in this video we're going to go over a list of common electrical formulas that you'll encounter when dealing with electricity and DC circuits so let's draw a simple circuit one with a battery and a resistor so here is the positive terminal of the battery this is the negative terminal and we're going to have a current flowing from a positive terminal to the negative terminal of the battery and of course that is conventional current we know that electron flow in Electra flows in the opposite direction now the first Formula you need to be familiar with is Ohm's Law which relates a voltage to current and resistance V equals IR V is the voltage measured in volts I is the current measured in amps and resistance is measured in ohms now the next equation that you need to be familiar with is power the power dissipated by resistor is going to be equal to the voltage across that resistor multiplied by the current flowing through the resistor you can also use this formula as well I squared r and also V squared over r so that's how you can calculate the power dissipated by a resistor or even the power delivered by a battery you could use any one of those formulas to get the answer now the next Formula you need to be familiar with is this the electrical work done is equal to power multiplied by time you can also think about work as energy being transferred so the amount of electrical energy being transferred to a resistor is power multiplied by time now we said that voltage is measured in the unit's volt current is measured in amps resistance is measured in ohms but power is measured in watts power electrical power is the rate at which electrical energy is being transferred so power is energy divided by time so it tells you how fast energy is being transferred to a device one horsepower is equal to 746 Watts so those are some details that you want to keep in mind one Watt is one joule per second so if you have a resistor that is dissipating 50 watts of electrical energy or rather 50 watts of power what that means is that it's converting 50 joules of electrical energy into heat every second so a resistor that's dissipated in 100 watts of power what it's really doing is it's converting 100 joules of electrical energy into heat per second so in two seconds it will have converted 200 joules of electrical energy into heat in three seconds 300 joules and so forth so that is the basic idea of electrical power it is the rate at which electrical energy is being transferred from one device to another per second now there are other ways in which you can calculate electrical energy so in addition to using this formula electrical energy is equal to the charge that's being transferred times the voltage electric charge is equal to the current multiplied by the time which means current is basically the rate at which charges flowing per unit time so what this means is that if you have a current an electric current of one amp what's really happening is that you have one column of charge flowing through a circuit or through a wire per second an electric current of 10 amps means that you have 10 coulombs of charge flowing through a wire every second so electric current is the rate at which electric charge is flowing and electric charge is basically the Quasi of charges that you have in any given material now if we replace Q with it we get this formula electrical energy is equal to voltage multiplied by the current multiplied by the time which you can also get that if you replace power with voltage times current so those are some common formulas that you want to be familiar with if you're dealing with simple circuits and you want to use Ohm's law and you want to calculate power or electrical energy now let's talk about the formulas that you need to know when dealing with a series circuit so let's say we have a battery and we have three resistors connected in series so what we have here is the voltage of the battery we'll call this resistor one this is going to be resistor 2 and resistor 3. now here are some things you need to know when dealing with a series circuit the current flowing from the battery we'll call it it current is the same current that's flowing through resistor one the current that flows through resistor one is called i1 and the current that flows through resistor two well that current is called I2 the current flowing through resistor 3 is called i3 now in a series circuit because it's only one path for the current to flow the current is the same i1 I2 i3 they're all equal to each other and they're equal to the current flown from the battery the total resistance of three resistors in series it's equal to the sum of those resistors now as regards to voltage if you want to calculate the voltage across resistor 1 it's equal to the current flowing through that resistor times the resistance so back to Ohm's law V equals IR if you want to calculate the voltage drop across resistor 2 it's equal to I2 times R2 and V3 is going to be i3 times R3 now the voltage of the battery is going to be equal to the sum of all the three voltage drops or the three voltages across the three resistors so VB is going to be equal to V1 plus V2 plus V3 this has to do with kirchhoff's voltage law which states that the sum of all the voltages in a loop will add a zero so a few more to move V1 V2 and V3 to the other side of the equation you'll get this positive VB the reason why it's positive because the battery adds energy to the circuit and then minus V1 V2 V3 the reason why they're negative is because they subtract energy from the circuit they absorb energy from it so they decrease the voltage whereas the battery increases the voltage of the circuit so we have a positive for VB but a negative for V1 V2 and V3 at least that's the way I like to think of it but this has to do with kirchhoff's voltage law but it is a lot easier to see it this way though so VB is going to be the sum of V1 V2 V3 when dealing with a series circuit now let's move on to a parallel circuit so we're going to have three resistors connected in parallel to each other which means they're connected across from each other VB is going to be the voltage of the battery and the current flowing from it we'll call it it the total current so in a series circuit we had just one path for the current to flow in a parallel circuit there are multiple paths for the current to flow so this is R1 R2 R3 and the current flowing through R1 we'll call that i1 this is going to be I2 and i3 now in a series circuit VB was the sum of V1 V2 V3 in a parallel circuit VB is going to be equal to V1 V2 and V3 because each resistor is connected across the same battery so therefore the same voltage will be applied to each of those resistors now in both series and parallel circuits V1 is still equal to i1 times R1 so the voltage across resistor 1 is equal to the current that's flowing through it times the resistance of resistor one V2 is going to be equal to I2 times R2 just as before and the same is true for V3 now in the series circuit the total resistance was the sum of the three resistors in a parallel circuit the reciprocal is true so 1 divided by the total resistance or the equivalent resistance is equal to 1 over R1 plus one over R2 plus 1 over R3 now in a series circuit because there was only one path of the current to flow the total current was equal to i1 which was equal to I2 which was equal to i3 for a parallel circuit it's going to be different the total current is going to be the sum of i1 I2 and i3 and let's talk about that because we talked about kirchhoff's voltage law it makes sense for us to talk about kirchhoff's current law so here we have the total current and here it's going to break off into i1 and let's call this IB actually let's call it i a so if we focus on this point here this Junction based on current I mean kirchhoff's current law the total current that is entering that Junction must equal the total current that is leaving that Junction so I T that's the current that's entering that point and i1 and I a that's the current that is leaving that point now if we focus on this point here I a is flowing into that point I2 is going towards resistor 2 and I3 goes towards resistor three so we have IA entering that Junction I2 and I3 they're leaving that point that's the current that's leaving that point so if we replace I a with I2 and I3 we get that the total current is equal to i1 plus I2 plus i3 so this formula arises from kirchhoff's current law so let me give you another example so let's say we have 20 amps of current flowing to this point we'll call it point a and let's say we have 18 amps of current going here we have 13 amps of current going this way and we have 25 amps of current going that way what is the current flow in in this branch and is it flowing towards point a or away from point A what would you say well first so let's give the currents variables let's call this i1 I2 i3 and I4 and this will be i5 now the sum of the currents entering and leaving that Junction should be equal to zero which is the same as saying the amount of currents entering that Junction and the amount of currents leaving will be equal to each other so we're going to use kirchhoff's current law to calculate I5 so the current that is flowing into the junction let's assign a positive value to those currents the currents that are leave in that point let's assign a negative value to it so i1 and I2 is going to be positive because the current is Flowing towards point a and I3 and I4 will be negative because the current is leaving from point A so we're going to have i1 plus I2 minus i3 minus I4 now I5 we don't know if it's entering or leaving but we're going to put a plus towards i5 so when we solve I5 if we get a positive answer that means the current is Flowing 2.8 if we get a negative answer that means the current is Flowing away from point A so i1 is 20. I2 is 18 I3 is 13 I4 is 25 and let's calculate i5 so 20 plus 18 is 38 and 13 and 25 that's also 38. so it looks like I5 is 0. but if we were to change this let's say if we made it 29. so we can get a value for I5 what would I5 be if I4 is 29 negative 13 minus 29. that's going to be negative 42. so this would be negative 4 which means I5 will be positive 4. so if I force 29 amps I5 is going to be 4 amps because it's positive current is Flowing towards 0.8 notice that if we were to add up these values what this tells us is that we have 42 amps of current flowing 2.8 and if we add up these values we would have 42 amps of current flowing away from point A which is how it should be the amount of electrical current that is Flowing to a point should equal the amount of current that is Flowing away from that point so that's the basic idea behind kirchhoff's current law so that's basically it for this video just want to give you a list of common electrical formulas that you'll use when solving DC circuits or when dealing with electricity in general thanks for watching by the way for those of you who want more example problems feel free to check out the links in the description section below I'm going to be posting more content as well as videos that are related to this video that you're currently watching so feel free to take a look at that when you get a chance