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Resistor Circuits 10,384 views

Teacher/Instructor Jonathan Osbourne
Jonathan Osbourne

PhD., University of Maryland
Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

Resistor circuits are resistor networks which combine resistors in series and resistors in parallel. Removing a resistor in parallel increases the current flow through the other resistor in parallel but decreases the overall current flow. Dimming a resistor in series within a parallel circuit kills an entire branch in parallel, reducing the overall current but increasing the current that goes through the other parallel branch.

So let's talk about resistor circuits and resistor networks. Often in Physics classes you'll see combinations of series and parallel together sometimes there'll be qualitative questions what will happen if this light bulb is unscrewed, what will happen if these are short circuited and sometimes there'll be quantitative questions what's the current through this resistor, how much power is that resistor using, what's the potential difference between this point and that point. So let's go through this and see if we can understand how it works. Most of it is actually quite simple once we are sure that we understand how series and parallel work okay. So let's consider this fairly simple circuit. I've go a 60 volt battery we can tell it's a battery because it's got a long side and a short side. Remember the long side is always positive and the short side is always negative. And then I've got a resistor network, I've got a 6 ohm resistor here and then I've got a branching remember that branching means parallel. So what we would say is that this 10 ohm resistor is connected in parallel to what? It's connected in parallel to the series combination of 7 and 8, see 7 and 8 are connected in series and then that combination is connected in parallel to the 10 ohm. And then what about the parallel combination, what would you say about the connection between this parallel combination and this 6 ohm resistor?

Well anybody who goes through the 6 ohm got to go through this parallel combination, so that means that we would say the 6 ohm resistor is connected in series to this parallel combination. Alright so let's do some qualitative stuff first and then we'll actually go through and solve this circuit completely and determine the current and the potential difference through every single resistance. Alright so some of the qualitative questions might go like this, suppose that we unscrewed the light bulb associated with the 10 ohm resistor? So we're going to kill the 10 ohm and then it'll ask something like this, what will happen to the 6 ohm resistor. If it's a light bulb will it get brighter, will it get dimmer if I unscrew down here? Alright now let's see what this means, if I unscrew down here then that means that I'm taking away a resistor in parallel. Now remember that when you add a resistor in parallel the overall resistance goes down, so when I unscrew this the overall resistance in this circuit is going to go up not down. So that means that the 6 ohm will get dimmer okay if I get rid of this 10.

What about the 7 and the 8 what will happen to them? Well this is a little bit more tricky because while the overall current did go down notice that now it has no choice it has to go through the 7 and the 8 right. So the overall current this current went down, but the current going through here, through the 7 and the 8 is going to go up. Alright so 7 and 8 brighter alright let's look at what will happen if we kill the 7 ohm. So now we're doing like that, this now is fully connected but we've unscrewed the 7 ohm alright so what's going to happen to the 8. Well the 8 was in series with that and that means that since no current can flow through the 7 none can flow through the 8. So the 8 ohm will go off alright so that kills that whole branch, so now notice that it's exactly the same thing. I'm killing a branch in parallel and that is going to reduce the overall current i but it will increase the amount of current that goes through the 10 so 6 dimmer and 10 brighter. Alright and that's the way that that's going to go.

Okay so now let's go ahead and deconstruct this circuit and just solve it, the way that we're going to do that is we're going to systematically combine the resistors in series or parallel whatever we need to do until we have a much simpler circuit and then we'll just find the currents and then we'll reconstruct the circuit. Alright so I'll write it first like this 60 there and then we've got 6 here this guy I left alone and that's my 10 and then here what I'm doing is I'm saying look these guys are in series so I'm just going to add them together. Well what does that give me, well it gives me 15 alright now I've got a 10 and a 15 in parallel so I'm going to add these 2 in parallel. And remember the way that you do that you do 10 to the negative 1 plus 15 to the negative 1 and then you do answer to the negative 1. You're using the calculator that's the easiest way to do it, that way I found that students make much less mistakes if they just type in this to the negative 1 that to the negative 1 and then answer to the negative 1.

Another way to do it is to write it as 10 times 15 over 10 plus 15, that will always give you the same answer alright but if you try to do it this way it won't work if you're trying to do 3 or more resistors in parallel whereas this will. Either way this turns out to be 6 so we'll end up with 6, 6 and 60 and now of course these 2 are connected in series so that means that really it's just 60 and 12 alright. So now what we're going to do is we're going to say well how much current is going through this 12 ohm effective resistor. Well if the potential difference across it is 60 then I must have 5 amps going there alright. Now let's go back through the reconstruction of the circuit because I think that it really helps you understand the parallel and series combinations as we go through and reconstruct a circuit. Alright so 5 amps was going through this 12 ohm resistor that was a series combination. Well in series current is the same so that means that I've got 5 amps going through both of these resistors alright.

Now I've got 5 amps going through this 6 ohm resistor, this 6 ohm resistor was the parallel combination of the 10 and the 15 over here. In parallel potential difference is the same, so what I need to do to reconstruct this parallel combination is I need to determine what's the potential difference across this resistor and of course I'll do that just by doing ir 5 times 6 is 30 so that means that I got to have 30 volts across the parallel combination. 30 volts across the 10, 30 volts across the 15 alright so that means that I've got 5 amps coming in here to have 30 volts across the 10, 3 amps to have 30 volts across the 15 2 amps alright. And then of course we'll go up and we'll reconstruct our 7 and our 8 of course that's a series combination and in series current is the same. So we've got to have 2 amps going through both of these guys alright.

Now what I usually do at this point is come over here and make a table like this alright so when I look at this table what's going on is that I've got each of the resistors written here these are the physical resistors that are actually in the first circuit the actual circuit and this is my effective resistance notice that I've got total here. That's going to be a nice little check of all of our Math and it'll check everything at once alright. So I've got all the potential differences I've got all the currents. Remember it was 5 amps and then it split into 2 and 3 the 5 amps are going through the total and the potential difference across the total is 60 notice that in each of these lines I could get delta v by doing i times r 6 times 5 is 30, 7 times 2 is 14, 8 times 2 is 16 and so on and so forth. Alright now for the check, what we're going to do is we're going to say look this effective resistor has to be the effective resistor it has to be the whole resistor network it has to do everything that, that resistor network does. So what we're going to is we're going to calculate how much power in watts each of these resistors is using up. Well to get the power we can always just do iv this times that so we'll get 150, we'll get 28, we'll get 32 and we'll get 90.

Now these are the amounts of power that are used by each of the physical resistors, if I add them all up I get 300 watts so that means that the resistor network is using 300 joules per second. What's the effective resistor using? Well what's 60 times 5? 300 and that gives us the check when we deconstruct the entire circuit and then reconstruct it back again we need to get powers that add to the total power that we obtained with our effective resistor and that's resistor circuits.