in this video we're going to talk about capacitors particularly when they're charging and discharging so let's say if we have an RC circuit that is a circuit containing a capacitor with a resistor and we're going to have an open switch initially let's say that the voltage of the battery is 12 volts this is the resistor R and here we have the capacitor represented by the symbol C now once the switch is closed charge will begin to flow current will flow from the positive terminal of the battery to charge the capacitor the longer side is the positive terminal now keep in mind though electrons they flow in the opposite direction electrons flow from the negative terminal and it enters the positive terminal now initially when it switch is open the voltage of the capacitor will be zero as soon as the switch is closed when T is zero the voltage will still be zero but the voltage across the resistor initially will equal the voltage of the value which is 12 now after some significant time has passed the capacitor will be be fully charged let's call it V final it's going to have a final voltage of 12 when the capacitor is fully charged the voltage across the resistance is going to be zero the reason being is once you once the capacitor is fully charged no current will flow in the circuit and if there's no current flowing through the resistor the voltage across that resistor will be zero so initially when a capacitor has a voltage of zero the voltage of the resistor will be 12 because it has to add up to the batter's voltage when the capacitor is fully charged to 12 volts the voltage across the resistor will be zero because at that point there's no current flowing through the circuit and keep in mind no current actually passes through the capacitor the current that flows through the circuit is simply used to pump charge from one plate of the capacitor to the other plate the equ equation that describes a capacitor trien is this equation it's equal to the EMF of the battery which in this problem is 12 * 1 minus E E is basically the inverse of the natural log function minus a Time / RC now the graph looks something like this this is this is the EMF of the battery which is 12 volts and the capacitor will charge progressively towards 12 volts but it starts from zero and so depending on the resistance and the value of the capacitance it will affect the shape of this graph but for the most part that's how it's going to look like it's going to increase towards 12 at which point it's just going to be a horizontal line so this is the voltage of the capacitor and on the x-axis you have the time now there's something called to which is the time constant and that's equal to the product of the resistance and the capacitance it's RC and so it has units ohms times fets now let's talk about the time constant let's make a table this is going to be the voltage at any time T that is the voltage of the capacitor so how much will the capacitor be charged after one time constant one time constant is basically it's equal to uh 1 RC two time constants is just 2 RC so what you need to do is replace t with one RC in the equation and RC will cancel so basically you just got to plug in one e to minus one if you type that in you should get 632 which is 63.2% now the maximum voltage of the capacitor is the voltage of the battery which is 12 63.2% of 12 is 7.58 volts so after one time constant the voltage of the capacitor will be 7.58 volts it's 63.2% of its maximum now what about after two time constants well if you type in 1 - e -2 you should see 86 5 or 86.5% 86.5% of 12 is 10.38 so after two time constants the voltage of the capacitor will be 10.38 now 1 minus E to the minus 3 will give you .95 or 95% which correlates to a voltage of 11.4 volts after the four time constants the capacity will be 98.2% charged and after five constants it's going to be 99.3% charge it's going to have a voltage of 11.92% charged at that point now what about discharging a capacitor let's say if the capacitor is fully charged and we connect it across the resistor and let's say it now has a voltage of 12 volts what type of graph will we have when a capacitor is discharging the graph will look like this it's going to start from its initial value of 12 and it's going to progressively decrease to zero the equation that describes this graph is this one V is equal to V initial that's the original voltage of the capacitor when T is zero that's 12 Vol time e raised to T RC now let's make another table vfc is the voltage of the Capac and on the xaxis we have time now after one time constant how much what percentage of the charge does the capacitor still have so notice this equation if we replace t with negative RC this is going to be e to 1 if you type in E to 1 you're going to again 368 or it's going to be 36.8% charged the voltage is now 4.42 volts that's 36.8% of 12 after two time Constance it's going to be 13.5% in the other table after two time constants it was 86.5% these two numbers have to add up to 100 so the voltage at this point is going to be 1.62 volts after three time constants it's going to be 5% charged 95% discharged so its voltage is now going to be6 volts after four time constants it's going to have 1.8% of its initial charge so the voltage is going to be 22 so it takes about five time constants for the capacitor to be 99% discharge where it's going to have 7% remaining of its original charge and so the voltage is going to be 08 so at that point we could say that the capacitor for the most part is discharged now let's work on an example problem so let's say we have an RC circuit and a Capac cap Itor is attached to the battery by means of a resistor and initially at T equals z let's say the voltage of the capacitor is zero and the capacitor is going to have a value of 500 microfarads and let's say that the resistor it's a one uh kiloohm resistor and the voltage of the battery is going to be 20 volts with this information how long will it take for the capacitor to reach 90% of its uh maximum charge so what is 90% of 20 volts 20 * 90% is 18 volts so we want to know how long will it take for the capacitor to be 90% charge or to reach a voltage of 18 volts since the capacitor is charging we can use this equation V is the voltage of the capacitor which is 18 volts the EMF of the battery is 20 and let's find out the value of RC which is one time constant R is 1 kilohm which is 1,000 ohms C is 500 microfarads which is 500 * 10- 6 fets now if we multiply these two 1,00 * 500 * 10- 6 this is equal to 0. five 0 five is the value of one time constant so a time constant is 05 seconds in this particular problem because sometimes you may need to know uh what the time constants are so let's replace RC with 0.5 and now our goal is to solve for team so the first thing we need to do is divide both sides by 18 I mean rather by 20 18 / 20 is9 so .9 is equal to 1 - e raised to T / .5 next subtract both sides by 1 9us 1 is .1 so we have .1 minus E raised to- T / .5 next multiply both sides by 1 this will change negative signs into a positive sign after that let's take the natural log of both sides so on the left we have the natural log of 0.1 and on the right we have the natural log of e to the T / .5 now a property of logs allows you to take the exponent and move it to the front so therefore what we now have is the natural log of 0.1 is equal to - t / .5 * Ln e Ln e is equal to one now let's just get rid of this stuff now let's multiply both sides by .5 this will cancel the 0 five on the right as well as the negative sign so T is equal to.5 times the natural log of 0.1 so if we type this in this will give us 1.15 seconds so that's the time it takes for the capacitor to reach 90% of its maximum value to reach a voltage of 18 Vols now is there an equation that can help us get this answer a lot faster if you rearrange the equation you can use this formula T is equal to negative RC times the natural log of 1 minus V / e where V is the volt Vol of the capacitor at any time T and this uh Epsilon symbol is basically the voltage of the battery now the way you should type it in should be like this R make sure you use a th000 don't use 1 kiloohm C is going to be 500 * 10- 6 and then multiply that by the natural log of 1us 18 volt / 20 if you type this in exactly the way you see it and this is a multiplication sign this will give you 1.15 seconds so we have the time and we also have the time constant as we said before one time constant is equal to rc which is5 seconds so using this information how many time constants does it take for the capacitor to be 90% charged how can we figure this out to find the number of time constants converting start with the time that you have which is 1.15 seconds over one and we know that one time constant is equivalent to 0.5 seconds so notice that the unit seconds will cancel given us the number of time constants so anytime you have to find the number of time constants which is we'll call it n it's equal to the time divided by to that's how you can find it so it's 1.15 ID 0.5 and so it takes about 2.3 time constants for the capacitor to be 90% charge now Does this answer make sense consider the first table that we went over earli in this video we said that it takes two time constants for the capacitor to be 86.5% charged and that it takes three time constants for the capacitor to be 95% charged so therefore 90% is between 86.5% and 90 which means that it should take somewhere between 2 and three time Constance to be 90% charged and 2.3 is between two and three so that answer does indeed make sense and so now you know how to find the number of time constants it takes to charge or even discharge a capacitor to a certain level and so we're going to stop here so that's it for this video thanks for watching and have a great day