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in this video we're going to look at another familiar pattern of resistors called parallel resistors and I've shown here two resistors that are in parallel this resistor is in parallel with this resistor and the reason is it shares nodes these two resistors share the same notes and that means they have the same voltage and they are called parallel resistors so if you share a node share the same node then you share the same voltage and you are in parallel that's what that word means now if we go look closer here we'll see some interesting things it's hooked up we have a battery here some voltage V and because there's a path a complete path round here we're going to have a current we're going to have a current I flowing in this circuit let's label these resistors let's call this one R one and this one are - those are our parallel resistors when the current reaches this point here when a current reaches this node it's going to split it's going to split into two different currents that current and that current we'll call that one I 1 because it goes through R 1 and we'll call this one I 2 and that goes through R 2 now we know any current that goes into a resistor comes out the other side otherwise it would collect inside the resistor and we know that doesn't happen this one comes here and they rejoin when they get to this node and flow back to the battery so the current down here is again I the same one as up here now what I want to do is I want to replace these two resistors with an equivalent resistor one that does the same thing and and by the same thing we mean causes the same current to flow in the main branch and so that's what's drawn over here here's a resistor here we're going to call this V again and we'll call this R parallel R P and this resistor causes the same current i2 flow here and now we're going to work out an expression for that we want to figure out how to how do we calculate RP in terms of the two parallel resistors here okay so let's go at what we know let's see what we know about this over here what we know is the voltages on the two resistors are the same we know there's two different currents assuming that these are two different valued resistors and now with just that information we can apply Ohm's law and we use our favorite thing Ohm's law which says that voltage on a resistor equals the current in the resistor times its resistance so let's write out an Ohm's law for r1 and r2 okay we know the voltage we'll just call the voltage V this is for R 1 V equals I 1 times R 1 and for R 2 we can write a similar equation which is V same V equals I to R 2 now there's one more fact that we know and that is that I 1 and I 2 add up to I and these are the three facts that we know about this circuit what I'm going to do now is come up with an expression for I 1 and I 2 based on these expressions plug them into this equation ok I can rewrite this equation as i1 equals V over R 1 I can write this one as i2 equals V over R 2 and now I'm going to plug these two guys into here let's do that I equals I 1 which is V over R 1 plus V over R 2 let me move the screen up a little bit and now we're going to continue here I just want to rewrite this a little bit I equals V times 1 over r1 plus 1 over r2 ok so here we have an expression it actually sort of looks like Ohm's law it has an i term a V term and this R term here let me go back up here here's our original Ohm's law I'm going to write this I'm going to solve for I in terms of V just to make it look look a little more obvious I can say I equals V over R and what I hope you see here is the similarity between this equation and this one down here so I have this R here and what's happening is this term is playing the role of that resistance so I'm going to bring this co patient down here and write it right down here times 1 over R I'm going to call this R P because what I want is for this expression and this expression I'm going to set those equal say my same V these guys are equal I can write it all I'll just write it over here one over RP equals 1 over r1 plus 1 over r2 this says we have a resistor we're going to call it R P or our parallel that acts like the parallel combination of r1 and r2 so this is the expression for a parallel resistor if you want to calculate a replacement for r1 and r2 in parallel you do this you do this computation and you get RP so let's do one of these for real here's an example here's an example where I've actually filled in some numbers for us so I have a 20 ohm resistor in parallel with a 60 ohm resistor driven by a 3 volt battery and what I want to do is I want to combine these two parallel resistors and find out what is the current right here find out what is the current that's my unknown thing here I know everything else about this so let's use our equation we said that 1 over R P was equal to 1 over R 1 plus 1 over R 2 and let's just fill in the numbers 1 over R P equals 1 over 20 plus 1 over 60 that equals let's just make 60 the common denominator so I have to multiply this 1 by 3 3 over 60 plus 1 over 60 and that equals 4 over 60 and so now I'm going to take the reciprocal here RP equals 60 over 4 or RP equals no no no no 15 ohms so what this is telling us is if we have two resistors in parallel 20 ohms and 60 ohms that is for the purposes of calculating the current here that's the same as 15 ohms it took to three volts just like that let's check what the current is current is AI equals V over R equals three volts over 15 ohms that's equal to 0.2 amps or you can say it's the same as 200 milliamps so we actually have now simplified our circuit from two resistors to one resistor and we were able to compute the current here which is 0.2 amps and I would invite you to check this by going back and computing this current up here to make sure it's the same and the way you would do that is you would calculate the voltage the voltage here is 3 volts 3 volts across 23 volts across 60 you'll get i1 and i2 and if you add those together you'll get the total I and it should come out the same as this and I think that's a good exercise for you to do to prove that the expression for a parallel resistor 1 over our parallel can be computed from 1 over r1 plus 1 over r2

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