RC Circuits Physics Problems, Time Constant Explained, Capacitor Charging and Discharging

WEBVTT
Kind: captions
Language: en

00:00:00.030
in this video we're going to talk about
00:00:03.139 00:00:03.149 capacitors particularly when they're
00:00:07.460 00:00:07.470 charging and discharging so let's say if
00:00:10.339 00:00:10.349 we have an RC circuit that is a circuit
00:00:13.249 00:00:13.259 containing a capacitor with a resistor
00:00:15.169 00:00:15.179 and we're going to have an open switch
00:00:17.689 00:00:17.699 initially let's say that the voltage of
00:00:21.380 00:00:21.390 the battery is 12 volts this is the
00:00:27.769 00:00:27.779 resistor R and here we have the
00:00:30.650 00:00:30.660 capacitor represented by the symbol C
00:00:34.360 00:00:34.370 now let's switch is closed charge will
00:00:39.470 00:00:39.480 begin to flow current will flow from the
00:00:44.090 00:00:44.100 positive terminal of the battery to
00:00:48.860 00:00:48.870 charge the capacitor the longer side is
00:00:51.410 00:00:51.420 the positive terminal now keep in mind
00:00:53.660 00:00:53.670 no electrons may flow in the opposite
00:00:56.060 00:00:56.070 direction electrons flow from the
00:00:58.279 00:00:58.289 negative terminal and it answers the
00:01:01.189 00:01:01.199 positive terminal now initially when a
00:01:06.109 00:01:06.119 switch is open the voltage of the
00:01:08.600 00:01:08.610 capacitor will be zero as soon as the
00:01:13.640 00:01:13.650 switch is closed when T is zero the
00:01:17.240 00:01:17.250 voltage will still be zero but the
00:01:19.160 00:01:19.170 voltage across the resistor initially
00:01:21.520 00:01:21.530 will equal the voltage of value which is
00:01:24.230 00:01:24.240 12 now after some significant time has
00:01:27.770 00:01:27.780 passed the capacitor will be fully
00:01:30.230 00:01:30.240 charged let's call it V final it's going
00:01:33.950 00:01:33.960 to have a final voltage of 12 when the
00:01:36.890 00:01:36.900 capacitor is fully charged the voltage
00:01:39.679 00:01:39.689 across the resistance is going to be
00:01:41.319 00:01:41.329 zero the reason mean is once you muscle
00:01:44.780 00:01:44.790 capacity fully charged no current will
00:01:47.330 00:01:47.340 flow in the circuit and if there's no
00:01:49.310 00:01:49.320 current flowing to the resistor the
00:01:50.990 00:01:51.000 voltage across that resistor will be
00:01:52.999 00:01:53.009 zero so initially when a capacitor has a
00:01:58.580 00:01:58.590 voltage of zero the voltage as a
00:02:00.499 00:02:00.509 resistor will be 12 because it has to
00:02:02.810 00:02:02.820 add up to the battery voltage when the
00:02:05.209 00:02:05.219 capacitor is fully charged to 12 volts
00:02:07.130 00:02:07.140 the voltage across the resistor will be
00:02:09.469 00:02:09.479 zero because I have absolute there's no
00:02:11.900 00:02:11.910 current flowing to the circuit
00:02:13.430 00:02:13.440 and keep in mind no current actually
00:02:16.550 00:02:16.560 passes through the capacitor the current
00:02:19.880 00:02:19.890 that flows through the circuit is simply
00:02:22.790 00:02:22.800 used to pump charge from one plate of
00:02:25.610 00:02:25.620 the capacitor to the other plate the
00:02:29.540 00:02:29.550 equation that describes a capacitor
00:02:31.580 00:02:31.590 charging is this equation that's equal
00:02:34.490 00:02:34.500 to the Engelmann of the battery which in
00:02:37.130 00:02:37.140 this problem is 12 times 1 minus e is
00:02:41.780 00:02:41.790 basically the inverse of the natural log
00:02:43.730 00:02:43.740 function minus the time divided by RC
00:02:49.210 00:02:49.220 now the graph looks something like this
00:02:59.160 00:02:59.170 this is the EMF of the battery which is
00:03:01.080 00:03:01.090 12 volts and the capacitor will charge
00:03:05.430 00:03:05.440 progressively towards 12 volts but it
00:03:11.729 00:03:11.739 starts from zero and so depending on the
00:03:15.750 00:03:15.760 resistance and the values of the
00:03:17.220 00:03:17.230 pathogens it will affect the shape of
00:03:19.979 00:03:19.989 this graph but for the most part that's
00:03:22.410 00:03:22.420 how it's going to look like it's going
00:03:24.420 00:03:24.430 to increase towards 12 at which point is
00:03:26.850 00:03:26.860 just going to be a horizontal line so
00:03:32.009 00:03:32.019 this is the voltage of the capacitor and
00:03:33.630 00:03:33.640 on the x-axis you have the time
00:03:37.280 00:03:37.290 now there's something called tau which
00:03:39.930 00:03:39.940 is the time constant and that's equal to
00:03:42.690 00:03:42.700 the product of the resistance and the
00:03:45.090 00:03:45.100 capacitance it's our scene and so it has
00:03:47.670 00:03:47.680 its own times farad's now let's talk
00:03:51.900 00:03:51.910 about the time constant let's make a
00:03:54.420 00:03:54.430 table
00:04:08.470 00:04:08.480 this is going to be the voltage at any
00:04:11.080 00:04:11.090 time team that is the voltage of the
00:04:13.420 00:04:13.430 capacitor so how much will the capacitor
00:04:18.099 00:04:18.109 be charged at the runtime constant one
00:04:23.320 00:04:23.330 time constant is basically it's equal to
00:04:29.400 00:04:29.410 run our scene two time constants it's
00:04:33.300 00:04:33.310 just two RC so what you need to do is
00:04:38.650 00:04:38.660 replace T with one RC in the equation
00:04:44.890 00:04:44.900 and RC will cancel so basically you just
00:04:49.330 00:04:49.340 got a plug in run e to the minus one if
00:04:52.510 00:04:52.520 you type that in you should get point
00:04:55.000 00:04:55.010 six three two which is sixty three point
00:04:58.450 00:04:58.460 two percent now the maximum voltage of
00:05:02.409 00:05:02.419 the capacitor is the voltage of the
00:05:04.540 00:05:04.550 battery which is twelve sixty three
00:05:06.670 00:05:06.680 point two percent of twelve is seven
00:05:09.430 00:05:09.440 point five eight volts so after run time
00:05:12.850 00:05:12.860 constant the voltage of the capacitor
00:05:15.040 00:05:15.050 will be seven point five eight volts the
00:05:18.450 00:05:18.460 63.2% of its maximum now what about
00:05:22.000 00:05:22.010 after two time constants well if we type
00:05:25.000 00:05:25.010 in 1 minus e to negative two you should
00:05:28.510 00:05:28.520 see 2865 or 86.5% 86.5% of twelve is ten
00:05:36.640 00:05:36.650 point three eight so after two time
00:05:39.279 00:05:39.289 constants the voltage of the capacitor
00:05:40.770 00:05:40.780 will be ten point three eight now one
00:05:44.650 00:05:44.660 minus e to the minus three will give you
00:05:48.190 00:05:48.200 point nine five one ninety five percent
00:05:51.900 00:05:51.910 which correlates to a voltage of eleven
00:05:55.060 00:05:55.070 point four volts after fourth time
00:05:59.050 00:05:59.060 constant the capacitor will be ninety
00:06:01.540 00:06:01.550 eight point two percent charged
00:06:07.070 00:06:07.080 after five constants it's going to be
00:06:11.740 00:06:11.750 99.3% charge it's going to have a
00:06:14.779 00:06:14.789 voltage of eleven point nine zero so it
00:06:19.490 00:06:19.500 takes about 5 time constants for a
00:06:22.969 00:06:22.979 capacitor to be fully charged
00:06:25.390 00:06:25.400 it's about 99 percent charge at that
00:06:28.129 00:06:28.139 point now what about this charge in a
00:06:30.890 00:06:30.900 capacitor
00:06:31.629 00:06:31.639 let's say if the capacitor is fully
00:06:33.770 00:06:33.780 charged and we connected across the
00:06:36.920 00:06:36.930 resistor let's say it now has a voltage
00:06:40.520 00:06:40.530 of 12 volts what type of graphs will we
00:06:43.550 00:06:43.560 have when a capacitor is discharging the
00:06:49.520 00:06:49.530 graph will look like this
00:06:50.959 00:06:50.969 it's going to start from its initial
00:06:52.999 00:06:53.009 value of 12 and it's going to
00:06:55.279 00:06:55.289 progressively decrease to 0 the equation
00:07:00.709 00:07:00.719 that describes this graph is this one Z
00:07:04.100 00:07:04.110 is equal to the initial that's the
00:07:06.709 00:07:06.719 original voltage of the capacitor from T
00:07:08.929 00:07:08.939 is zero at 12 volts times e raised to
00:07:12.740 00:07:12.750 the negative T over RC now let's make
00:07:19.070 00:07:19.080 another table
00:07:27.489 00:07:27.499 vfc is the voltage of the capacitor and
00:07:30.939 00:07:30.949 on the x-axis we have time now after one
00:07:36.499 00:07:36.509 time constant
00:07:39.070 00:07:39.080 how much will percentage of the charge
00:07:42.469 00:07:42.479 does the capacitor still have so that is
00:07:46.999 00:07:47.009 this equation if we replace T with
00:07:49.640 00:07:49.650 negative R seen this is going to be e to
00:07:56.360 00:07:56.370 negative 1 if you type in each negative
00:07:58.249 00:07:58.259 one you're going to get 0.368 what's
00:08:01.939 00:08:01.949 going to be 36.8% charged the voltage is
00:08:06.309 00:08:06.319 now four point four two volts that's
00:08:10.700 00:08:10.710 thirty six point eight a set of twelve
00:08:13.390 00:08:13.400 after two time constants it's going to
00:08:17.809 00:08:17.819 be thirteen point five percent then the
00:08:22.219 00:08:22.229 other table after two time constants it
00:08:24.409 00:08:24.419 was eighty-six corn-factor sir these two
00:08:26.929 00:08:26.939 numbers have to add up to 100
00:08:28.869 00:08:28.879 so the voltage at this point is going to
00:08:31.219 00:08:31.229 be one point six two volts
00:08:35.500 00:08:35.510 after three time constants it's going to
00:08:38.930 00:08:38.940 be five percent charge 95 percent
00:08:42.949 00:08:42.959 discharged
00:08:44.410 00:08:44.420 so it's voltage is now going to be point
00:08:48.259 00:08:48.269 6 volts after four time constants it's
00:08:54.439 00:08:54.449 going to have one point eight percent of
00:08:56.269 00:08:56.279 its initial charge so the voltage is
00:08:58.850 00:08:58.860 going to be point twenty two so it takes
00:09:02.930 00:09:02.940 about 5 time constants for the capacitor
00:09:06.259 00:09:06.269 to be 99 percent discharge where it's
00:09:09.949 00:09:09.959 going to have 27 percent remaining of
00:09:13.069 00:09:13.079 the original truck and so the voltage is
00:09:15.470 00:09:15.480 going to be point zero eight so at that
00:09:17.629 00:09:17.639 point you can say that the capacitor for
00:09:19.879 00:09:19.889 the most part is discharged now let's
00:09:23.269 00:09:23.279 work on an example problem so let's say
00:09:26.480 00:09:26.490 we have an RC circuit and the capacitor
00:09:29.600 00:09:29.610 is attached to the battery by means of a
00:09:32.269 00:09:32.279 resistor and initially at T equals zero
00:09:35.840 00:09:35.850 let's say the voltage of the capacitor
00:09:37.040 00:09:37.050 is there
00:09:39.450 00:09:39.460 and the capacitor is going to have a
00:09:41.850 00:09:41.860 value of 500 micro frats and let's say
00:09:48.269 00:09:48.279 that the resistor it's a 1 kilo ohm
00:09:51.750 00:09:51.760 resistor and the voltage of the battery
00:09:56.010 00:09:56.020 is going to be 20 volts with this
00:10:01.230 00:10:01.240 information how long will it take for
00:10:04.829 00:10:04.839 the capacitor to reach 90% of its
00:10:09.329 00:10:09.339 maximum charge so what is 90% of 20
00:10:15.780 00:10:15.790 volts 20 times 90% is 18 volts so we
00:10:24.510 00:10:24.520 want to know how long will it take for
00:10:27.090 00:10:27.100 the capacitor to be 90% charge or to
00:10:30.810 00:10:30.820 reach a voltage of 18 volts
00:10:35.120 00:10:35.130 since the capacitor is charging we can
00:10:38.190 00:10:38.200 use this equation
00:10:45.710 00:10:45.720 z is the voltage of the capacitor which
00:10:49.400 00:10:49.410 is 18 volts the EMF of the battery is 20
00:10:53.710 00:10:53.720 and let's find out the value of RC which
00:10:59.150 00:10:59.160 is one time constant R is 1 kilo ohm
00:11:03.470 00:11:03.480 which is a thousand ohms C is 500 micro
00:11:08.720 00:11:08.730 farad which is 500 times 10 to minus six
00:11:12.920 00:11:12.930 farad now if we multiply these two a
00:11:18.100 00:11:18.110 thousand times 500 times 10 to the minus
00:11:21.560 00:11:21.570 6 and this is equal to 0.5 0.5 is the
00:11:27.860 00:11:27.870 value of one time constant so a time
00:11:31.310 00:11:31.320 constant is 0.5 seconds in this
00:11:33.770 00:11:33.780 particular problem because sometimes you
00:11:36.860 00:11:36.870 need to know what the time constants are
00:11:39.730 00:11:39.740 so let's replace RC with point five and
00:11:43.630 00:11:43.640 now our goal is to solve for team
00:11:52.630 00:11:52.640 so the first thing we need to do is
00:11:54.770 00:11:54.780 divide both sides by 18 I mean rather by
00:11:58.340 00:11:58.350 20 18 divided by 20 is 0.9 so point 9 is
00:12:04.250 00:12:04.260 equal to 1 minus e raised to the
00:12:07.010 00:12:07.020 negative T divided by point 5 next
00:12:10.120 00:12:10.130 subtract both sides by 1 point 9 minus 1
00:12:14.330 00:12:14.340 is negative point 1 so we have negative
00:12:17.660 00:12:17.670 point 1 minus e raised to the negative T
00:12:20.770 00:12:20.780 divided by point 5 next multiply both
00:12:24.350 00:12:24.360 sides by negative 1 this will change
00:12:26.600 00:12:26.610 negative signs into a positive sign
00:12:29.980 00:12:29.990 after that let's take the natural log of
00:12:32.990 00:12:33.000 both sides so on the Left we have the
00:12:38.750 00:12:38.760 natural log of point 1 and on the right
00:12:42.110 00:12:42.120 we have the natural log of e to the
00:12:45.050 00:12:45.060 negative T divided by point 5 now a
00:12:49.070 00:12:49.080 property of logs allows you to take the
00:12:51.860 00:12:51.870 exponent and move it to the front so
00:12:54.970 00:12:54.980 therefore what we now have is the
00:12:58.370 00:12:58.380 natural log of point 1 is equal to
00:13:01.670 00:13:01.680 negative team divided by 0.5 times ln e
00:13:06.790 00:13:06.800 ln e is equal to 1 now let's just get
00:13:15.200 00:13:15.210 rid of this stuff
00:13:22.250 00:13:22.260 now let's multiply both sides by
00:13:26.290 00:13:26.300 negative 0.5 this will cancel the point
00:13:30.950 00:13:30.960 5 on the right as well as the negative
00:13:33.110 00:13:33.120 sign so T is equal to negative 0.5 times
00:13:40.250 00:13:40.260 the natural log of 0.1 so if we type
00:13:43.670 00:13:43.680 this in this will give us 1 point 1 5
00:13:52.160 00:13:52.170 seconds so that's the time it takes for
00:13:56.720 00:13:56.730 00:13:59.930 00:13:59.940 maximum value to reach a voltage of 18
00:14:02.720 00:14:02.730 volts now is there an equation that can
00:14:06.050 00:14:06.060 help us get this answer a lot faster if
00:14:09.100 00:14:09.110 you rearrange the equation you can use
00:14:12.230 00:14:12.240 this formula T is equal to negative RC
00:14:16.150 00:14:16.160 times the natural log of 1 minus V
00:14:25.460 00:14:25.470 divided by E where Z is the voltage of
00:14:29.300 00:14:29.310 the capacitor at any time T and this
00:14:33.920 00:14:33.930 epsilon symbol is basically the voltage
00:14:37.310 00:14:37.320 of the battery now the way you should
00:14:42.290 00:14:42.300 type it in should be like this R make
00:14:44.660 00:14:44.670 sure you use a thousand don't use one
00:14:46.400 00:14:46.410 kilo C is going to be 500 times 10 to
00:14:50.360 00:14:50.370 the minus 6 and then multiply that by
00:14:52.730 00:14:52.740 the natural log of 1 minus 18 volts
00:14:57.260 00:14:57.270 divided by 20 if you type this in
00:15:00.650 00:15:00.660 exactly the way you see it and this is a
00:15:03.020 00:15:03.030 multiplication sign this will give you
00:15:05.780 00:15:05.790 one point one five seconds so we have
00:15:10.940 00:15:10.950 the time and we also have the time
00:15:14.960 00:15:14.970 constant as you said before one time
00:15:18.800 00:15:18.810 constant is equal to RC which is point
00:15:23.030 00:15:23.040 five seconds so using this information
00:15:26.530 00:15:26.540 how many time constants does it take for
00:15:31.160 00:15:31.170 the capacitor to be 90% charge
00:15:34.269 00:15:34.279 how can you figure this out to find the
00:15:38.960 00:15:38.970 number of time constants convert it
00:15:42.370 00:15:42.380 start with the time that you have which
00:15:45.290 00:15:45.300 is 1.15 seconds over 1 and we know that
00:15:49.819 00:15:49.829 one time constant is equivalent to point
00:15:53.840 00:15:53.850 five seconds so notice that the unit
00:15:57.410 00:15:57.420 seconds will cancel giving us the number
00:15:59.930 00:15:59.940 of time constants so anytime you have to
00:16:03.110 00:16:03.120 find the number of time constants which
00:16:04.610 00:16:04.620 is we'll call it n it's equal to the
00:16:07.220 00:16:07.230 time divided by tau that's how you can
00:16:12.860 00:16:12.870 find it so it's 1.15 divided by point
00:16:16.160 00:16:16.170 five and so it takes about two point
00:16:22.550 00:16:22.560 three time constants for the capacitor
00:16:25.340 00:16:25.350 to be 90% charged now does this answer
00:16:28.340 00:16:28.350 make sense
00:16:31.750 00:16:31.760 consider the first table that we went
00:16:34.340 00:16:34.350 over early in this video we said that it
00:16:37.639 00:16:37.649 takes two time constants for the
00:16:40.970 00:16:40.980 capacitor to be 86.5% charged and that
00:16:47.240 00:16:47.250 it takes three time constants for the
00:16:51.230 00:16:51.240 capacitor to be 95% charge
00:16:55.360 00:16:55.370 so therefore 90% is between 86.5% and 90
00:17:01.720 00:17:01.730 which means that it should take
00:17:04.640 00:17:04.650 somewhere between two and three time
00:17:06.470 00:17:06.480 constants to be 90% charged and 2.3 is
00:17:11.360 00:17:11.370 between 2 and 3 so that answer does
00:17:14.059 00:17:14.069 indeed make sense and so now you know
00:17:17.179 00:17:17.189 how to find the number of time constants
00:17:19.490 00:17:19.500 it takes to charge or even discharge the
00:17:22.549 00:17:22.559 capacitor to a certain level and so
00:17:26.090 00:17:26.100 we're going to stop here so that's it
00:17:28.189 00:17:28.199 for this video thanks for watching and
00:17:30.289 00:17:30.299 have a great thing
Office location
Engineering company LOTUS®
Russia, Ekaterinburg, Lunacharskogo street, 240/12

Phone: +7 343 216 77 75

E-mail: info@lotus1.ru

Sales phone

Russia: +7 343 216 77 75

WhatsApp: +79122710308