Alternating Current: Phasors and Power Factor

In a direct current circuit, a resistor's current is the voltage divided by the resistance, and that is the end of it. Alternating current has a second question hiding inside it, and the whole of this lecture comes out of that question. Here is a sinusoidal source, and the load it is driving is a coil. A big motor winding, say, or a transformer. Time runs across, and both volts and amps are read up the side. The yellow curve is the voltage the source puts across that coil. It is a sine wave, of amplitude V m, going round at angular frequency omega. And here is the current the coil draws. Same shape, same frequency, and that is the first thing worth noticing: a sinusoidal source in a linear circuit gives you sinusoidal everything, at one single frequency. But look where it sits. The voltage reaches its peak here, and the current does not reach its own peak until later, over here. That gap is a quarter of a cycle. Ninety degrees. The current is running late, and in the language of the trade we say it lags the voltage by an angle phi. For a pure coil that lag is exactly ninety degrees. For a real motor, which has resistance in its windings as well, it will be somewhere between zero and ninety, and finding it will be our job shortly. So where does the lag come from? From one relation, and it is the defining property of an inductor. The voltage across a coil is L times the rate of change of the current through it. Read that carefully, because it is not a statement about how much current is flowing. It is a statement about how fast that current is changing. Now put the two together on the picture. Here is the instant the current is climbing at its very steepest. Look at the red tangent: that is the biggest slope anywhere on the green curve. And at that same instant, the yellow voltage is at its maximum. Steepest current, largest voltage, at the same moment. Now take the instant the current is at its own peak. Its slope there is zero: the red tangent is flat. And the voltage at that instant is passing through zero. So the voltage tracks the slope of the current, and the slope of a sine is a cosine, which is a quarter cycle ahead of it. The lag is not a quirk of the coil. It is calculus. Write it out. Take the current as I m sine of omega t minus ninety degrees. Differentiate it, multiply by L, and what comes back is a sine of omega t with no shift at all, which is the voltage we started with. And hold on to that factor out front, omega L. It has the units of ohms, and it is going to become half of the answer. Before we go on, let me tell you where this ends, so that every step has somewhere to be walking towards. First, we replace those two sinusoids by two arrows on a turning wheel. That picture is called a phasor diagram, and it is the single most useful thing in this subject. Second, the angle between those arrows turns Ohm's law into something that works for coils and capacitors as well. That is impedance. Third, we multiply voltage by current and find that the power splits in two. One part does work. The other part sloshes back and forth twice every cycle and delivers nothing at all. And fourth, we put one capacitor across the load, watch the current arrow swing back into line, and deliver exactly the same useful power with noticeably less current in the wires. That last one is worth real money to a factory, and we will see exactly why.

Two sinusoids at the same frequency carry only three pieces of information between them: one amplitude, a second amplitude, and the angle by which one trails the other. Everything else is the same frequency ticking along underneath. So here is a picture that keeps exactly those three things. A yellow arrow, whose length is the amplitude of the voltage. And a green arrow, shorter, whose length is the amplitude of the current. The green one is set behind the yellow one by the phase angle phi, and that angle is fixed. It is built into the picture and it will not change. Drop each arrow tip onto the vertical axis. Those two dashed projections are the heights we are going to plot, on the time axes over here. Now spin both arrows, counterclockwise, at the angular frequency omega. They turn together, rigidly, like two spokes of one wheel. Watch what the projections trace out as the wheel turns. The yellow height rises, falls, goes negative, comes back. That is a sine wave, and here it is being drawn instant by instant. Keep going for a second turn. The green projection does the same thing, but it arrives at each landmark a quarter of a cycle later, because its arrow is a quarter of a turn behind. So the two waveforms we started this lecture with are the shadows of two arrows on one turning wheel. That is the whole content of the phasor idea, and everything else follows from it. And now the move that makes it useful. Both arrows turn at the same rate, forever. So the turning tells us nothing we did not already know. What carries all the information is the two lengths and the angle between them, and those never change. So stop the wheel. Photograph it at one instant and throw the rotation away. What is left is this. Line the voltage arrow up along the reference direction and call its angle zero. We are free to do that, because only the difference between the two angles ever mattered. The current arrow then sits below it, at minus phi. For an inductive load it is always below: lagging. Two arrows, frozen. That is a phasor diagram, and each arrow is a phasor: a length and an angle, standing in for a whole sinusoid. One more piece of bookkeeping before we use it. From here on, when I write V or I as a length, I mean the root mean square value, not the peak. Two forty volts from a wall socket is an r m s number, and using r m s throughout keeps the power formulas free of stray factors of one half. So we have replaced two moving curves by two still arrows, and lost nothing. Now let us put those arrows to work and get an Ohm's law that a coil obeys.

A real motor winding is a resistance and an inductance in series, and one current flows through both of them. So let us draw that current first, and hang everything else off it. Across the resistance, voltage and current stay in step. So the resistor's voltage phasor lies flat along the current, and its length is R times I. Across the coil it is different, and we did the calculus for this in the opening. The coil's voltage leads its current by ninety degrees, so its phasor stands square to the current, and its length is omega L times I. That quantity omega L is called the inductive reactance, X sub L, and like resistance it is measured in ohms. The j in front of it is the algebra's way of saying: turn it ninety degrees. The voltage law still holds, provided we add the phasors as vectors rather than as numbers. So the source voltage is the resistor's voltage and the coil's voltage joined head to tail. And there is the whole geometry of an inductive load in one triangle. The source voltage is the hypotenuse, and it sits ahead of the current by an angle phi. That angle is what we are about to name. Factor the current out of both terms. What is left multiplying it is R plus j X sub L, and that combination is the impedance, bold Z. It is a complex number, and its two parts are exactly the two sides of the triangle. Its magnitude is the square root of R squared plus X squared. And its angle is the arctangent of X over R, which is the same phi we have been carrying since the first waveform. The reactance sets the phase; the resistance holds it back. So Ohm's law survives into alternating current unchanged in form. Voltage phasor equals impedance times current phasor. The only thing that has grown is that all three are complex, and the angles carry the timing. Let us put numbers on it, because these numbers will follow us to the end of the lecture. A motor on a two hundred and forty volt supply, with eight ohms of winding resistance and six ohms of reactance. The triangle is now measured in ohms rather than volts, which is the same picture divided through by the current. Eight along, six up. So the magnitude of the impedance is the square root of sixty four plus thirty six, which is the square root of a hundred: ten ohms. Now the current is two hundred and forty volts over ten ohms. Twenty four amps. That is the number a clamp meter on the supply lead would read. And the phase angle is the arctangent of six over eight, which is thirty six point eight seven degrees. The current lags the voltage by that much. Impedance ten ohms at thirty seven degrees, current twenty four amps lagging by thirty seven degrees. Hold on to both. In the next section we multiply the voltage by the current, and that angle is about to cost somebody money.

Power at any instant is voltage times current at that instant. No phasors, no averages: just the product of the two numbers, moment by moment. Here are our two waveforms again, the voltage in yellow and the current lagging it by thirty seven degrees in green. And here in magenta is their product, the instantaneous power flowing into the motor. Two things about it. First, it oscillates at twice the supply frequency, because it is a product of two sinusoids: two humps for every one cycle of the voltage. Second, and this is the interesting one, it goes negative. Here, and here again half a cycle later. Negative power means energy flowing out of the motor and back into the supply. The coil's magnetic field is storing energy on one part of the cycle and handing it straight back on the next. Nothing is consumed by that exchange. It just sloshes. Let us separate the sloshing from the working. Expand the product of the two sines with the usual identity, and it falls into exactly two terms. Look at the first term. It is V I cosine phi, multiplied by one minus cosine of two omega t. That bracket never goes negative: it swings between zero and two, and its average is one. So that term has an average value of V I cosine phi, and it is always into the load. That is energy actually leaving the supply and not coming back. The second term is V I sine phi times sine of two omega t. That is a pure sinusoid at double frequency, sitting symmetrically about zero. Its average over a cycle is exactly nothing. Here are those two terms drawn separately. The blue one is the working term. Notice that it never dips below the axis: it is one-directional, always from supply to load. Its average is this dashed level, and that average is what we call the real power P. V times I times cosine phi, measured in watts. That is the number that turns the shaft and heats the windings. And here is the red one. Equal time above the axis and below it, twice every cycle. Energy out, energy back, out, back. Its average is zero. Over any whole number of cycles it delivers not one joule. But its amplitude is not zero, and that amplitude has a name: reactive power Q, V I sine phi, measured in var. And the plain product of the voltage and current magnitudes, with no angle in it at all, is the apparent power S, in volt amps. That is what a voltmeter and an ammeter multiplied together would tell you. The ratio of the real power to the apparent power is the power factor, and by the algebra it is simply cosine phi. A load whose current is in step has a power factor of one. Our motor, lagging thirty seven degrees, has cosine of thirty seven, which is zero point eight. Now put our numbers in, and draw the three powers as a triangle. It is the impedance triangle from the last section, scaled up by the current squared, so it has the same shape and the same angle. Two forty volts times twenty four amps is five point seven six kilovolt amps of apparent power. That is the whole hypotenuse. Multiply by the power factor, zero point eight, and the real power is four point six one kilowatts. That is the useful part, and it is the horizontal side. The vertical side is the reactive power, three point four six kilovar, and the power factor is zero point eight lagging. So here is the puzzle the customer raises. The reactive power delivers no energy at all. Its average is zero. Why should anybody be charged for it? And the answer is not in the energy. It is in the current. The utility does not send a phase angle down its cables. It sends twenty four amps. Losses in a transmission line go as the current squared times the line resistance. Not as the useful power. As the current, squared. A customer taking the same four point six kilowatts at unity power factor would draw only nineteen point two amps. Compare the squares: twenty four squared over nineteen point two squared is one point five six. Fifty six percent more loss in the cables, and fifty six percent more conductor and transformer capacity needed to serve the same useful load. That is why the tariff has a power factor clause. The wires and the transformers are sized by the hypotenuse, and the customer with the poor power factor is asking for a bigger hypotenuse to get the same horizontal side. Which raises the obvious question. If the trouble is that the current arrow is tilted, can we tilt it back?

Here is the motor's phasor picture as we left it. Voltage along the reference, current lagging by thirty seven degrees, twenty four amps long. Now think about what a capacitor does. Its current is C times the rate of change of its voltage, which is the mirror image of the coil relation. So where the coil's current lagged by ninety degrees, the capacitor's current leads by ninety. It is reactive current pointing the other way. Put one in parallel with the motor. Both see the same supply voltage, because that is what parallel means, and the supply now has to provide the sum of the two currents. The motor's current is unchanged. It has no idea the capacitor is there: same voltage across it, same impedance, same twenty four amps at the same angle. That is the crucial point and it is worth pausing on. What changes is what the supply cable carries, which is the motor's current plus the capacitor's current, added as phasors, head to tail. So watch what happens to that magenta line current as we make the capacitor bigger. The motor's current has a vertical component: twenty four times the sine of thirty seven, which is fourteen point four amps of lagging reactive current. The capacitor's current points straight up, so it cancels part of that vertical component. The line current shortens, and its angle to the voltage closes up. A bigger one takes out more. The magenta line current is getting shorter and flatter, while the green motor current has not moved at all. And at fourteen point four amps of capacitor current, the two vertical components cancel exactly. The line current lies flat along the voltage. What is left in the supply cable is only the horizontal component: twenty four times cosine thirty seven, which is nineteen point two amps. The phase angle is now zero and the power factor is one. The current is back in step with the voltage. And to size the capacitor, we ask for the reactive power it must supply: two forty volts times fourteen point four amps, which is three point four six kilovar. Exactly the reactive power the motor was drawing from the supply, and now drawn from the capacitor instead. That is the physical picture worth keeping. The coil and the capacitor pass the sloshing energy back and forth to each other, locally, twice a cycle. The supply is no longer part of that exchange. So here is the ledger, before and after one capacitor. Line current down from twenty four amps to nineteen point two. Real power unchanged at four point six one kilowatts, because the motor is doing exactly the same job. Apparent power down from five point seven six to four point six one kilovolt amps, and the power factor up from zero point eight to one. And the line loss goes as current squared, so nineteen point two squared over twenty four squared is zero point six four. Thirty six percent less heat wasted in the cables, for the same work done. That is why the capacitor bank sits in the factory switchroom and not at the power station. Correcting at the load takes the reactive current out of every metre of cable between them. So look back at what four ideas did. Two sinusoids became two arrows. The angle between the arrows became an impedance. That angle split the power into a part that works and a part that only sloshes. And one capacitor sent the sloshing part back to the load that wanted it, leaving the wires to carry only the work. Every one of those steps came out of one small observation at the very start: that the voltage across a coil follows the slope of its current, and not its size.