Electrical Engineering: Ch 8: RC & RL Circuits (31 of 65) General Strategy of Solving RC Circuits

welcome to electron line now let's start reviewing the Kjetil methodology for solving RC circuits here we have a typical first order RC circuits with a few voltage sources a capacitor a few resistors and a time equals zero the switch goes from the left to the right and then the way to solve it is to follow the following rules first what we want to find is 1 find the voltage across the capacitor at T equals 0 that's before any changes were allowed to happen that's what at the time that the switch will switch over at that very moment what's the voltage across the capacitor next we want to know what the voltage will be when time has gone by a lot of time has gone by not quite infinity of course because that takes forever but a very long time has gone by after the switch switched from this position to this position so let's just kind of notice that this is what it will look like and then we want to know when the current stops flowing what will be the voltage across the capacitor and we want to calculate what we call the time constant tau it stands at a time constant which is R times C once you have those values then you want to use the equation where the voltage across the capacitor as a function of time is equal to the voltage across a capacitor at time equals infinity when the large amount of time has gone by plus the transient phase where we take the voltage across capacitor at time equals zero minus the voltage across the capacitor at time equals infinity or very large times e to the minus T over RC so let's use that technique on this particular function right here at this particular circuit so let's see we want to find out the voltage across capacitor right here when time is equal to zero just before the switch moves over to the different position well let's see here we have a 20 volts over here let's assume that this is zero volts at this point this point here is 20 volts and then you can see that the current will flow around these or through these resistors that means that the voltage across the capacitor here will be equal to the voltage across this resistor and so we need to know what percentage or a portion of the voltage will drop across this resistor this resistor right there since those then will be in serious once a current no longer flows to the capacitor but you then say that the voltage across the three kilo ohm resistor is going to be equal to twenty volts times the ratio of the three kiloohm resistor divided by the total resistance going from there to there which is 2 kilo ohms plus 3 kilo ohms and so that would be 3/5 or 12 volts that means if we have 12 volts across this resistor we'll have a 12 volt drop across the capacitor for step number one we could say that V across the capacitor at time equals zero which is the same as the voltage across the three kilo ohm resistor is going to be equal to 12 volts now the switch moves from this position to that position now we can see that there's no longer any current flowing through the capacitor on account of this but now have current flowing through the capacitor I guess in this direction right here on a kind of this voltage supply but of course that will happen for a while until the capacitor fills full of charge and then the no longer will be any current flowing through this particular path right here meaning there's no current going through the resistor meaning there's no voltage drop across the resistor which means that the voltage across the capacitor must equal to the voltage across the source that means that the voltage across the capacitor a time equals infinity of course when it approaches infinity or when a large time has gone by that will then be equal to 50 volts the last thing we need to do is find the time constant we know that the time constant is equal to the resistance times the capacitance and of course that will be after the switch moves over from here that will be the transient period where current flows through a capacitor and through this resistor until this fully charged we can then say that the time constant will be this resistor times this capacitor so be 5 kilo ohms multiplied times 0.4 millions okay that's five thousand times Oh point zero zero zero four so that would be equal to five thousand ohms multiplied times zero point zero zero zero four ferrets and that would be equal to let's see here this V zeros cancel this that would be equal to two seconds so the time constant is equal to two seconds in this case looks about right so now they have this information now we can go ahead and use the equation for point two now we can say that the voltage across capacitor as a function of time is equal to the voltage across the capacitor and infinity which we have right here would be fifty volts plus the voltage across capacitor at zero time T equals zero that would be equal to twelve volts minus the voltage across capacitor again an infinity would be fifty volts multiplied times e to the minus T over tau would be two seconds and so the equation becomes 50-38 and of course our problem it but volts on there so fifty volts volts - thirty-eight volts times e to the minus T over two seconds that would be the function for the voltage across the capacitor let's do one more thing here let's say what would be the voltage across capacitor after one second so the voltage across capacitor when T is equal to one second is equal to well here we have 50 volts - thirty-eight volts multiplied times e to the minus now T will become one second divided by two seconds so be e to the minus one-half and let's see what that's equal to so we have point five negative e to the X and times 38 and that would be equal to fifty volts minus twenty-three volts as I would be equal to twenty seven volts so I think example the voltage across capacitor on this particular circuit when time is equal to one second would be equal to 27 volts as an example of how to use that equation and that's how it's done

Channel: Michel van Biezen

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Keywords: ilectureonline, ilectureonline.com, Mike, Mike van Biezen, van Biezen, ilecture, ilecture online, Electrical Engineering, RC & RL Circuits, Ch 8, Chapter 6, First Order, Voltage, Current, Differential Equation, Resistor, Inductor, General Equation, Voltage Divider, Parallel, Equilvalent, Circuit, Load Resistor, The Derivative of a Unit Step Function, Input Function, Unit Impulse Function, Step, Function, Natural, Forced, Response, Transient, Steady, State

Id: N0vcIGaXsn4

Length: 6min 58sec (418 seconds)

Published: Thu Nov 09 2017