Video Lectures created by Tim Feiegenbaum at North Seattle Community College.
Okay, we're in section 6.4. Circuit analysis of AC resistive circuits. AC circuits composed of resistors can be analyzed using Ohm's Law and Kirchhoff's Law just as we did with DC circuits. Care must be taken that correct formulas are used for peak voltages and currents, average voltages and currents, RMS voltages and currents. Typically, power ratings are RMS values in AC circuit. With that introduction, the rest of this session is going to be purely circuit analysis and calculations.
We will start off with a little discussion about RMS and peak power. Here we have a circuit and this is 100 volts. This is 100 volts and this is in RMS. Remember this is an AC calculation, AC calculations concerning Ohm's Law will be just the same as we did we DC. Remember we said that E/R=I so, in this case, we have 100volts RMS/10K and if we solve for that we would find that that equals 10MA. When we did power we said that Voltage times current equals power, (E*I=P) and we also said that Voltage squared over R (E2/R=P). We could use either formula.
At any rate, let's do calculate the power. If we have 100 V*10ma we are going to get 1Watt. If we took 1002 divided by 10k (1002 /10k =1watt). Now, this is the standard way that RMS power is calculated and the reason is that, remember when we talked about the comparison of peak to RMS. Remember we had a peak value and we said that the RMS equivalent would be 0.707 of that and that value represents what the heat dissipation would be if it was measured in Dc. So RMS is always a little bit lower than the peak. So, let's consider what the power would be if we were evaluating this in peak. First of all, we need to convert this to peak value so we would say 100 volts divided by 0.707 (100v/0.707) and that would yield 141 volts. So we are starting with a higher value and so we are going to have a higher voltage or higher current drives.
We say again 141 volts in this case divided by 10Kohm, we are going to have 14.1 milliamps of current instead of the 10 milliamps we had over here. 141v/10k=14.1ma. If we were to calculate power now and if we said 141*14.1 milliamps, then we would, it would be about 1.98watts. So when you consider peak power, it is going to yield a much higher, almost twice as much than the equivalent RMS. As I said RMS is pretty much the standard. Sometimes, you'll see peak ratings with manufacturers of audio equipment so they try to make their product look a little better so be very careful when you evaluate the power of amplifiers and such. Because in this particular situation, is the, let's amplify it 100 times and so here you got 100watts amplifier rated in RMS then there is another amplifier that says it's 198watts, but this is in peak. Technically if you do the math, they are the same power. This is related to RMS versus peak power.
The next circuit we are going to look at, is right out of your textbook and we are going to be looking at this, in fact, we are going to look at this on three different screens. The first screen we will evaluate peak values and the next will be RMS and the final will be average. This starts out with a 25 RMS voltage source here and since we are doing peak we need to first of all convert over to peak so we would take our 25 volts divided by 0.707 (25volts/0.707) and that would yield 35.36 volts peak. Since we have a combination of series and parallel here, we should evaluate the resistance of the two in parallel. So let's draw a little-simplified circuit here, and we'll say this is R1 here, and then we'll have just one resistance here to represent R2 in parallel with R3.
So we need to do a little math here. Remember the calculations for parallel circuits? We would say one over… In this case, it would be one over 470 plus one over 680. If we did this calculation, the resulting value is about 278. And this would be our equivalent simplified circuit. If we wanted to know, because one of our goals is to calculate all the currents in these circuits, so you know, what are the voltage troughs at each point, so we could do that. Given these two values, we could as well… There are a number of ways we could do this. We could take the total voltage divided by these two components together and we can get the total current and then we could calculate what are the voltage dropped so we can do it that way.
Another way is to do a voltage divider, which is what I'm going to do and we will say 278/278 +100 and this is just during a proportion here to find out what proportion of this compared to the total resistance will this drop. So in this case, the proportion here is going to be 0.735 and if we wanted to know the voltage drop across this component, remember this is both piece components and we could go 0.735 times the total voltage which is 35.36 and that would yield real close to 26 volts peak and so that would be the value across these two components.
We could use Kirchoff's Law to determine the voltage drop here and remember Kirchhoff's Law says that the sum of the voltages in the circuit will equal the source voltage. We note that the circuit voltage is 35.36. If we subtract the 26 that's dropped here, that would leave what's left over here and that is going to be about 9.35 volts, something like that. We'll have 9.35 across this component and about 26volts across these two. Then we need to calculate the current drops and so we now know we have 26 volts across this component.
We could calculate the current across this, I simply say taking 26 volts across 278 and we would know the total current through here and that would actually give us a current here as well. But we do need to know the current individual drop so let's just use Ohm's law and we will say 26 volts across 470 and that's going to yield 55milliamps and then the same things 26 volts across 680 Ohms. That is going to yield about 38 milliamps. Again we could add those together and that would give us this but just for the sake if calculation lets say 9.35 divided by 100 and that would yield what, 93.5 (9.535/100=93.5 milliamps) and if you add this together you should get that.
I did a little bit of rounding here, so you get the 0.5 there but these two together is going to roughly equal this.
Those are the basic calculations using peak. And remember we started out with RMS we converted to peak, we did the voltage divider here and we calculated the voltage drops and currents in this circuit. This was in peak. Your text does mention the values for peak to peak and remembers that, let's see, where can I draw it?
Here's a... this is an AC signal, remember with the peak value is from here to here. The peak to peak is going to be the from peak to peak. The peak is going to be double what we see here. If we had done these calculations with the peak to peak, everything in this circuit would be double what it currently is, except for this ratio right here. The next one we're going to look at, we're going to the same circuit, only we're going to do it in RMS. With RMS, we start out with the 25 volts, we do not need to convert. We can just go directly with that value and let's see. Let's consider our equivalent circuit.
We had our one then we had the parallel of our two and our three here. We already did the voltage dividers so let's just take our value 0.735*25 and that is going to yield 18.37 volts. That would be our voltage across R2. Again if we use Kirchhoff's law and we took our initial 25 volts minus 18.37, that would yield our voltage across R1, which is going to be about 6.62, something like that. That would be our voltage across R1. If we calculated our currents, again we have 18.37 is our voltage across this leg, divided by 470. That should yield about 39 milliamps (mA) and 18.37 across the 680 component, would yield 27. Again these two together should equal this value.
If we took, let's see, 6.62 across the 100 ohms here, and that would yield about 66 milliamps. This would equal the sum of both of these. This is the RMS calculations and I skipped some of the things because I knew we did them already in the peak calculations. Given the same circuit and doing the average power, and again I'm not going to do all of them, I'm going to do the... Remember, first of all, let's talk about what the average is. Remember from our previous discussion, the average value is going to be 0.637 times the peak. In this case, we're, remember we started out… This is the circuit we initially started with, with 25 volts. We need to convert that to peak, so 25/0.07, and that's going to yield the value 35.36. If we take the 0.637 times 35.36 we will get 22.48. This represents our average voltage.
Again, I'm not going to repeat all those calculations but we would start out with this as our source as 22.48. We would just do everything the way we had previously done at calculating our currents or parallel voltage drops and all of the calculations. We will just start with this value: 22.48. In this lesson, we have looked at AC analysis for resistive circuits. We've looked at average then we looked at RMS and we looked at peak. And the very beginning we started out with a discussion comparing the RMS power to peak power.
Video Lectures created by Tim Fiegenbaum at North Seattle Community College.
In Partnership with Future Electronics
by Aaron Carman
by Jeff Child