in this video we're going to talk about how to use norton's theorem to calculate the current flowing through the load resistor so let's go ahead and work on this example so the first thing we need to do is calculate the norton's resistance once we do that and once we calculate the norton's current we're going to draw an equivalent circuit that will look like this so this right here is the norton's current this is the norton's resistance and this is the load resistor and then we'll use the circuit to calculate the current flowing through the load resistor which we'll call il so let's begin by calculating the norton's resistance the norton's resistance is basically the same as the thevenin's resistance we would find it the same way so we're going to replace the independent voltage source with a short circuit and the independent current source will be replaced with an open circuit so we don't need to write anything there and we're going to disconnect the load resistor so we need to find the equivalent resistance across points a and b by the way this should be not a 5 ohm resistor but an 8 ohm resistor so let me just correct that so we have an 8 ohm resistor a 3 ohm resistor and another 3 ohm resistor so these two resistors are in series so that's 11. now we have an 11 ohm resistor in parallel with a 3 ohm resistor so to calculate the equivalent resistance is going to be 1 over 11 plus 1 over 3 raised to the minus 1. so that is going to be i got 2.357 ohms for the norton's resistance which is the same as the thevenin's resistance so that's the first part now we need to calculate the norton current to do that i'm going to disconnect the load resistor so keep in mind rl is equal to 6. we'll get back to it later so right now we have an open terminal across the 3 ohm resistor so we're going to call this point b point a and point c so now we need to perform a nodal analysis at point c so we're going to use kirchhoff's current law so we have a current flowing into point c in that direction let's call that i1 the 7 amp current we're going to call it i2 and the current flowing from point c to point a let's call that i3 so we're going to assign a positive current value for any current that is going towards point c so i1 and i2 are entering that junction so those will be positive currents i3 is leaving point c so we're going to assign that a negative value now keep in mind according to ohm's law the current is the voltage divided by the resistance so i1 is going to be the voltage across the 8 ohm resistor which is the potential difference between 100 and vc over the 8 ohm resistor i2 is simply the 7 amp current source i3 is going to be the voltage across the 3 ohm resistor which is vc minus the potential at b so that's zero and these two resistors are in series because there's no current flowing through this open circuit so from c to b we have a total resistance of six and this is going to equal zero so now we need to solve for v c to get rid of the fractions let's multiply everything by 24. so we have 24 divided by 8 which is 3. so this is going to be 3 times 100 minus vc and then 24 times 7. let's see what that's going to be so that's going to be positive 168 and then 24 times negative vc divided by 6 that's going to be negative 4 vc and that's going to equal 0. so let's distribute the three so we're gonna have 300 minus three vc plus 168 minus 4vc which equals zero so combining like terms we're going to have 468 minus 7vc is equal to 0. moving this to the other side we have 468 is equal to positive 7vc so dividing both sides by 7 we get that the voltage potential at point c or rather the electric potential to be more specific it's 468 divided by seven and so that's going to be 66.857 volts so now that we have the potential at point c we could use that to calculate the potential at a keep in mind the potential at a is equal to the thevenin voltage once we have the thevenin voltage we can use that to calculate the norton's current now notice that in this branch what we have here is a voltage divider let's call this r2 and let's call this r3 so to calculate the potential at a which is basically the thevenin voltage that's going to be the potential at c times r3 divided by r2 plus r3 so that's our voltage divider equation so vc as we said before 66.857 r3 is three our two is three so that's three over six which is one half so va is half of vc so half of 66.857 is 33.4285 so that is the thevenin voltage now the norton current is going to be the thevenin voltage divided by the norton resistance so that's 33.4285 volts divided by 2.357 ohms so the norton current is 14.183 amps so now that we have the norton's current we could use this circuit to calculate the current flowing through the load resistor so this circuit forms a current divider circuit it's equivalent rather to a current divider circuit and to calculate the current flowing through rl is going to be the norton current times rn over rn plus rl so the norton current that's 14.183 amps the norton resistance is 2.357 and keep in mind rl was six you can rewind the video if you want to check that so it's going to be 14.183 times 2.357 divided by 8.357 so the current flowing through the load resistor is 4 amps so that is the answer now let's confirm the answer to make sure that it makes sense so we calculated a current of four amps flowing through the load resistor so if we set b at a potential of 0 volts 4 times 6 will give us the potential at a which is 24 volts now 24 volts divided by 3 ohms tells us that we have an 8 amp current flowing through the 3 ohm resistor 8 amps plus 4 amps gives us a current of 12 amps so we have 12 amps flowing through the 3 ohm resistor and 3 times 12 is 36 so there's a 36 voltage drop across the 3 ohm resistor which means the potential at c is 36 volts higher than the potential at a so 24 plus 36 will give us a potential of 60 volts at point c now we have seven amps of current flowing in this direction and 12 going this way that means that there has to be 5 flowing through the 8 ohm resistor because 5 and 7 will make up to 12. and 5 times 8 is 40 adding 40 to 60 will give us a potential of 100 at this point which we can call point d and that is in agreement with the voltage source so everything here makes sense in the circuit so that's basically it so now you know how to solve a circuit using norton's theorem