Suppose you manage a factory that uses many electric motors....

Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with a component of current that is 90° out of phase with the voltage as well as with current in phase with the voltage. The electric company charges you an extra fee for “reactive volt-amps” in addition to the amount you pay for the energy you use. You can avoid the extra fee by installing a capacitor between the power line and your factory. The following problem models this solution. In an $R L$ circuit, a $120-mathrm{V}$ (rms), $60.0-mathrm{Hz}$ source is in series with a $25.0-mathrm{mH}$ inductor and a $20.0-Omega$ resistor. What are (a) the rms current and (b) the power factor? (c) What capacitor must be added in series to make the power fact tor 1$?$ (d) To what value can the supply voltage be reduced if the power supplied is to be the same as before the capacitor was installed?

Transcript

00:01 Here we're looking at what is meant by the power factor in a circuit.

00:07 If you're familiar with phasers, you are familiar with the fact that there is a capacitive and inductive reactants associated with capacitors and inductors.

00:22 And that is kind of like a generalized resistance, except it pulls the voltage and current out of.

00:31 Phase with each other.

00:33 So one would be doing a cosine while the other one is doing a sign.

00:39 And inductive reactants for both capacitor and an inductor is frequency dependent.

00:47 And we'll start off with just a resistor and inductor in a circuit.

00:53 But the power factor comes from the relationship for power in an ac circuit is a product of the rms voltage that's being supplied by the generator times the rms current through in this case the series circuit.

01:12 And then there is a cosine of phi.

01:17 That cosine of phi is the power factor.

01:22 And it is simply a measure of how far out of phase your voltage and your current are.

01:30 And typically it means that you are not using power efficiently in your devices.

01:39 But the way we determine phi is if we make a little phaser diagram, if you have an rlc circuit, what is true is the resistance is like a resistance, if you will.

01:59 Yeah, it is a resistance that lies along the cosine action.

02:03 Which is sometimes called the real axis.

02:08 Inductive reactants is 90 degrees ahead of that on this phaser diagram.

02:17 Capacitive reactants would be 90 degrees behind the resistance.

02:24 And we add those together like a vector would be added to get kind of the generalized resistance in the circuit called the imbiased.

02:35 Impedance.

02:36 It will be in oms.

02:37 All of these are in oms.

02:40 And that is equal to the square root of the r along the real axis squared, like an x component, plus the inductive minus capacitive reactances because they're once positive, once negative long, the imaginary axis or y -axis.

03:02 And we square those to get z.

03:06 What the phase measures is the angle that the z vector, if you want to think about it as a z -vector, makes with the real axis.

03:19 So phi measures, the tangent of phi comes from the inductive minus the capacitive reactants, kind of like a y component divided by the resistance.

03:37 And so that phase angle being equal to zero means that there is nothing out of phase with the current.

03:49 And it means that there are no inductors or capacitors in the circuit or they equal each other out.

03:58 So that's a lot to say.

04:01 In the case that there's just a resistor, an inductor, of course there is going to be a phase angle.

04:09 And we can calculate that phase angle by first calculating the inductive reactants, which is the omega, the 2 pi times the frequency, times the inductance itself.

04:27 So that is 2 pi times 60 hertz times the inductance, which is 25 millie, henry, 10 to the minus 3.

04:37 And that would all come out in oms.

04:40 So let's see, that inductive reactants when we calculate it is 9 .42 oms.

05:02 So our phaser would have 20 oms along the real or x -axis and 9 .42 oms along the y -axis or imaginary axis.

05:16 And that really just means that the voltage in the generator is out of phase with the current by that angle.

05:28 Okay, so we can now find the tangent of phi would be in this case the inductive reactants over the resistance or 9 .42 over 20.

05:44 And working that out, the phase angle turns out to be, let's.

05:52 Go put it in degrees just because that's a little easier to think about.

05:57 It turns out to be 25 .2 degrees.

06:05 And that means that our power factor is the cosine of that and that is equal to .924.

06:22 So yeah, the power, let's see, that's not quite right.

06:32 I'm getting .905.

06:33 Okay, so let's get that a little bit more accurately by keeping a few more secure.

06:42 Significant figures in the cosine.

06:53 0 .905 sounds good.

06:56 Ok, so let's just figure out the power that's used by the circuit.

07:04 And to do that, we want to figure out the rms current.

07:12 So to do that, you actually kind of need something that looks like oms law, that v -rms is equal to i -rms times the generalized resistance z.

07:28 And we configure z out...

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