Power in AC Circuits

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. An industrial load takes ten amperes from two hundred volt mains. Power is voltage times current, so two hundred times ten gives two thousand watts, right? But the true power it uses is only one thousand two hundred watts. So where did the other eight hundred watts go? By the end of this video, you will know exactly. Power in A C circuits comes up often in JEE, so let's begin.

0:32Average power and the power factor

Alright, here is the figure from our series LCR video. The black arrow is the source voltage V, the grey arrow is the current I, and phi, the Greek letter, is the angle between them. So tell me, is the power in A C still simply V times I? Not quite, and here is the answer. P, the average power, equals V rms times I rms times cos phi. V rms, the capital V with the small letters r m s, is the root mean square voltage, the value an A C meter reads, and I rms is the same for the current. The extra factor is cos phi. Okay? And what if a question gives peak values? V naught, the V with a small zero, is the peak voltage, and I naught is the peak current. For a sinusoid, the rms value is the peak over root two, so the power becomes one half, times V naught, times I naught, times cos phi. Don't forget that half, right? Now, V rms times I rms on its own is called the apparent power. It is measured in volt-ampere, the capital V A on screen, not in watts. Cos phi converts it into the true power, in watts. Clear? So what decides cos phi? In the impedance triangle on the right, R is the base and Z, the impedance, is the hypotenuse. So cos phi equals R over Z, as the figure writes at the bottom. This cos phi is called the power factor. With me? Now think about this. Our circuit has a resistor, an inductor and a capacitor. Which of them actually uses up the power? It is only the resistor, yes, exactly! Put V rms equal to I rms times Z, and cos phi equal to R over Z; the double arrow means which gives. The Z cancels, leaving I rms squared times R, nothing else. So only the resistor uses power. Got it?

2:28Two parts of the current

Here is another way to see cos phi. Split the current I into two parts. The first part, I cos phi, is in phase with the voltage, so it does the work. Okay? And the second part is at ninety degrees to the voltage; the small circle on screen means degrees. Does this part do any work? It does none at all, that is correct! This part, I sine phi, is called the wattless component. It flows in the circuit, but it does no work. Right? Now look at the figure: the phasors for a resistor, an inductor and a capacitor, each on its own. In the inductor, V leads I by ninety degrees, and in the capacitor, I leads V by ninety degrees. So, how much power does a pure inductor use? It uses zero power, yes! For a pure inductor or a pure capacitor, phi is ninety degrees, so cos phi is zero and the power is zero. The whole current is wattless. Current flows, and still no power is used. Got it? And at the other end is a pure resistor. In the first picture, V and I are in phase, so phi is zero and cos phi is one. The same happens at resonance, where the whole circuit behaves just like a pure resistor. Clear?

3:47Half-power frequencies

This graph is from our resonance video: the current I against omega, the angular frequency. Look at the taller, sharper curve. At omega naught, the omega with a small zero, we have resonance, and the current reaches its peak, I max, with the small word max below. So the power is largest there, okay? Now look at the dashed grey line: I max over root two. On each side of the peak, the current falls to this level at one frequency. What happens to the power at these two frequencies? It becomes exactly half, yes! Power is I rms squared times R, and root two squared is two. So the power is P max over two, half the power at resonance. That is why these are called the half-power frequencies. Right?

4:36Example 1: power factor and power for R 40 ohm, Z 50 ohm

Now, our first example. Here, R is forty ohms and Z, the impedance, is fifty ohms, and the circuit carries four amperes from a two hundred volt source. The capital omega is the unit ohm, capital A means amperes, and capital V means volts. Find the power factor and the power. Pause the video and try it first. Are you done? Okay, let's check it together. The power factor is cos phi, which is R over Z, remember? Forty over fifty gives zero point eight. Okay? Now the power. Two hundred volts times four amperes is eight hundred, so the power is eight hundred watts, right? No, no, careful, eight hundred is only the apparent power, in volt-ampere. These A C values are rms values, as meters read. So P equals two hundred times four times zero point eight, which gives six hundred and forty watts; the capital W means watts. Clear? Now check it with the other formula, I rms squared times R, okay? Four squared is sixteen, and sixteen times forty gives six hundred and forty watts again. It is the same answer. Got it?

5:50Example 2: true and apparent power, wattless current, pf 0.6

Now, remember our puzzle? Here it is as a real question. The load takes ten amperes from two hundred volt mains at a power factor of zero point six, lagging. Lagging means the current lags behind the voltage. Find the true power, the apparent power and the wattless current. Pause the video and try it first. Are you done? Okay, let's check. True power is V rms times I rms times cos phi, so two hundred times ten times zero point six. That gives one thousand two hundred watts, the number from our puzzle. With me? And the apparent power is just two hundred times ten, which is two thousand volt-ampere. So, mystery solved! The two thousand at the start was never the true power. It was the apparent power, in volt-ampere, and the power factor brings the true power down to one thousand two hundred watts. So no eight hundred watts were lost anywhere; that eight hundred was never real power at all, okay? Now the wattless current. Ten times zero point six is six, so the wattless current is six amperes, right? No, no, wait, that is the trap! Ten times cos phi is the working part, in phase with the voltage. The wattless part needs sine phi, which is the square root of one minus the square of zero point six. That is the root of zero point six four, which gives zero point eight. With me? So the wattless current is I sine phi, ten times zero point eight, which gives eight amperes. Eight amperes flowing in the line, and doing no work at all! See how much the power factor matters?

7:28Quick revision: P = Vrms Irms cos phi, wattless I sin phi

So, quick revision. Average power is V rms times I rms times cos phi, which equals I rms squared times R, and the power factor, cos phi, is the resistance over the impedance. V rms times I rms alone is only the apparent power, while the current I rms times sine phi is wattless. Pure inductors and capacitors use no power, while a pure resistor, or a circuit at resonance, has a power factor of one. Now go and try the practice problems on this page. All the best!