[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.71,0:00:05.76,Default,,0000,0000,0000,,>> In a circuit that's being driven\Nby a sinusoidally varying voltage,
Dialogue: 0,0:00:05.76,0:00:11.43,Default,,0000,0000,0000,,the current that's flowing in that circuit\Nis also going to be sinusoidal.
Dialogue: 0,0:00:11.43,0:00:16.08,Default,,0000,0000,0000,,The power delivered to\Nthis resistive load which is equal to
Dialogue: 0,0:00:16.08,0:00:20.64,Default,,0000,0000,0000,,the voltage times the current\Nis also going to be varying.
Dialogue: 0,0:00:20.64,0:00:23.48,Default,,0000,0000,0000,,That means, at some points when\Nthe voltage and current are at
Dialogue: 0,0:00:23.48,0:00:26.59,Default,,0000,0000,0000,,maximum values you're going to be\Ndelivering a lot of power to the load,
Dialogue: 0,0:00:26.59,0:00:30.23,Default,,0000,0000,0000,,and at times when the voltage\Nor the current is zero,
Dialogue: 0,0:00:30.23,0:00:33.41,Default,,0000,0000,0000,,at that instant, you won't be\Ndelivering any power to the load.
Dialogue: 0,0:00:33.41,0:00:35.42,Default,,0000,0000,0000,,So then becomes interesting to say,
Dialogue: 0,0:00:35.42,0:00:36.80,Default,,0000,0000,0000,,well is there a way of determining what
Dialogue: 0,0:00:36.80,0:00:41.06,Default,,0000,0000,0000,,the average power is being\Ndelivered over one cycle?
Dialogue: 0,0:00:41.06,0:00:44.48,Default,,0000,0000,0000,,Is there are some effective quantity\Nthat we can use to
Dialogue: 0,0:00:44.48,0:00:46.13,Default,,0000,0000,0000,,calculate the effective power or
Dialogue: 0,0:00:46.13,0:00:48.70,Default,,0000,0000,0000,,the average power that's\Nbeing delivered to the load?
Dialogue: 0,0:00:48.70,0:00:51.62,Default,,0000,0000,0000,,We might say, well, let's\Njust take the average of it.
Dialogue: 0,0:00:51.62,0:00:57.04,Default,,0000,0000,0000,,Let's start with a voltage source\Nor some sinusoidally varying thing.
Dialogue: 0,0:00:57.04,0:01:00.20,Default,,0000,0000,0000,,Calculate its average value.
Dialogue: 0,0:01:00.20,0:01:02.63,Default,,0000,0000,0000,,The average value of say
Dialogue: 0,0:01:02.63,0:01:07.11,Default,,0000,0000,0000,,some sinusoidally varying voltage\NV of t. If V of t is
Dialogue: 0,0:01:07.11,0:01:14.66,Default,,0000,0000,0000,,equal to V_m cosine Omega t,
Dialogue: 0,0:01:14.66,0:01:17.33,Default,,0000,0000,0000,,then, the average voltage is defined as one
Dialogue: 0,0:01:17.33,0:01:20.24,Default,,0000,0000,0000,,over t times the integral from 0 to t,
Dialogue: 0,0:01:20.24,0:01:26.18,Default,,0000,0000,0000,,V_m cosine Omega t, dt.
Dialogue: 0,0:01:26.18,0:01:29.07,Default,,0000,0000,0000,,Without going to bother\Ndoing this integration,
Dialogue: 0,0:01:29.07,0:01:31.13,Default,,0000,0000,0000,,we know that when you\Nintegrate you are actually
Dialogue: 0,0:01:31.13,0:01:33.66,Default,,0000,0000,0000,,adding up the area under the curve.
Dialogue: 0,0:01:33.66,0:01:36.50,Default,,0000,0000,0000,,In this case going from zero to t,
Dialogue: 0,0:01:36.50,0:01:43.38,Default,,0000,0000,0000,,we're going over one period\Nand adding up the area.
Dialogue: 0,0:01:43.38,0:01:46.00,Default,,0000,0000,0000,,It turns out that in a sine wave,
Dialogue: 0,0:01:46.00,0:01:52.87,Default,,0000,0000,0000,,there's just as much area above\Nthe axis positive area as there is area
Dialogue: 0,0:01:52.87,0:02:00.30,Default,,0000,0000,0000,,under the line and thus\Nthe average voltage is zero.
Dialogue: 0,0:02:00.30,0:02:02.08,Default,,0000,0000,0000,,Well, that doesn't help us\Na whole lot because we know
Dialogue: 0,0:02:02.08,0:02:03.88,Default,,0000,0000,0000,,there's power being delivered to
Dialogue: 0,0:02:03.88,0:02:06.43,Default,,0000,0000,0000,,that resistive load but calculating
Dialogue: 0,0:02:06.43,0:02:10.33,Default,,0000,0000,0000,,the average voltage of the average current\Nisn't going to do anything for us.
Dialogue: 0,0:02:10.33,0:02:13.93,Default,,0000,0000,0000,,Instead, we calculate what\Nis known as the root means
Dialogue: 0,0:02:13.93,0:02:17.70,Default,,0000,0000,0000,,squared or the effective voltage.
Dialogue: 0,0:02:17.70,0:02:23.32,Default,,0000,0000,0000,,To do that we start by taking\Nour voltage or current.
Dialogue: 0,0:02:23.32,0:02:26.40,Default,,0000,0000,0000,,We're going to calculate it for\Nboth the voltage and the current.
Dialogue: 0,0:02:26.40,0:02:30.28,Default,,0000,0000,0000,,You square that quantity.
Dialogue: 0,0:02:30.28,0:02:32.21,Default,,0000,0000,0000,,Here's the voltage squared.
Dialogue: 0,0:02:32.21,0:02:34.31,Default,,0000,0000,0000,,You'll notice that what was negative is
Dialogue: 0,0:02:34.31,0:02:37.10,Default,,0000,0000,0000,,now been flipped over\Nso it is now positive.
Dialogue: 0,0:02:37.10,0:02:40.04,Default,,0000,0000,0000,,The shape is somewhat\Ndifferent but it's still
Dialogue: 0,0:02:40.04,0:02:43.56,Default,,0000,0000,0000,,ranges from in this case zero to one.
Dialogue: 0,0:02:43.56,0:02:48.89,Default,,0000,0000,0000,,We then average this squared value
Dialogue: 0,0:02:48.89,0:02:50.75,Default,,0000,0000,0000,,and then to bring it back to
Dialogue: 0,0:02:50.75,0:02:53.75,Default,,0000,0000,0000,,the appropriate amplitude\Nor the appropriate units,
Dialogue: 0,0:02:53.75,0:02:55.76,Default,,0000,0000,0000,,we take the square root of it.
Dialogue: 0,0:02:55.76,0:02:58.13,Default,,0000,0000,0000,,In other words, the RMS value or
Dialogue: 0,0:02:58.13,0:03:02.69,Default,,0000,0000,0000,,this effective value of the voltage\Nis found by squaring the voltage,
Dialogue: 0,0:03:02.69,0:03:04.68,Default,,0000,0000,0000,,calculating the average of the square,
Dialogue: 0,0:03:04.68,0:03:06.78,Default,,0000,0000,0000,,and then taking the square root of that.
Dialogue: 0,0:03:06.78,0:03:08.93,Default,,0000,0000,0000,,Let's go ahead and show\Nyou what we mean here.
Dialogue: 0,0:03:08.93,0:03:12.42,Default,,0000,0000,0000,,If V of t is equal to V_m cosine Omega,
Dialogue: 0,0:03:12.42,0:03:17.22,Default,,0000,0000,0000,,then V squared of t is equal to V_m
Dialogue: 0,0:03:17.22,0:03:22.53,Default,,0000,0000,0000,,squared cosine squared of\NOmega t. We're going to take
Dialogue: 0,0:03:22.53,0:03:24.23,Default,,0000,0000,0000,,that and we're going to\Ncalculate the average of
Dialogue: 0,0:03:24.23,0:03:27.02,Default,,0000,0000,0000,,that which is done by integrating from
Dialogue: 0,0:03:27.02,0:03:33.04,Default,,0000,0000,0000,,0 to t. Let's bring the V_m squared\Nout in front since it's a constant.
Dialogue: 0,0:03:33.04,0:03:39.21,Default,,0000,0000,0000,,Cosine squared of Omega t dt.
Dialogue: 0,0:03:39.77,0:03:42.15,Default,,0000,0000,0000,,We have to avail ourselves of
Dialogue: 0,0:03:42.15,0:03:47.18,Default,,0000,0000,0000,,a trigonometric identity that\Nsays that the cosine squared of
Dialogue: 0,0:03:47.18,0:03:56.40,Default,,0000,0000,0000,,Theta is equal to 1.5 times 1\Nplus the cosine of 2 Theta.
Dialogue: 0,0:03:56.40,0:03:57.89,Default,,0000,0000,0000,,So to perform this integral,
Dialogue: 0,0:03:57.89,0:04:00.53,Default,,0000,0000,0000,,the trick is to instead of trying\Nto integrate the cosine squared,
Dialogue: 0,0:04:00.53,0:04:05.68,Default,,0000,0000,0000,,we use this identity and substitute\Nin then for this we have,
Dialogue: 0,0:04:05.68,0:04:10.83,Default,,0000,0000,0000,,V_m squared over T times\Nthe integral from 0 to
Dialogue: 0,0:04:10.83,0:04:19.37,Default,,0000,0000,0000,,t. We'll bring the 1.5 out in front also,
Dialogue: 0,0:04:19.37,0:04:24.84,Default,,0000,0000,0000,,times 1 plus the cosine of
Dialogue: 0,0:04:24.84,0:04:32.43,Default,,0000,0000,0000,,2 Omega t dt.
Dialogue: 0,0:04:32.43,0:04:35.50,Default,,0000,0000,0000,,Now, when we integrate\Nthis over the period,
Dialogue: 0,0:04:35.50,0:04:38.20,Default,,0000,0000,0000,,this term we're actually\Nintegrating over two periods.
Dialogue: 0,0:04:38.20,0:04:40.72,Default,,0000,0000,0000,,In the same way that when you\Nintegrate it over one period,
Dialogue: 0,0:04:40.72,0:04:42.94,Default,,0000,0000,0000,,the positive area cancel the negative area,
Dialogue: 0,0:04:42.94,0:04:49.63,Default,,0000,0000,0000,,this term here likewise gives us\Na zero contribution to this integral.
Dialogue: 0,0:04:49.63,0:04:52.42,Default,,0000,0000,0000,,When you integrate one with respect to t,
Dialogue: 0,0:04:52.42,0:05:01.90,Default,,0000,0000,0000,,we adjust t and so we have\NV_m squared over 2T times
Dialogue: 0,0:05:01.90,0:05:05.67,Default,,0000,0000,0000,,t evaluated from 0 to
Dialogue: 0,0:05:05.67,0:05:13.62,Default,,0000,0000,0000,,T. That then is just V_m squared over 2.
Dialogue: 0,0:05:13.62,0:05:17.01,Default,,0000,0000,0000,,So we squared it,
Dialogue: 0,0:05:17.01,0:05:19.96,Default,,0000,0000,0000,,we calculated the average.
Dialogue: 0,0:05:19.96,0:05:23.48,Default,,0000,0000,0000,,Now all is left to do is to\Ntake the square root of it and
Dialogue: 0,0:05:23.48,0:05:26.96,Default,,0000,0000,0000,,we've have in V_rms is
Dialogue: 0,0:05:26.96,0:05:34.55,Default,,0000,0000,0000,,equal to the square root\Nof V_m squared over 2.
Dialogue: 0,0:05:34.55,0:05:38.57,Default,,0000,0000,0000,,If we take that square root of\Ntwo out from under the radical,
Dialogue: 0,0:05:38.57,0:05:41.78,Default,,0000,0000,0000,,it comes out as 1 over
Dialogue: 0,0:05:41.78,0:05:46.22,Default,,0000,0000,0000,,the square root of 2 times the\Nsquare root of V_m squared.
Dialogue: 0,0:05:46.22,0:05:48.96,Default,,0000,0000,0000,,Well, that's just V_m.
Dialogue: 0,0:05:48.96,0:05:51.62,Default,,0000,0000,0000,,Thus, we see that the RMS value,
Dialogue: 0,0:05:51.62,0:05:57.24,Default,,0000,0000,0000,,this effective voltage is equal\Nto 1 over the square root of
Dialogue: 0,0:05:57.24,0:06:04.12,Default,,0000,0000,0000,,2 times V_m where V_m was\Nthe maximum value, our peak value.
Dialogue: 0,0:06:04.31,0:06:07.87,Default,,0000,0000,0000,,Well, 1 over the square root of\N2 that's approximately equal
Dialogue: 0,0:06:07.87,0:06:12.48,Default,,0000,0000,0000,,to 0.707 times V_m.
Dialogue: 0,0:06:12.48,0:06:16.75,Default,,0000,0000,0000,,So the effective voltage is something\Non the order of pretty close to being
Dialogue: 0,0:06:16.75,0:06:24.04,Default,,0000,0000,0000,,about 70 percent of\Nthe peak value 0.707 V peak.
Dialogue: 0,0:06:24.04,0:06:26.96,Default,,0000,0000,0000,,Now then, what does that do for us?
Dialogue: 0,0:06:26.96,0:06:30.82,Default,,0000,0000,0000,,With this effective or this RMS value,
Dialogue: 0,0:06:30.82,0:06:36.22,Default,,0000,0000,0000,,we can now calculate an average power\Ndelivered to a resistive load is equal
Dialogue: 0,0:06:36.22,0:06:43.52,Default,,0000,0000,0000,,to V_rms times Pi_rms.
Dialogue: 0,0:06:43.52,0:06:46.76,Default,,0000,0000,0000,,The same expression that we've\Nbeen used to using all along.
Dialogue: 0,0:06:46.76,0:06:52.40,Default,,0000,0000,0000,,In other words, this RMS voltage and\NRMS current are the effectively the
Dialogue: 0,0:06:52.40,0:06:58.12,Default,,0000,0000,0000,,same as what you would have if\Nyou had a DC source driving it.
Dialogue: 0,0:06:58.12,0:07:00.56,Default,,0000,0000,0000,,To say that even yet a different way,
Dialogue: 0,0:07:00.56,0:07:06.62,Default,,0000,0000,0000,,a DC source with a value of V_rms will\Ndeliver the same amount of power to
Dialogue: 0,0:07:06.62,0:07:15.94,Default,,0000,0000,0000,,a resistive load as an AC signal\Nwith V_m voltage peak.