the supply, 00:17:33.226 --> 00:17:36.257 and now drawn from the capacitor instead. 00:17:36.960 --> 00:17:39.655 That is the physical picture worth keeping. 00:17:39.851 --> 00:17:43.299 The coil and the capacitor pass the sloshing energy back and forth 00:17:43.299 --> 00:17:46.238 to each other, locally, twice a cycle. 00:17:46.551 --> 00:17:49.477 The supply is no longer part of that exchange. 00:17:50.781 --> 00:17:54.822 So here is the ledger, before and after one capacitor. 00:17:55.618 --> 00:17:59.613 Line current down from twenty four amps to nineteen point two. 00:18:00.065 --> 00:18:01.319 Real power unchanged 00:18:01.319 --> 00:18:06.347 at four point six one kilowatts, because the motor is doing exactly the same job. 00:18:07.144 --> 00:18:12.018 Apparent power down from five point seven six to four point six one kilovolt amps, 00:18:12.101 --> 00:18:15.236 and the power factor up from zero point eight to one. 00:18:15.778 --> 00:18:17.923 And the line loss goes as current squared, 00:18:18.006 --> 00:18:23.174 so nineteen point two squared over twenty four squared is zero point six four. 00:18:23.370 --> 00:18:28.050 Thirty six percent less heat wasted in the cables, for the same work done. 00:18:28.661 --> 00:18:31.689 That is why the capacitor bank sits in the factory switchroom 00:18:31.772 --> 00:18:33.816 and not at the power station. 00:18:34.129 --> 00:18:37.298 Correcting at the load takes the reactive current out of every metre 00:18:37.298 --> 00:18:38.959 of cable between them. 00:18:39.663 --> 00:18:42.157 So look back at what four ideas did. 00:18:42.240 --> 00:18:45.086 Two sinusoids became two arrows. 00:18:45.282 --> 00:18:48.139 The angle between the arrows became an impedance. 00:18:48.935 --> 00:18:54.079 That angle split the power into a part that works and a part that only sloshes. 00:18:54.276 --> 00:18:58.419 And one capacitor sent the sloshing part back to the load that wanted it, 00:18:58.502 --> 00:19:01.046 leaving the wires to carry only the work. 00:19:01.842 --> 00:19:05.965 Every one of those steps came out of one small observation at the very start: 00:19:06.068 --> 00:19:10.910 that the voltage across a coil follows the slope of its current, and not its size.