00:00:02.240 so now I have my two capacitor equations 00:00:04.82000:00:04.830 the two forms of this equation one is I 00:00:06.95000:00:06.960 in terms of V and the other is V in 00:00:10.19000:00:10.200 terms of I and we're going to basically 00:00:12.29000:00:12.300 look at this equation here and do a 00:00:14.33000:00:14.340 little exercise with it to see how it 00:00:15.77000:00:15.780 works 00:00:20.10900:00:20.119 I'm going to draw a little circuit here 00:00:34.06000:00:34.070 and it's going to have a current source 00:00:36.00900:00:36.019 and a capacitor the value of the 00:00:38.68000:00:38.690 capacitor is 1 micro farad and the value 00:00:43.45000:00:43.460 of our current source we'll call it big 00:00:45.70000:00:45.710 I and it's going to actually look like a 00:00:47.74000:00:47.750 pulse like that that will go from 0 to 3 00:00:53.97900:00:53.989 milliamps and then back to 0 over here 00:00:57.67000:00:57.680 and the amount of time it takes that's 00:01:00.25000:01:00.260 going to be this time here is going to 00:01:02.64900:01:02.659 be 3 milli seconds and the question I 00:01:08.77000:01:08.780 want to answer is what is what is V of T 00:01:13.74900:01:13.759 right here we're going to use this 00:01:16.99000:01:17.000 integral equation to figure that out 00:01:19.91900:01:19.929 so over here is where we're going to put 00:01:22.30000:01:22.310 our answer this is going to be T and 00:01:26.55900:01:26.569 this will be I here's I and our plot 00:01:34.44900:01:34.459 right down here will be T and here's V 00:01:39.24900:01:39.259 of T will plot I this way on here in 00:01:45.30900:01:45.319 time it's 0 and then it goes up to some 00:01:48.79000:01:48.800 value and then it goes over so it's a 00:01:52.38900:01:52.399 pulse of current and we said that that 00:01:55.35900:01:55.369 was zero this is three milliamps and 00:01:59.24900:01:59.259 this is three milliseconds and now what 00:02:04.02900:02:04.039 we want to do is we want to find V of T 00:02:09.21000:02:09.220 one of the things we have to do we have 00:02:11.94900:02:11.959 to make an assumption about V naught 00:02:13.72000:02:13.730 here and we're going to assume for our 00:02:15.97000:02:15.980 problem here that V naught equals zero 00:02:19.21000:02:19.220 volts and what that means is there's no 00:02:21.28000:02:21.290 charge there's zero charge stored on 00:02:23.32000:02:23.330 this capacitor when we start the 00:02:25.03000:02:25.040 experiment so now that let's look at 00:02:28.08900:02:28.099 three different time periods let's look 00:02:29.64900:02:29.659 at the period before the pulse during 00:02:31.66000:02:31.670 the pulse and after the pulse so we'll 00:02:34.50900:02:34.519 break the problem into three parts part 00:02:38.41000:02:38.420 one is before and we'll do that by just 00:02:41.32000:02:41.330 looking up here and we decided that V 00:02:43.75000:02:43.760 dot was zero in the before state so put 00:02:47.37900:02:47.389 a little zero 00:02:47.87000:02:47.880 there we decided that I is zero so that 00:02:53.09000:02:53.100 means that the term inside the integral 00:02:54.71000:02:54.720 is zero and what that means zero plus 00:02:58.55000:02:58.560 zero is equal to V of T so before the 00:03:02.78000:03:02.790 pulse before the pulse the capacitor 00:03:05.93000:03:05.940 equation tells us the voltage is zero 00:03:11.02000:03:11.030 okay now let's go during the pulse let's 00:03:13.61000:03:13.620 do the second period of time let's do 00:03:15.20000:03:15.210 let's do during the pulse so now we have 00:03:17.60000:03:17.610 to be a little more careful we have V 00:03:19.64000:03:19.650 equals one over C integral from now time 00:03:25.64000:03:25.650 equals zero to time equals some time T 00:03:28.76000:03:28.770 and what's I of what's the what's I 00:03:33.26000:03:33.270 during the pulse what's sitting right 00:03:35.15000:03:35.160 here I is a constant so we write in 00:03:38.75000:03:38.760 three milliamps D tau 00:03:48.77000:03:48.780 Plus don't forget the starting voltage 00:03:50.90000:03:50.910 what's our starting voltage well we look 00:03:53.09000:03:53.100 right here and the starting voltage is 00:03:54.77000:03:54.780 zero plus zero so now we can solve this 00:03:59.63000:03:59.640 V equals 1 over C times 3 milliamps 00:04:06.19900:04:06.209 comes out times the integral from 0 to T 00:04:13.72900:04:13.739 of D tau and that equals let's plug in C 00:04:21.31900:04:21.329 this time three milliamps divided by 1 00:04:29.30000:04:29.310 microfarad times what does this evaluate 00:04:34.94000:04:34.950 to the integral from zero to T of D tau 00:04:37.58000:04:37.590 is just T let's do a little bit of 00:04:44.03000:04:44.040 arithmetic here to reduce this milliamps 00:04:46.58000:04:46.590 is 10 to the minus 3 and microfarads is 00:04:49.64000:04:49.650 10 to the minus 6 so let me move this up 00:04:57.19000:04:57.200 we'll keep our plots on there and what 00:05:01.13000:05:01.140 we end up with is V equals 3 10 to the 00:05:08.06000:05:08.070 minus 3 10 to the minus 6 so that's 3 00:05:09.98000:05:09.990 times 10 to the third or 3,000 times 00:05:15.26000:05:15.270 time and what's that that's the equation 00:05:19.27900:05:19.289 that's the equation of a line and it has 00:05:22.40000:05:22.410 a slope of 3000 what's the unit's here 00:05:24.86000:05:24.870 this is 3000 volts per second so I'll go 00:05:29.96000:05:29.970 over here almost we'll sketch this in 00:05:31.55000:05:31.560 this is going to be aligned that's a 00:05:32.90000:05:32.910 straight ramp with a constant slope like 00:05:35.96000:05:35.970 that it's going to be happening all 00:05:38.81000:05:38.820 during this pulse and we can ask what's 00:05:41.60000:05:41.610 this value right here what's that 00:05:45.32000:05:45.330 voltage right there okay let's work that 00:05:47.42000:05:47.430 out T is 3 milliseconds so let's plug in 00:05:53.27000:05:53.280 3 milliseconds right here so V at 3 00:05:56.65900:05:56.669 milliseconds equals 3000 00:06:01.55000:06:01.560 times three milliseconds and that equals 00:06:08.13000:06:08.140 three times three is nine and three 00:06:11.34000:06:11.350 thousand times this is an exponent of 00:06:14.19000:06:14.200 minus three so it's going to be nine 00:06:17.76000:06:17.770 volts so this value right here right 00:06:21.42000:06:21.430 there is nine volts so that says the 00:06:26.85000:06:26.860 voltage on our capacitor during the 00:06:29.67000:06:29.680 pulse during the current pulse rises in 00:06:32.13000:06:32.140 a straight line up to nine volts and we 00:06:35.49000:06:35.500 got that from this integral that we did 00:06:37.41000:06:37.420 the capacitors integrating the current 00:06:39.84000:06:39.850 adding up the current its integrating 00:06:41.88000:06:41.890 this pulse to get an ever-rising voltage 00:06:45.32000:06:45.330 okay so now we've solved the capacitor 00:06:48.21000:06:48.220 equation during a pulse and let's now go 00:06:51.03000:06:51.040 back to let's go back now to what 00:06:53.73000:06:53.740 happens after the pulse and we'll do 00:06:55.86000:06:55.870 that over in the corner over here so now 00:07:00.45000:07:00.460 let's solve what happens but now let's 00:07:02.43000:07:02.440 solve what happens after after the 00:07:05.07000:07:05.080 current pulse what happens to this 00:07:06.42000:07:06.430 voltage from here on does it go down 00:07:08.31000:07:08.320 does it go up does it go straight 00:07:09.81000:07:09.820 sideways let's find out and let's use 00:07:11.91000:07:11.920 our capacitor equation to do this so now 00:07:15.93000:07:15.940 what we're doing is we're going to 00:07:17.28000:07:17.290 define we're going to do a new integral 00:07:18.87000:07:18.880 and we're going to start with we're 00:07:21.15000:07:21.160 going to start at time equals three 00:07:24.51000:07:24.520 milliseconds and what is our voltage at 00:07:28.29000:07:28.300 three milliseconds what's V naught so V 00:07:30.84000:07:30.850 naught in this case equals nine volts 00:07:35.81000:07:35.820 that's for this period of time after 00:07:38.64000:07:38.650 three milliseconds and we'll use our 00:07:42.08000:07:42.090 equation again V is one over C times the 00:07:46.74000:07:46.750 integral of I of tau D tau plus V naught 00:07:58.25000:07:58.260 all right so let's and it goes from time 00:08:03.37000:08:03.380 it goes from time three milliseconds to 00:08:07.46000:08:07.470 time T so time T now is out here 00:08:12.08000:08:12.090 somewhere 00:08:12.53000:08:12.540 on the timescale and we're we know the 00:08:15.89000:08:15.900 voltage right here we filled it in 00:08:17.24000:08:17.250 there's there's the voltage right there 00:08:19.31000:08:19.320 we're going to fill that in and we're 00:08:21.08000:08:21.090 going to work out what's the voltage out 00:08:22.64000:08:22.650 here after three milliseconds so to 00:08:27.29000:08:27.300 solve this integral we look and see well 00:08:29.03000:08:29.040 what's I of towel what's I of towel 00:08:30.98000:08:30.990 during this time let's look at our chart 00:08:32.63000:08:32.640 the pulse is at zero 00:08:34.64000:08:34.650 oh look so this term this whole term 00:08:37.70000:08:37.710 right here this entire term right here 00:08:41.80000:08:41.810 is zero and what does V naught V naught 00:08:46.76000:08:46.770 we decide it was right there V naught is 00:08:48.83000:08:48.840 nine volts that's where we started from 00:08:51.29000:08:51.300 and so the answer here after after the 00:08:54.83000:08:54.840 pulse goes away is V equals nine volts 00:08:59.53000:08:59.540 and I can sketch that in up here and 00:09:02.00000:09:02.010 basically it just goes straight sideways 00:09:03.92000:09:03.930 at the new voltage so that's how we 00:09:08.81000:09:08.820 solve a capacitor problem a really 00:09:11.09000:09:11.100 simple one and it happens that we were 00:09:13.49000:09:13.500 integrating a current pulse and what we 00:09:17.03000:09:17.040 got was a voltage ramp
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