in this video we're gonna go over Kirchhoff's voltage law abbreviated KVL now his law states that if you have a closed circuit the voltages around that circuit must add up to zero now some voltages are positive and others have a negative value so you need to determine which one are positive and which ones will be negative now say if we have a resistor and let's say this is at 20 volts and this is at 10 volts you know that current naturally flows from the direction of high potential to low potential in a resistor so it flows from positive to negative now a resistor consumes energy and any time the current flows from a high electric potential to a low electric potential the energy of the charges that are flowing in that circuit is actually being reduced energy is being transferred from the charges to the resistor so therefore because the resistor decreases the energy of the charges we're gonna sign a negative voltage to it because it reduced the energy per unit charge and that's what a vote is one volt is one Joule per Coulomb so voltage is basically the energy change per unit charge which is basically per one Kalume now a battery does something different in the battery current can flow from low potential to high potential so this could be zero volts and this could be 12 and when you see this whenever the current is flowing from a low potential to a high potential that means that the energy of the charges is increasing and so this is going to create a positive voltage because the battery is transferring energy to the charges and so it increases the voltage of the charges but a resistor will decrease the voltage of the chargers now let's see if you remember so let's say if we have a resistor and a current is flowing in that direction will we assign a positive voltage or a negative voltage to the resistor so keep this in mind a resistor always creates a voltage drop it always consumes energy from the charges so always assign a negative v4 resistor now what if we have a battery that's listed this way and there's a current flowing in this direction this battery is it increase in the energy of the charges or is it decrease in the energy of the charges and compare it to this situation so what if we reverse the polarity now let's say the current is still flowing in the same direction so which one should we assign a positive voltage and which one should we assign a negative voltage now keep this in mind any time the current flows from when we put that in blue a high potential to a low potential the energy of the charges is being reduced so it's gonna be a negative voltage now any time it flows in the opposite direction that is from a low potential to a high potential then the voltage is increase in the energy of the charges is increasing so looking at this diagram the current is flowing from positive to negative so it's going from a high potential to a low potential so this battery reduces the energy of the charges now for this one it's going for a low potential to a high potential so it's increasing the energy of the charges so we should assign a positive V to this value and a negative V to this battery now let's work on some practice problems so let's say if we have a battery and it's connected to four resistors in a circuit this is the positive terminal and this is the negative terminal and let's say it's a 12-volt battery and this is going to be 8 ohms 10 ohms and 12 ohms so this is R 1 R 2 R 3 so write an expression that highlights Kirchhoff's voltage law show that the sum of the voltages around a circuit must add to zero now the current is going to leave the positive term of the battery so it's going to be flowing in this direction so as it flows through R 1 the voltage will decrease but the battery notice that the current flows from the negative terminal to the positive terminal of the battery so the battery supplies the energy to the circuit so I'm going to put positive VB for the voltage of the battery and then negative V 1 that's the voltage drop across R 1 and R 2 also creates a voltage drop and R 3 will also create a voltage drop and based on kerckhoffs of voltage law the sum of all the voltages in a circuit must add up to zero now let's use that equation to calculate the current in a circuit so the voltage of the battery is 12 volts now I know that V is equal to IR and so the current is the same in a circuit because there's only one path for the current to flow and so what we have is a series circuit so V 1 is going to be I times R 1 and R 1 is 8 ohms so we can say that V 1 is 8 times I V 2 is I times R 2 or I times 10 so that's going to be 10 times I V 3 is going to be 12 times the current that flows through it which is I so now we can add these like terms negative 8 minus 10 that's negative 18 minus 12 that's negative 30 so this is gonna be negative 30 I and now let's move this to that side so we have 12 V is equal to 30 I and then let's divide both sides by 30 so the current that flows in this circuit is 12 volts divided by 30 ohms and that's equal to 0.4 amps now let's say the electric potential at this point is 0 volts let's calculate the electric potential at these points so what is the electric potential at Point a B and C so notice that the potential difference of the battery is 12 volts and so since this is a positive sign a has to be higher than this point so at Point a the voltage is or the potential is 12 volts now what is the potential at point B so we need to calculate the voltage drop across R 1 so we can use V is equal to I 1 so we have the current which is point 4 amps and let's multiply by R 1 which is 8 so Point 4 times 8 is 3 point 2 so the voltage is going to be it's going to drop by 3.0 so if we take 12 volts and subtract it by 3 point 2 this will give us a potential of eight point eight volts at this point now if you want to show the work here's what you can do so we're going to start with this equation Ohm's law the potential difference is equal to the current flowing through the resistor times the resistor itself now the change in the potential is VB minus VA and with a resistor we know that there's a voltage drop so we need to put a negative sign so this is gonna be negative I times r1 so to calculate the potential at point B it's going to be the potential at Point a minus I times r1 so the potential at a is 12 and then subtract that by a current of 0.4 times the resistance of 8 and that will give you eight point eight now let's do the same thing for Point C so we could start with this equation now this is gonna be VC minus VB and that's equal to negative I r2 so the potential at C is gonna be the potential at B minus IR - so I'd be it's eight point eight and the current is point four and R 2 is 10 so if you type in eight point eight minus point four times ten this will give you four point eight volts and so that's the potential at C now if you take the current which is point four amps and multiply by R 3 which is 12 ohms you should get the difference between these two values so point four times twelve is four point eight and that's the potential difference across r3 and so that's how you can solve this particular circuit using kvl kirchoff's voltage law now let's work on another example let's say if we have this circuit this is going to be a 50 ohm resistor which we'll call r1 and r2 is a 30 ohm resistor so here we have a 12-volt battery and here we have a a full battery go ahead and use Kirchhoff's voltage law to calculate the current in the circuit now the first thing needs to determine is the direction of the current in a circuit because we have two batteries now the 12 volt battery wants to send current in this direction so that is the clockwise direction and the Eightfold battery also wants to send current in the clockwise direction so therefore these two batteries they don't oppose each other they support each other because they want to send current in the same direction so now that we know the direction of the current let's set up the equation so let's call this vb1 and vb2 so the current flows from low potential to high potential for VB one so therefore the VB one increases the energy of the charges so we're going to assign it a positive voltage the resistor consumes energy from the circuit so we're gonna sign that a negative voltage now current flows from the negative terminal to the positive terminal so therefore the battery increases the energy of the circuit so that's going to be plus VB two and then this resistor will consume energy from the circuit so that's going to be negative V two so VB one is 12 we can replace V 1 with I times R 1 so that's I times 50 or simply 50 i vb 2 is positive 8 and V 2 that's gonna be I times 30 or 30 I now let's combine like terms 12 plus 8 is 20 and then negative 50 I minus 30 I that's negative 80 I so moving this term to that side we have 20 is equal to 80 I so if we divide both sides by 80 20/80 will give us a current of 0.25 amps so now that we have the current flowing in a circuit let's calculate the potential at every point so let's say this is point a b c and d and let's say that the potential at point a is zero volts calculate the electric potential at every other point now we can see that the voltage across the battery is 12 and the plus side indicates that the top side is higher than a bond side so the potential at point B has to be 12 volts now the resistor R 1 is going to drop the voltage to calculate the voltage drop it simply I times R 1 so the current is point 25 amps multiplied by resistance of 50 ohms and so the voltage drop there is 12 point 5 volts so 12 minus 12 point 5 this is going to be negative point 5 volts at Point C now you could set it up the way we did in the last problem you could say that the voltage drop is I times R 1 both a negative sign because it decreases the voltage and VC minus VB is equal to negative I at times R 1 so you could say that VC is VB minus I times R 1 so VB is 12 I is point 25 and resistance is 50 ohms so twelve minus point twenty-five times fifty this will give you negative point five volts so that's potential at Point C now what about the potential at point D so we have a battery whoo that's eight volts and this is the positive side which means that this side is going to be eight 'fl it's higher than that sign so if we take a negative point five and then add eight volts to it this will give a seven point five so that's the potential at Point D now let's confirm that a is indeed zero so let's use this equation the voltage drop across r2 is going to be I times r2 and so VA minus VD because we're going from D to a that's equal to this value so VA is gonna be VD minus IR tune so VD is seven point five the current is point 25 and r2 is 30 so if you type in seven point 5 minus point 25 times 30 this will give you zero the potential at Point a and so it goes to show that kirchoff's voltage law does work the sum of all the voltages in the closed circuit will always add up to zero let's try another example but one with a lot more stuff in it so this is gonna be a 50 volt battery here are the positive terminals and here are the negative terminals this is gonna be 30 ohms so we'll call it r1 and then this is going to be 70 ohms which we'll call r2 and this is gonna be a 10 volt battery and also this is gonna be a 20 volt battery so first we need to determine the direction of the current so let's call this battery 1 and battery 2 and battery 3 so battery 1 wants to create a current that flows in this direction away from the positive terminal battery 3 wants to create a current in this direction and that is a counterclockwise current this is a clockwise current so those two batteries oppose each other and this battery and wants to generate a counterclockwise current so these two batteries are in support with each other and they're against this one however 50 is greater than the sum total of 10 and 20 so therefore the current is going to be in this direction now let's go ahead and calculate the current so this battery will create a positive contribution to the circuit it's going to increase the energy of the circuit so we're going to say it's positive 50 now the current is flowing in a different direction for this battery it's going from high potential to low potential so this battery will create a voltage drop so therefore we're going to assign it a negative value and here are the current flows from high potential to low potential but here flows from low to high so that's a positive contribution but here since it flows from high to low it's going to be a negative contribution so this battery will decrease the energy of the circuit now let's write an equation so the first battery we'll add 50 volts or 50 joules per Coulomb of charge to the circuit r1 is going to consume energy from the circuit so the voltage drop will be the current that flows through it times r1 and r1 is 30 so this is going to be 30 times I now this battery reduces the energy of the circuit so that's gonna be negative 20 volts and this one also reduces the energy of the circuits that's gonna be negative 10 volts and this one also consumes the energy from the circuit so that's gonna be negative 70 times I this is supposed to be r2 and so all of this is equal to zero so now let's combine like terms so we have 50 minus 20 minus 10 so that's equal to positive 20 and then we have negative 30 minus 70 so that's gonna be negative 100 I and now let's move this to that side and so 20 V is equal to 100 times I so the current is going to be 20 divided by 100 and so in this example we have a current of point 2 amps now let's call this point a B C D and E let's calculate the potential at each point and let's say that a is at zero volts so B has to be fifty volts higher because that battery provides energy to the circuit so this is going to be at fifty volts now let's calculate the voltage drop across the thirty-one resistor so it's going to be a current of 0.2 amps times 30 ohms so the voltage drop across this resistor is negative six so if the potential is 50 at B then that point C is gonna be 6 volts less it's gonna be forty four volts now this battery will reduce the energy of a circuit by 20 joules per Coulomb or by 20 volts so the potential at D is gonna be 44 minus 20 which is 24 volts now this battery also reduces the energy of the circuit so it's gonna decrease the energy by 10 volts so now the voltage is going to be 14 volts and to calculate the voltage drop across r2 it's going to be the current multiplied by r2 so negative point 2 times 70 that's negative 14 so that's the voltage drop across this resistor and 14-14 will bring us back to zero so now you know how to calculate the electric potential at every point in a circuit the key to understanding how to solve these types of problems is to know which device will increase the energy of the circuit and which device will decrease the energy of the circuit so resistors always consume energy so they will always decrease the energy of the circuit so therefore they will always have a voltage drop so you should always apply a negative V value to it now a battery can either increase or decrease the energy of the circuit and for this you need to look at the direction of the current so if the current is flowing from a low potential to a high potential then the battery is increase in the energy of the circuit so in that case we need to apply a positive voltage to this battery because it's add in voltage to the circuit now if the current is flowing from a high voltage or high potential to a low potential then this battery is reduced in the energy of the circuit and so we need to apply a negative voltage to it so hopefully you understand that concept and it's gonna help you to solve more complicated problems than the ones that we covered in this video this is simply a basic introduction