in this video we're going to talk about how to use kirchhoff's junction rule and loop rule to calculate the current in a complex circuit so let's start with this problem we have a resistor in series with two other resistors that are parallel to each other and let's say this resistor has a value of 3 ohms and let's call it r1 this is going to be r2 and it has a value of 4 ohms and r3 has a value of 12 ohms so if you want to try this problem feel free go ahead and pause the video if you want to work it out so calculate the current flowing through each resistor using kirchhoff's rules so let's say that the current that's flown through the first resistor we're going to call it i1 and the current flowing through the second resistor let's call it i2 and the current flowing through the third resistor we're going to call it i3 now using kirchhoff's junction rule or his current law let's focus on this junction here so notice that we have a current i1 which flows into the junction and i2 leaves a junction i3 also leaves it to go this way now according to his junction rule the current that enters the junction is equal to the total current that leaves the junction so the only current that's entering the junction is i1 the other two currents are leaving the junction so i1 has to equal the sum of i2 plus i3 so that's one equation that we have so far we're going to use it later now the next thing that we need to talk about is kirchhoff's voltage law which is important when using the loop rule so that law states that the sum of all the voltages around the loop or around the closed circle must add to zero now when going around the loop you need to determine which elements will create a positive voltage contribution to the circuit and which one will create a voltage drop or a negative contribution to the voltage of the circuit so anytime you travel from a high potential to a low potential there's going to be a voltage drop and it makes sense because you're going from high to low now if you're going from a low potential to let's say a high potential if you're traveling in that direction then the voltage is increasing so there should be a voltage lift now here's the question for you let's say the current is flowing in this direction and let's say as you're traveling around the circuit you're going through the loop and it's in the same direction as the current will this be associated with a positive voltage or a negative voltage now keep in mind current flows from high potential to low potential so we're going in this direction we're going from a positive to negative so therefore this will be associated with a voltage drop so when using the loop rule if you're going in a direction of the current that's associated with a voltage drop now if you're going opposite to the direction of the current then you're traveling from a low potential to a high potential which is what we have in this situation so that's going to be associated with an increase in voltage or voltage lift because you're going towards a higher potential if you're going towards a lower potential that's associated with a voltage drop and that makes sense so if you're traveling towards a negative sign you should be associated with voltage drop if you're traveling towards let's say a positive sign we're going in that direction that should be associated with a positive voltage or voltage lift so now let's say if you have a resistor and this part is positive and this part is negative so if you're traveling in this direction using the loop rule should you apply a positive voltage contribution or negative voltage in the voltage law equation now we know that current flows from high potential to low potential so notice that we're going against the current and we're traveling towards the positive side so this is going to be associated with a voltage lift now let's say if we have a resistor and the current is flowing in this direction and using the loop rule we're traveling in this direction will this be associated with a voltage drop or voltage lift now we know that current flows from positive to negative so we're traveling towards a low potential so this will be associated with a voltage drop now let's focus on a battery so let's say this is positive and this is negative and i'm going to draw another battery in this direction so let's say if we're using the loop rule and we're going in this direction should we apply a positive voltage or negative voltage in the voltage law equation is this a voltage lift or a voltage drop now we're going towards the positive sign so this is going to be associated with a voltage lift now what if we're going in this direction using the loop rule so notice that we're going towards the negative sign and so that should be associated with a voltage drop so hopefully that makes sense and understanding this will be very useful when solving these types of problems because if you get let's say a negative sign wrong the whole question is gone it's going to be very hard to find the right answer all it takes is one single mistake to make a long problem one big headache so now that we've finished with all the sign conventions let's go back to this problem so let's focus on the first loop we'll call this loop one now we're going in this direction so we're going towards the positive side of that battery so we're going from low potential to high potential so that's going to be associated with a voltage lift so that's going to be positive 24 volts now keep in mind when using kirchhoff's voltage law the sum of all the voltages must add up to zero now some of these voltages will be positive some will be negative now when you're moving across a resistor the voltage drop will be the current times the resistor so just keep that in mind as well now current flows from high potential to low potential so then this is going to be the positive side of the resistor and that's the negative side and this is going to be the positive side of r2 and that's the negative side of r2 because current is flowing in that direction and the same is true for r3 so now using loop rule we're going in this direction through r1 so we're going in the direction of the current towards a lower potential so because we're going towards the lower potential that's going to be a voltage drop so the voltage drop across r1 is v1 and v1 is equal to the current that flows through it times r1 and r1 is three so then this voltage drop is simply the current times three ohms which we can write as uh three i one now as we follow the direction of the loop we're going through r2 towards a low potential so that's going to be another voltage drop and so that's going to be minus i2 times the resistance of 4. so we could say four i2 and so now we've completed loop one so we're going to set this equal to zero now what i'm going to do is take these two terms and move it to this side so we have the equation 24 is equal to 3i1 plus 4i2 now let's move on to the second loop so we're going to go in this direction that's going to be loop two so initially we're going up through r2 towards the positive sign we're going against the current and so since we're going from negative to positive that's going to be a voltage lift and so the voltage lift is going to be i2 times the resistance of 4. so it's 4 i2 and then following the loop we're going in this direction through r3 so we're going towards the negative sign we're going from positive to negative so that's going to be a voltage drop so the current is i3 the resistance is 12 so that's going to be 12 i3 and so we have a negative sign for a voltage drop but a positive sign for a voltage lift and this is equal to zero now notice the two equations that we have we have two equations and we have three variables so in order to solve a system with three variables we need three equations and that's why this equation is important so now let's combine those three equations so what i'm going to do in this case is i'm going to replace i1 with i2 plus i3 because it's equal to that so 24 is equal to 3 times i2 plus i3 plus 4i2 and so 24 is equal to 3i2 plus 3i3 plus 4i2 and now we can combine like terms so far we have 24 is equal to 7 i2 plus 3 i3 now notice that these two equations have the same variables i2 and i3 so i'm going to solve by elimination which means that i want to change this to negative 3 i3 so that it can cancel with positive 3 i3 so i'm going to multiply this equation by 1 4 which is equivalent to dividing everything by 4. so 0 divided by 4 is 0 and then 4 i2 divided by 4 is 1i2 and then negative 12 i3 divided by 4 is negative 3 i3 now let's add these two equations so these will cancel 24 plus 0 is 24 and 7 plus 1 is 8. so now let's divide both sides by 8 and 24 divided by 8 is 3. so i2 is equal to 3 amps that's the first answer now let's find a second answer so let's use this equation here to calculate i3 using i2 so we have 0 is equal to i2 minus 3 i3 so i'm going to move this to that side so 3 i3 is equal to i2 so 3i3 is equal to i2 is equal to 3 amps so let's replace it with 3 and then let's divide both sides by 3. so 3 divided by 3 is 1. therefore i3 is equal to 1 amp and so that's the answer for that and now we can calculate i1 i1 is simply the sum of i2 and i3 so i1 is equal to 3 plus 1 so that's 4 amps so now we have all the answers that we need so let's analyze the circuit to make sure that everything makes sense so let's redraw the original circuit so that's the positive terminal this is the negative terminal and this is 24 volts this was three ohms and we have a current i1 flowing through it so that's a current of 4 amps and then this resistor is 4 ohms and i2 was flowing through that so i2 is 3 amps and then this resistor is 12 ohms and i 3 was flown through that which is a current of one amp and it makes sense if four amps is flowing into this junction four amps should leave it we have three going here and one going there so the numbers add up now let's make sure that the potentials add up as well so let's assume that this is zero volts so that's the potential anywhere along this portion of the wire now here we have a difference of 24 volts across that battery so the potential at let's say this point is 24 volts it's 24 volts higher than this point now what is the potential at let's say point a now let's calculate the voltage drop across this resistor v equals ir so it's the current that flows through it times the resistance four times three gives us a voltage drop of twelve and twenty-four minus twelve tells us that the potential at this point is twelve volts now if we take 12 volts because 12 minus 0 gives us a voltage of 12 volts across this resistor 12 volts divided by 4 ohms will give us the current of 3 amps and 12 volts divided by 12 ohms will give us a current that flows through that resistor which is one amp so all the numbers make sense so it's good to check your work to see if you have the right answer now let's try another problem if you want to you can pause the video and work on it so this time we're going to have a second battery in this problem and so here's the positive terminal and here is the negative terminal and let's say on this side we have a 30 volt battery and on this side a 10 volt battery so because the voltage of this battery is higher chances are that the current will flow in this direction as opposed to that direction now even if you use the wrong arrow and guess it incorrectly the current will simply be negative instead of positive so if you get a negative current it just means you have to reverse the arrow and then it's going to be positive so really doesn't matter what direction you choose the current to be you can still get the right answer now we're going to say that this resistor is a 2 ohm resistor we'll call it r1 and this is going to be a 5 ohm resistor which we'll call r2 and this will be a 3 ohm resistor and that's r3 now once again this current will be i1 and the current flowing through this resistor we're going to call i2 now like before we used to call this i3 but this time we're going to do something different now in this junction we have i1 flowing in i3 leaving the junction and i2 also leaving it so based on kirchhoff's current law or junction rule the current that enters into the junction must equal the current that leaves the junction so i1 which enters the junction must add up to i2 and i3 so if we wish to solve for i3 i3 is simply the difference between i1 and i2 now why is this important instead of writing i3 as the current flowing through this branch we could express it as i1 minus i2 and so instead of dealing with three variables i1 i2 and i3 we can deal with two variables i1 and i2 now let's say if we have two junctions let's say this is junction one and junction two now we can say this is current one current two current three current four and current five now you're gonna need five equations to solve these five missing variables however we could simplify the process if we do it this way so i3 is going to be the difference between i1 and i2 so we can simply write it as i1 minus i2 now we're going to have a new i3 which is different from this one so now the current that goes in this direction is going to be this current minus this one so it's i1 minus i2 minus i3 and instead of having five different currents we only have three different variables so we have i1 i2 i3 instead of i1 all the way to i5 and so for more complicated examples you can make the process a lot easier if instead of writing this as i3 you can write it as i1 minus i2 and so from this point on that's what i'm going to do for any future problems that i solve in this video so if you don't get this pause the video rewind and review this topic one more time until you understand it now let's go ahead and finish this problem so let's start with loop one so initially we're going in this direction towards the positive part of the battery so that's going to be a voltage lift of 30 and based on the direction of the current this is going to be positive and this is going to be negative so remember the current flows from high potential to low potential so now as we travel this way towards resistor we're going in a direction of i1 towards the lower potential so that's going to be a voltage drop of 2 times i1 so i1 times the resistance of 2 and now following the loop we're going to go in this direction that's in the direction of the current towards the low potential so that's another voltage drop of i2 times the resistance of 5. so that's going to be negative 5i2 and so that's equal to zero now i'm going to take these two terms and move it to the other side so now i have this equation 30 is equal to 2i1 plus 5i2 now let's save this equation for later now let's focus on loop two so here we're going against i2 towards a higher potential so that's going to be a voltage lift of 5 times i2 so we can write that as positive 5 i2 next as we follow the loop we need to go through this battery towards the higher potential so that's another voltage lift of positive 10 and then we need to travel in this direction that is in the direction of this current as it goes this way towards the lower potential so that's going to be a voltage drop and so that's going to be negative 3 times this current which is i1 minus i2 so because we have a single branch the current that flows in this branch equals the current that flows through this resistor and that's i1 minus i2 which would have been i3 now all of this is equal to zero so keep that in mind now let's distribute the negative three so zero is equal to five i two plus ten minus three i one plus three i two and so let's go ahead and combine like terms and so we have zero is equal to ten minus three i one and five plus three is eight so that's gonna be plus eight i two and i'm gonna take these two and move it to the other side so now i have positive 3 i 1 minus 8 i 2 that's equal to 10. so this is the second formula that we need now let's do some algebra so let's try to get rid of i1 the least common multiple of 2 and 3 is 6. so i'm going to multiply this by positive three and this equation by negative two so three times thirty is ninety two i one times three is six i one and five i two times three is fifteen i two now negative 2 times 10 is negative 20 and then 3i1 times negative 2 that's negative 6i1 and negative 8i2 times negative 2 is positive 16i2 so now let's add up these two equations these will cancel 90 plus negative 20 is positive 70 and 15 plus 16 is 31. so i2 is equal to 70 divided by 31. so i2 is 2.258 amps now let's calculate i1 let's use this formula so 3i1 minus 8i2 or a times two point two five eight is equal to ten negative eight times two point two five eight that's negative eighteen 18.064 so let's add 18.064 to both sides so this is going to be 3i1 that's going to be equal to 28.064 and so i1 is 28.064 divided by 3. and so it's about 9.3 amps so now that we have i1 let's calculate the potential everywhere along the circuit so i'm going to redraw the circuit at this point and let's see if these two answers make sense so let's call this point a point b point c which is the same as that point and point d now let's say that point a is the ground let's say it's at zero volts so given the potential at point a calculate the potential at b c and d now we know this is 30 and this is 10 and this is a 2 ohm resistor this is a 5 ohm resistor and this is a 3 ohm resistor so go ahead and try it so going from a to b we have a voltage lift of 30 volts so potential at b is going to be 30 volts now going from b to c we know that the current flowing through the 2 ohm resistor is i1 and that's a current of 9.355 amps so to calculate the voltage drop of this resistor is the current times the resistance so let's take 9.355 multiplied by 2. so that gives us a voltage drop of 18.71 volts so we know this is positive and this is negative so as we go from b to c the voltage is going to decrease by 18.71 so 30 minus 18.71 that means that the electric potential at point c is 11.29 volts now as we go from let's say c to a the potential drops to zero so we can calculate the current in the 5 ohm resistor by taking the voltage across the 5 ohm resistor which is the potential difference between these two points so that's 11.29 and then if we divided by the resistance of 5 ohms that will give us a current of 2.258 amps which is in agreement with this answer i2 so that means that we're on the right track now let's move from c to d so as we travel in this direction we're going towards the positive terminal and so that's a voltage lift of 10 volts so it's going to be 11.29 plus 10 so the potential id is 21.29 volts now if we go from d to a the potential difference or the voltage across the 3 ohm resistor is 21.29 volts so if we take 21.29 and divided by 3 that will give us the current flowing in this resistor which is zero 7.09 and then six eupeden so we can round that to seven point zero nine seven volts i mean amps now the current flowing through the three ohm resistor is the same as i1 minus i2 so we could confirm this answer by subtracting these two currents so if you take 9.355 and subtract it by 2.258 this will give you the same answer of 7.097 amps and so the numbers are all in agreement with each other so we know these answers have to be correct and so that's it for this problem let's move on to the next one now let's work on a more complicated example so feel free to try if you want to so this is going to be a 20 volt battery this is going to be a 3 ohm resistor and this is going to be a 2 ohm resistor and then we have a 4 ohm resistor and a 12 volt battery and then this is going to be an 18 volt battery a 10 ohm resistor and a 6 volt battery and a 6 ohm resistor so go ahead and calculate the current flowing through the circuit and we're going to say that the potential at this point is zero volts so relative to that point calculate the potential at every other point in the circuit go ahead and try it now the first thing i like to do is try to predict the direction of the currents in this circuit so let's focus on this loop now the 20 volt battery wants to shoot out a current in that direction and so that current wants to travel in this direction in the clockwise direction and the 12 volt battery wants to shoot out a current in this direction which is going counterclockwise however this battery is stronger so therefore comparing those two circuits i believe the net current in this part will be going in that direction now looking at this loop the 12 volt battery wants to send a current going in this direction and these two batteries they add up to make a 24 volt battery the combined effect of those two batteries will want to send the current going in this direction as opposed to that direction and this voltage wants to send a current going through this direction so therefore it makes sense that the current in this branch is going in that direction towards the left because the 24 volt battery wants to send current in this direction which is greater than the 12 volt battery and a 20 volt battery also wants to send current in that direction which is greater than the 12 volt battery so i believe there's going to be a net current going in this direction now if we compare the current flowing in this circuit the 20 volt battery wants to send a current going in this direction the 24 volt battery wants to send a current going in this direction so i think there's going to be a current flow in this direction now let's assume that we have a current going in this direction which we'll call i1 and there's a current going in this direction which we'll call i2 now we can say that there's a current going in this direction which is i1 minus i2 if we're wrong we're going to get a negative answer for this result which means that the current is going this way and it's going to be i2 minus i1 instead but we can solve it this way so we could define the current as going in that direction and then change it later it doesn't matter how you define it as long as the math makes sense so i'm going to define the current goal in this way which is going to be i1 minus i2 and let's see what happens in this example now before we begin there's something else that we need to talk about perhaps you've noticed that there's two junctions and we really don't need to take into consideration this junction because the current that flows through the 2 ohm resistor is the same current that flows in that part of the circuit so this is also i1 and this is i2 and the current that flows here is the difference between i1 and i2 so we get the same currents in that junction so we don't have to worry about it so that's why we only need to take into consideration one of the two highlighted junctions so now let's focus on the top loop so let's define the loop going in that direction we'll call it loop one i'm going to use a different color so in loop one we're going towards a higher potential and so that's going to be a voltage lift of 20 and i1 is going in this direction so therefore this is going to be positive negative positive negative we're assuming that i1 goes in that direction and the current always flows from the high potential to a low potential so as we flow through the the 3 ohm resistor as we follow the loop we're going in the direction of the current towards a negative potential or lower potential so that's going to be a voltage drop across the 3 ohm resistor and that's going to be negative 3 times i1 now as we follow the loop through the 2 ohm resistor we're still going in a direction of i1 so that's another voltage drop which is negative 2 times i1 and then we're going in this direction through the 12 volt battery but towards the negative terminal so we're going from positive to negative so that's going to be a voltage drop of 12. and then we're going to travel through the 4 ohm resistor from positive to negative and so the current in that branch is i2 as opposed to i1 and that's a voltage drop so that's going to be minus 4 i1 and so we've covered all the elements in loop 1. so now let's combine like terms so we have 20 minus 12 which is eight and then we have negative three i one minus two i one and that's negative five i one this is supposed to be i2 i don't know why i have i1 here because it was i2 times 4 so then this is going to be -4 i2 now let's take these terms and move it to this side so now we have this equation 8 is equal to five i one plus four i two now let's focus on the next loop so this is going to be loop two so let's start with this direction so we're going towards the negative terminal of the battery so from positive to negative that's a voltage drop of 18 volts and then we're following this resistor from negative to positive so we're going from low to high potential against i2 so that's a voltage lift of 4 times i2 and then as we follow loop 2 we're traveling through the 12 volt battery from negative to positive or towards the higher potential so that's a voltage lift of 12 volts and then we're going to follow through this resistor which is in the direction of this current which we define at speed that way so we're going from positive to negative in the direction of the current so that's a voltage drop and that's gonna be negative six times i one minus i2 next we're going to go through this six volt battery from positive to negative so we're going towards the lower potential and that's gonna be a voltage drop of six and then since the current is going in this direction this is going to be the positive terminal of the resistor and the negative terminal of the resistor so we're going from high potential to low potential in the direction of the current so that's a voltage drop of 10 times i1 minus i2 and all of this is equal to zero now let's simplify so we have negative 18 plus 12 which is negative six plus another negative six so that's negative 12 and then plus 4 i2 now let's distribute so that's going to be negative 6 i1 plus 6 i2 and then let's distribute the negative 10 so that's going to be negative 10 i1 plus 10 i2 and that's equal to zero so now let's combine like terms so we have four i2 plus six i2 which is 10 i2 plus another 10 i2 so far we have negative 12 plus 20 i2 and then we have negative 6 minus 10 so that's negative 16 i1 that's equal to 0. now let's take the negative 12 move it to the other side so it becomes positive so on one side of the equation we're going to have negative 16 i1 plus 20 i2 it's on the left side of this equation on the right side of this equation we have positive 12 but i've reversed the equation again so now it's on the left side again so now let's focus on those two equations let's cancel i2 so all i need to do is multiply the first equation by negative 5. so it becomes negative 40 which is equal to negative 25 i1 minus 20i2 and then i'm going to rewrite the other equation right beneath it so if we add these two equations we can cancel those two terms negative 40 plus 12 is negative 28 and then that's equal to negative 25 minus 16 which is negative 41 times i1 so the value for i1 is negative 28 divided by negative 41. so i1 is equal to 0.6829 amps now let's calculate i2 so i'm going to use this formula before i multiply it by negative 5. so 8 is equal to 5 times i1 and i1 is 0.6829 and then plus 4 times i2 so 5 times 0.6829 that's equal to 3.4145 so 8 minus 3.4145 that's 4.5855 and that's equal to 4i2 so to calculate i2 we need to divide both sides by 4. so i2 is 4.5855 divided by 4. and so i2 is 1.14 amps now let's analyze the circuit completely so we said that this was three ohms this was 2 ohms and this is 6 ohms and this is a 4 ohms here we have a 20 volt battery a 12 volt battery an 18 volt battery and this is 10 ohms and this is a 6 volt battery so at this point what i want you to do is to calculate the potential everywhere so let's call this point a and this is going to be point b point c point d point e and then let's make this f g and h calculate the potential at each of those points now let's talk about the currents notice that i1 is positive which means that there is a current flowing in this direction and not in that direction and so that current is 0.6829 amps now i2 is also positive which we define it to be a current going in this direction so we don't have a current go in that direction now what about i1 minus i2 i1 minus i2 was going in this direction now if this result is negative that means there is a current going in this direction which is i2 minus i1 so let's see which one is the correct direction so if we take i1 minus i2 0.6829 minus 1.1464 that's going to give us a negative answer negative 0.4635 so therefore this direction is not correct so this is the correct direction if we do i2 minus i1 it's going to give us positive 0.4635 so now we know the correct direction of the current so we have this current which is 0.4635 amps so now we can place the appropriate signs across each resistor so across the 3 ohm resistor i1 is flowing in that direction so this is going to be positive and this is going to be negative now i1 is also flowing in this direction so this is going to be the negative part of the resistor and this is the positive part i2 is flown in this direction so this is going to be the positive terminal of the resistor and this is the negative terminal i2 minus i1 flows in this direction so this should be positive and this should be negative and it's also flowing in this direction which means this is negative and this is positive so now we have the appropriate signs now let's calculate the potential everywhere so let's go from a to b as we travel from a to b we're going from low potential to high potential so that's gonna be a voltage lift of 20 volts so the potential at b is 20 volts which i'm going to highlight in green now let's go from b to c so i1 flows from positive to negative so as we go from b to c this is going to be a voltage drop of 3 times i1 so that's going to be 3 times 0.6829 which is 2.0487 so 20 minus that number that's going to give us a potential at c which is 17.95 volts so now let's calculate the potential at d so we're going from c to d so we're going towards a lower potential so that's going to be a voltage drop and it's i1 times 2 so 2 times 0.6829 that's 1.3658 17.95 minus that number that's going to give us a potential at point d of 16.58 volts now let's move from point d to e so we're going from positive to negative so that's going to be a voltage drop of 12 volts so 16.58 minus 12 that's going to give us a potential at point e which is 4.58 volts now we don't need to go to a because we already know the potential at a so we have the potential difference between points e and a so that's a voltage of 4.58 minus 0 which is 4.58 volts if we take the voltage across the 4 ohm resistor and divide it by 4 that will give us the current flowing through the resistor and so that current is 1.145 amps and so this tells us that wonder right track because this matches i2 now this is slight difference because there was some rounding in this problem i mean this is not exactly 17.95 it's like 17.951 but however this is close enough 1.146 and 1.145 you could round that and say it's about 1.15 if you want to now let's start back at d and let's go towards f so as we travel from d to f is this a voltage lift or voltage drop notice that we're going against the current and we're going towards a higher potential from low to high so that's a voltage lift so to calculate the voltage across the 6 ohm resistor it's going to be 6 ohms times 0.4635 that's the current flowing through it and so the voltage across that resistor is 2.781 volts so as we go from d to e i mean d to f that's a voltage lift of that value so we're going to start with 16.58 and then add 2.781 to it so the potential at f is 19.36 volts so now let's go from f to g so we're going from positive to negative and that's going to be a voltage drop of 6 volts so 19.36 minus 6 is 13.36 so that's the potential at g now let's go from a to h as we travel in this direction we're going from negative to positive so that's a voltage lift of 18 volts so that's the potential at h and current is going to flow from a high potential to a low potential 18 is greater than 13.36 so the current is flowing in this direction which is an agreement with this current which goes in that direction so now let's calculate the current across the 10 ohm resistor so first we need to calculate the potential difference between points h and g so 18 minus 13.36 that's a voltage of 4.64 volts and if we divide that by 10 that will give us a current of 0.464 amps notice that it's approximately equal to i2 minus i1 which is 0.4635 amps so that means that we're on the right track so everything is correct in this problem now we do have some rounded answers but it all adds up now this is going to be the final problem that we're going to work on so let's say this is a 2 ohm resistor and here we have an 8 ohm resistor and a 4 ohm resistor this one's going to be a 3 ohm resistor five and six on the left we have a 20 volt battery and this is gonna be a six volt battery and there's an eight volt a 12 volt and the last one's a 16 volt battery so go ahead and calculate the current flowing through every branch of the circuit and let's define this as point a b c d this is going to be e f g h and that point a the potential will be zero volts so relative to that potential calculate the potential at every point now let's define the current that flows through i2 as i1 and the current that flows through the 8 ohm resistor that's going to be i2 now the current flowing through the 3 ohm resistor that's going to be i1 minus i2 and through the 5 ohm resistor we're going to call that i3 now the current that flows in this branch which is basically the same as through the 6 ohm resistor that's going to be i1 minus i2 minus this current here which is i3 so let's start with the first loop we'll call that loop one so as we go in this direction we have a voltage lift of 20 volts now based on the direction of the current this is going to be positive this is going to be negative and for the 8 ohm resistor this is positive and this is negative as it follows i2 and for the 4 ohm resistor i1 is flowing in this direction so this is going to be positive and negative so as we flow in this direction that is in a direction of i1 that's going to be a voltage drop so that's negative 2 i1 and then as we follow the loop through the 8 ohm resistor in the direction of i2 that's going to be negative 8 i2 that's another voltage drop and then as we go through the six volt battery we're going from a high potential to a low potential so that's going to be a voltage drop of six volts and then following i1 which goes through the 4 ohm resistor that's another voltage drop of negative 4 i1 so let's combine like terms we have 20 minus 6 which is 14 and then we can combine negative two i one minus four i one and that's going to be negative six i one and then minus eight i two so i'm going to take these two terms move it to that side and so that gives me fourteen is equal to six i one minus eight i mean plus eight i two now i'm going to divide everything by two so it becomes seven is equal to three i one plus four i two now let's move on to the next loop so let's call this loop two so let's go in this direction so we have a voltage lift of 6 volts and then as we flow through the 8 ohm resistor that's going to be a voltage lift going from negative to positive against i2 so that's going to be plus 8i2 and then we need to travel in this direction in the direction of i1 minus i2 so that's the voltage drop and so that's going to be -3 times i1 minus i2 now as we flow through the 5 ohm resistor in a direction of i3 that's going to be a voltage drop of 5 i3 and then let's go through the 8 volt battery so we're going towards the lower potential so that's a voltage drop of 8 volts and so that should equal 0. so now let's simplify the expression we have six minus eight which is negative two and then we have plus eight i2 and let's distribute the negative three so that's negative three i1 plus three i2 minus 5i3 so we could combine these two so right now we have negative 2 and then -3 i1 8i2 plus 3i2 that's 11i2 and then minus 5 i3 so i'm going to move the negative 2 to the other side so this is the second equation that we have negative 3 i1 plus 11i2 minus 5i3 that's equal to 2. so let's save that equation for later and now let's move on to the third loop so as we travel in this direction we have a voltage lift that is 5 i3 it's going against i3 as we go up so that's going to be positive 5 i3 and then we're traveling in this direction towards the positive terminal of the battery so that's the voltage lift of 16 volts and then we're following the direction of this current so that's a voltage drop that's going to be negative 6 times i1 minus i2 minus i3 and then we're traveling in this direction towards the positive terminals that's a voltage lift of 12 volts now let's combine like terms so 16 plus 12 is 28 and then we need to distribute the negative six so it's negative six i one plus six i two and then plus six i three and that is equal to zero and so we have 28 minus six i one plus six i two now let's combine these two so 5 plus 6 that's 11. and so now we have our third equation now what i'm going to do is simply take everything other than a 28 and move it to the other side so we can write this as 28 is equal to 6i1 minus 6i2 minus 11i3 now let's line up the three equations that we have so the first one is three i one plus four i two and that's equal to seven the second one negative three i i1 plus 11 i2 minus 5i3 that's equal to 2. now for the last one it's 6i1 minus 6i2 minus 11i3 and that's equal to 28. so we have three equations with three variables so we need to use the system of equations process to get the answer and if you're not sure how to do this you can check out one of my videos entitled system of equations with three variables i just type in organic chemistry tutor as you come up i have a 26 minute video that gives you some examples on how to solve these types of problems so go ahead and try this if you want to now the best way to approach this particular situation is that we need to combine equations two and three and our goal is to cancel i3 because that's going to give us an equation in terms of i1 and i2 which we can combine that with equation one to either solve for i1 or i2 so what i'm going to do is i'm going to multiply the second equation by a negative 11 and so that's going to give me positive 55 i3 and i'm going to multiply the third equation by positive 5 which will give me negative 55 i three so negative 3 i1 times negative 11 that's positive 33 i1 and then this times negative 11 that's going to be negative 121 i2 and negative 5i 3 times negative 11 that's positive 55 i3 and 2 times negative 11 is negative 22. now let's multiply the third equation by five so six i one times five that's going to be thirty i one and then negative six i two times five that's negative thirty i two negative eleven i three times five is negative fifty five by three and twenty eight times five we know twenty times five is a hundred eight times five is forty so this is going to be one forty now we could cancel these two terms 33 plus 30 is 63 and negative 121 plus negative 30 that's negative 151 times i2 and 140 minus 22 140 minus 20 is 120 and then 120 minus 2 that's 118. so now we need to combine this equation with this one and so let's focus on canceling i1 so i'm going to multiply the first equation by negative 21 because 3i1 times negative 21 is negative 63 i1 and then 4i2 times negative 21 4 times 20 is 80 and then 1 times 4 is 4 so this is going to be negative 84 i2 and then 7 times 21 7 times 2 is 14 so 7 times 20 is one forty and then seven times one is seven so this is going to be negative one forty seven now negative one fifty one 84 that's negative 235 and 118 minus 147 that's negative 29. so i2 is equal to negative 29 divided by negative 235 and so i2 is 0.1234 amps so that's the first answer that we need now let's use the first equation before we multiply it by 7 to calculate i1 so it's going to be 3 i1 plus 4i2 or 4 times 0.1234 and that's going to equal 7. so 4 times two three four that's point four nine three six so seven minus that number that's equal to 6.5064 and that's equal to 3i so to calculate i i mean i1 that is need to divide it by 3. so i1 is equal to 6.5064 divided by 3 which comes out to be 2.1688 amps so now that we have the value for i1 we need to calculate i3 and i'm going to use the third equation before multiplying it by five so this is going to be six i one or six times two point one six eight eight minus six i two and then minus eleven i three and that's equal to twenty eight so 6 times 2.1688 that's 13.0128 and then 6 times 0.1234 that's 0.7404 now let's combine those two numbers so 13.0128 minus 0.7404 that's 12.2724 so we need to subtract both sides by that number so 28 minus 12.2724 that's 15.7276 and now let's divide both sides by negative 11 and this will give us i3 so i3 is negative 1.4298 amps because it's negative we need to reverse the direction of i3 now let's redraw the circuit so this was the 2 ohm resistor and this was 8 ohms and then this was 3 ohms 5 ohms 6 ohms and this was 4 ohms and this was a 20 volt battery here we had a 6 volt battery and this is a 8 volt battery and that's a 12 volt battery and then over here we have a 16 volt battery now i1 the current that flows through this branch that's in the right direction so this is 2.1688 amps and so this is positive and this side of the resistor has to be negative now i2 is flowing through this resistor and because i2 is positive we have the right direction and i2 is 0.1234 amps which we can write that later but we know this is the positive terminal of the resistor and this is going to be the negative terminal of the resistor now i1 also flows through the 4 ohm resistor in that direction so therefore this side is positive and this side is negative now i1 minus i2 if we subtract these two currents that's going to give us a positive value and i1 minus i2 flows through the 3 ohm resistor and so that's 2.1688 minus 0.1234 and so this current is 2.0454 amps so this is going to be the positive side of the resistor and this is the negative side now i3 is a negative value before we define i3 flown this way but in actuality it's flowing in that direction and so therefore this should be positive and this should be negative now the current that flows in this direction is i1 minus i2 minus i3 so that's 2.1688 minus i2 which is 0.1234 minus negative 1.4298 so the current that should be flowing in this branch is 3.4752 now it's flowing through this resistor so this is the positive side of this resistor and this is the negative side so now that we have all the appropriate signs let's calculate the potential at every point so let's calculate it at well we're gonna assume point a is zero this is b c d e f g and h so if the potential is 0 at a what is it going to be at b so going in this direction we have a voltage lift of 20 volts so the potential at b is 20. now going from b to c we have a voltage drop because we're going towards a lower potential and so it's going to be 20 minus ir or minus 2.1688 times 2. so the potential at c is 15.66 volts now let's go this way so we're going towards a higher potential and the current through the 4 ohm resistor is i1 which is this value so it's going to be 0 plus because you're going towards the higher potential plus 4 times 2.1688 so the potential at h is 8.675 volts now let's go up to this point and that should be labeled a new point so let's call this point i so as we go from h to i we have a voltage lift we're going towards a positive terminal so a voltage lift of 6 volts and so 8.675 plus 6 that's going to give us 14.675 so that's potential at i so now we could confirm the value of i2 so first let's calculate the potential difference or the voltage across the 8 ohm resistor so it's the difference between the electric potentials at c and i so it's going to be 15.66 minus 14.675 so that's a voltage of 0.985 and then divided by the 8 ohm resistor this will give you a current of 0.1231 amps which is very close to this current so that means that the information that we have in this loop is correct now let's move on from c to d so we have a voltage drop and since we're going towards the negative terminal so to calculate the potential at d it's going to be the potential at c 15.66 minus the voltage drop so minus 2.0454 times the resistance of 3. and so the potential at d is going to be 9.5 2 4 volts now let's go to h starting from there let's go to j i mean not j but g so that's a voltage increase of eight volts so eight point six seven five plus eight that's going to give us 16.675 for the electric potential at g so now let's calculate the potential difference between d and g to confirm the value of i3 so notice that we're going from a high potential of 16 volts to a low potential of like 9.5 volts and so does it make sense that the current is going in this direction so 16.675 minus 9.524 that's a voltage of 7.151 if we divide that by 5 that will give us a current of i3 which is 1.4302 amps and keep in mind i3 is negative 1.4 going in this direction but once we reverse it it becomes positive 1.4 it's no longer negative once we reverse it and so 1.4298 is approximately 1.43 so this answer is acceptable now let's confirm the current flowing through this resistor which should equal this number because if you think about it i3 is 1.4298 and we have this current which is 2.05 2.0454 so these two currents should add up to this one so we add 2.0454 plus 1.4298 that will give you 3.4752 because this these two currents are the currents going into the junction and this is the current going out of the junction and so based on kirchhoff's junction rule the current that flows into a junction must equal the current that flows out of a junction now we have the potential at d so let's calculate the potential at e so going from d to e we have a voltage lift of 16 volts so 9.524 plus 16 that's going to give us a potential of 25.52 volts now let's calculate the electric potential at f so we have it from g so let's go from g to f and so going from a high potential to a low potential so that's going to be a voltage drop of 12 volts so 16 minus 12 is 4. so the potential is going to be 4.675 volts at f so the potential difference between e and f that's 25.524 minus 4.675 and so that's a potential difference of 20.849 so that's the voltage across the 6 ohm resistor if we divide it by 6 that will give us a current flowing in this direction which is 3.475 or four eight if you want to round it and so that's very close to this answer so all the values make sense in this problem so these answers are indeed correct so now you know how to use kirchhoff's junction rule and loop rule to calculate all the currents in the circuit and now you know how to calculate the electric potential at any point in a circuit but you need to be given a reference value first so relative to that reference value you know how to calculate the potential at any point so that's it for this video thanks for watching and have a good day you