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# Current through resistor in parallel: Worked example

## Video transcript

so we have an interesting circuit here the goal of this video is to figure out what is the current that flows through the six ohm resistor pause this video and see if you can work through it so the way that I am going to tackle it is first simplify the circuit take these two resistors in parallel and think about what the equivalent resistance would be and we have seen that before 1 over the equivalent resistance is going to be equal to 1 over 6.0 ohms plus 1 over 12 point 0 ohms 1/6 is the same thing as 2 over 12 so 2 12 s plus 1 12 is 3 12 + 3 12 you could view that it's the same thing as 1 over 4.0 s and if so 1 over the equivalent resistance is equal to 1 over 4 ohms well that means that the equivalent resistance is 4 ohms and so we can simplify our circuit now where we replace these two resistors in parallel with one resistor of the equivalent resistance and that is going to be equal to a 4 ohm resistor now the next thing we could do is we could figure out what the current is through this part of the circuit which would be the same thing as the current right over there we could call that I sub 1 and we can just use Ohm's law for that we would have I sub 1 would be equal to our voltage drop which is 24 volts 24 volts divided by the equivalent resistance of these two resistors in series and when you have resistors in series you just add them up to figure out the equivalent resistance so this would be divided by 2.0 ohms plus 4 ohms plus 4.0 ohms 24 divided by 2 + 4 24 divided by 6 is 4 and since we're dealing with two significant digits it'll be 4.0 and we're talking about current so this is 4.0 amps or 4.0 and peers now how do we use that information to calculate this current right over here we can call that I sub 2 now one way to think about it is what is going to be your voltage drop from this point to this point if you know the voltage drop from that point to that point and if you know that the voltage dropped from this point all the way down here is then we can figure out what the voltage drop from here to here is going to be so let's do that so the voltage drop across this first resistor remember your change in voltage is just equal to your current times your resistance and so this is going to be your current is going to be 4 amperes times your resistance is 2 ohms times 2 ohms which is going to be equal to 8.0 volts and so if the voltage difference between that point and that point is 24 volts which we know from this voltage source but if we drop eight volts as we go to this point well then the difference between this point and this point or this point and this point right over here this has got to be a 16 volt drop if our Delta V across the 6 ohm resistor is equal to 16 volts well then we can use Ohm's law again to figure out I 2 I sub 2 is going to be equal to our drop in voltage so 16 volts divided by this resistance 6 ohms and so what is this going to be equal to 16 divided by 6 is 2 + 4 6 or 2 and two thirds or two point six six six six six six and if we round to two significant digits you're going to have two point seven amps and so we just figured out what we wanted to figure out this right over here is two point seven amps or two point seven amperes but we can keep analyzing it for fun I encourage you to figure out what that current is now the current I sub three and use the exact same technique and one thing that you should feel very comfortable of is that this current that is flowing through the first resistor that 4 ampere current that current gets split between I 2 and I 3 and so I 2 and I 3 should add up to the original 4 amps so just thinking about it that way if you do the same type of analysis we just did you should get 1 point 3 amps for I 3 because 2.7 plus one point three is going to be equal to four amps
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