WEBVTT 00:00:00.710 --> 00:00:05.760 >> In a circuit that's being driven by a sinusoidally varying voltage, 00:00:05.760 --> 00:00:11.430 the current that's flowing in that circuit is also going to be sinusoidal. 00:00:11.430 --> 00:00:16.079 The power delivered to this resistive load which is equal to 00:00:16.079 --> 00:00:20.645 the voltage times the current is also going to be varying. 00:00:20.645 --> 00:00:23.480 That means, at some points when the voltage and current are at 00:00:23.480 --> 00:00:26.590 maximum values you're going to be delivering a lot of power to the load, 00:00:26.590 --> 00:00:30.230 and at times when the voltage or the current is zero, 00:00:30.230 --> 00:00:33.410 at that instant, you won't be delivering any power to the load. 00:00:33.410 --> 00:00:35.420 So then becomes interesting to say, 00:00:35.420 --> 00:00:36.800 well is there a way of determining what 00:00:36.800 --> 00:00:41.060 the average power is being delivered over one cycle? 00:00:41.060 --> 00:00:44.480 Is there are some effective quantity that we can use to 00:00:44.480 --> 00:00:46.130 calculate the effective power or 00:00:46.130 --> 00:00:48.695 the average power that's being delivered to the load? 00:00:48.695 --> 00:00:51.620 We might say, well, let's just take the average of it. 00:00:51.620 --> 00:00:57.035 Let's start with a voltage source or some sinusoidally varying thing. 00:00:57.035 --> 00:01:00.200 Calculate its average value. 00:01:00.200 --> 00:01:02.630 The average value of say 00:01:02.630 --> 00:01:07.110 some sinusoidally varying voltage V of t. If V of t is 00:01:07.110 --> 00:01:14.655 equal to V_m cosine Omega t, 00:01:14.655 --> 00:01:17.330 then, the average voltage is defined as one 00:01:17.330 --> 00:01:20.240 over t times the integral from 0 to t, 00:01:20.240 --> 00:01:26.185 V_m cosine Omega t, dt. 00:01:26.185 --> 00:01:29.070 Without going to bother doing this integration, 00:01:29.070 --> 00:01:31.130 we know that when you integrate you are actually 00:01:31.130 --> 00:01:33.665 adding up the area under the curve. 00:01:33.665 --> 00:01:36.500 In this case going from zero to t, 00:01:36.500 --> 00:01:43.380 we're going over one period and adding up the area. 00:01:43.380 --> 00:01:46.000 It turns out that in a sine wave, 00:01:46.000 --> 00:01:52.870 there's just as much area above the axis positive area as there is area 00:01:52.870 --> 00:02:00.300 under the line and thus the average voltage is zero. 00:02:00.300 --> 00:02:02.080 Well, that doesn't help us a whole lot because we know 00:02:02.080 --> 00:02:03.880 there's power being delivered to 00:02:03.880 --> 00:02:06.430 that resistive load but calculating 00:02:06.430 --> 00:02:10.330 the average voltage of the average current isn't going to do anything for us. 00:02:10.330 --> 00:02:13.930 Instead, we calculate what is known as the root means 00:02:13.930 --> 00:02:17.695 squared or the effective voltage. 00:02:17.695 --> 00:02:23.315 To do that we start by taking our voltage or current. 00:02:23.315 --> 00:02:26.405 We're going to calculate it for both the voltage and the current. 00:02:26.405 --> 00:02:30.275 You square that quantity. 00:02:30.275 --> 00:02:32.210 Here's the voltage squared. 00:02:32.210 --> 00:02:34.310 You'll notice that what was negative is 00:02:34.310 --> 00:02:37.105 now been flipped over so it is now positive. 00:02:37.105 --> 00:02:40.040 The shape is somewhat different but it's still 00:02:40.040 --> 00:02:43.565 ranges from in this case zero to one. 00:02:43.565 --> 00:02:48.890 We then average this squared value 00:02:48.890 --> 00:02:50.750 and then to bring it back to 00:02:50.750 --> 00:02:53.750 the appropriate amplitude or the appropriate units, 00:02:53.750 --> 00:02:55.760 we take the square root of it. 00:02:55.760 --> 00:02:58.130 In other words, the RMS value or 00:02:58.130 --> 00:03:02.689 this effective value of the voltage is found by squaring the voltage, 00:03:02.689 --> 00:03:04.685 calculating the average of the square, 00:03:04.685 --> 00:03:06.785 and then taking the square root of that. 00:03:06.785 --> 00:03:08.930 Let's go ahead and show you what we mean here. 00:03:08.930 --> 00:03:12.420 If V of t is equal to V_m cosine Omega, 00:03:12.420 --> 00:03:17.220 then V squared of t is equal to V_m 00:03:17.220 --> 00:03:22.530 squared cosine squared of Omega t. We're going to take 00:03:22.530 --> 00:03:24.230 that and we're going to calculate the average of 00:03:24.230 --> 00:03:27.020 that which is done by integrating from 00:03:27.020 --> 00:03:33.035 0 to t. Let's bring the V_m squared out in front since it's a constant. 00:03:33.035 --> 00:03:39.210 Cosine squared of Omega t dt. 00:03:39.770 --> 00:03:42.150 We have to avail ourselves of 00:03:42.150 --> 00:03:47.180 a trigonometric identity that says that the cosine squared of 00:03:47.180 --> 00:03:56.400 Theta is equal to 1.5 times 1 plus the cosine of 2 Theta. 00:03:56.400 --> 00:03:57.890 So to perform this integral, 00:03:57.890 --> 00:04:00.530 the trick is to instead of trying to integrate the cosine squared, 00:04:00.530 --> 00:04:05.675 we use this identity and substitute in then for this we have, 00:04:05.675 --> 00:04:10.830 V_m squared over T times the integral from 0 to 00:04:10.830 --> 00:04:19.370 t. We'll bring the 1.5 out in front also, 00:04:19.370 --> 00:04:24.840 times 1 plus the cosine of 00:04:24.840 --> 00:04:32.430 2 Omega t dt. 00:04:32.430 --> 00:04:35.495 Now, when we integrate this over the period, 00:04:35.495 --> 00:04:38.200 this term we're actually integrating over two periods. 00:04:38.200 --> 00:04:40.720 In the same way that when you integrate it over one period, 00:04:40.720 --> 00:04:42.940 the positive area cancel the negative area, 00:04:42.940 --> 00:04:49.630 this term here likewise gives us a zero contribution to this integral. 00:04:49.630 --> 00:04:52.420 When you integrate one with respect to t, 00:04:52.420 --> 00:05:01.900 we adjust t and so we have V_m squared over 2T times 00:05:01.900 --> 00:05:05.670 t evaluated from 0 to 00:05:05.670 --> 00:05:13.620 T. That then is just V_m squared over 2. 00:05:13.620 --> 00:05:17.010 So we squared it, 00:05:17.010 --> 00:05:19.965 we calculated the average. 00:05:19.965 --> 00:05:23.480 Now all is left to do is to take the square root of it and 00:05:23.480 --> 00:05:26.960 we've have in V_rms is 00:05:26.960 --> 00:05:34.550 equal to the square root of V_m squared over 2. 00:05:34.550 --> 00:05:38.570 If we take that square root of two out from under the radical, 00:05:38.570 --> 00:05:41.780 it comes out as 1 over 00:05:41.780 --> 00:05:46.220 the square root of 2 times the square root of V_m squared. 00:05:46.220 --> 00:05:48.955 Well, that's just V_m. 00:05:48.955 --> 00:05:51.615 Thus, we see that the RMS value, 00:05:51.615 --> 00:05:57.240 this effective voltage is equal to 1 over the square root of 00:05:57.240 --> 00:06:04.120 2 times V_m where V_m was the maximum value, our peak value. 00:06:04.310 --> 00:06:07.870 Well, 1 over the square root of 2 that's approximately equal 00:06:07.870 --> 00:06:12.480 to 0.707 times V_m. 00:06:12.480 --> 00:06:16.750 So the effective voltage is something on the order of pretty close to being 00:06:16.750 --> 00:06:24.035 about 70 percent of the peak value 0.707 V peak. 00:06:24.035 --> 00:06:26.955 Now then, what does that do for us? 00:06:26.955 --> 00:06:30.815 With this effective or this RMS value, 00:06:30.815 --> 00:06:36.220 we can now calculate an average power delivered to a resistive load is equal 00:06:36.220 --> 00:06:43.520 to V_rms times Pi_rms. 00:06:43.520 --> 00:06:46.760 The same expression that we've been used to using all along. 00:06:46.760 --> 00:06:52.400 In other words, this RMS voltage and RMS current are the effectively the 00:06:52.400 --> 00:06:58.120 same as what you would have if you had a DC source driving it. 00:06:58.120 --> 00:07:00.560 To say that even yet a different way, 00:07:00.560 --> 00:07:06.620 a DC source with a value of V_rms will deliver the same amount of power to 00:07:06.620 --> 00:07:15.940 a resistive load as an AC signal with V_m voltage peak.