 Okay, so I told you I would look at this problem, basically you have three light bulbs hooked up in a situation, no, it's just two light bulbs, I thought it was three, okay, it's just two, well that makes it a little bit easier, that's fine. So there's some battery, it's three volts right there, and then we have two light bulbs like that. And so the light bulbs I've drawn in the conventional resistor style, since that's what they are, but we could really think of it like here's the normal wire, and then the filament's like a really thin wire of different material like that, that's all it is, it's just a really thin wire made of tungsten, okay. And so in this case we're going to assume the contribution to the change in potential due to these thicker wires is negligible, all that matters is the thinner wire, is this two bulbs, yeah, okay. So the first question says, you know what, if you measured the current in these different locations what would you get, okay, so if I measured the current, let's say right here, right here, right here, and right here, okay. Well let me call this I1, I2, and I3. So we have the current, first of all we have conservation of current. So if this were a water pipe and there's water going through here and it splits, some of it goes through one light bulb and some of it goes through the other, and it comes back, these two have to be the same, you know, think about what if they weren't, what if this had 8 electrons per second and this was 10 electrons per second at that point, then where would those extra electrons be coming from? I don't know, there's no other thing that would make that not the same, okay. It's one loop, you have to have the same electrons per second, which is current. So these two would have the same value of current. The question asked about a compass, if you have the same current it's going to make the same magnetic field and deflect the compass the same amount. Well right here we have a node and so at a node we also know that the current coming in has to be the current coming out. So that means that the current in two has to be less than one and the current in three has to be less than one, and also since this is symmetrical the current in two and three has to be the same as each other, but they're both less than wire one. Okay, that was the first part, the first real part. The second part asked about charges, I'm not going to do that. Let me do the next part that you probably had trouble with and that was they give the filament length and the filament radius and they asked for the electric field here, they called this A, B, and C, so I called it E1, E2, and E3, they want the electric field in these three locations, I think that's what it was, three bulbs, wait this is three bulbs, oh in the three bulbs they had two circuits, okay they had another circuit like this, okay and this is what we'll call another one. What about the deflection in this one, is the deflection going to be the current deflection due to this current going to be more or less than that one, that's a good question, if this is the same battery then think of it in terms of the loop rule, right, if I do E.dl along that whole thing then this piece is going to have to have the same current as that because I could do a loop this way and I could add up the voltages which I'm going to do in a second and I would see that this current is going to have to be the same as that, that means that the current here is going to be greater than the current here because this current is going to be the sum of those two currents so it's going to be greater than just one bulb by itself, okay but let's find the electric field, I'm going to do this one first and then I'll find the electric field in those two cases next, so right here let's say that they, this is really like that and this is four millimeters long and then they give a radius of six times ten to the negative six, r equals six times ten to the negative six meters, okay so in this case let me apply the loop rule, so the loop rule says that as I add up the change the potential around this whole thing it has to come up to zero so I get zero equals, if I go across there I'm going to get a plus emf and then I'm going to have, the electric field is going to be going down that way so in here I have an electric field so the, and the rest I'm going to assume is negligible so here I'm going to get minus e times a length because that's a constant electric field over the length of that filament that's what I get so how would I find the electric field? Well I'll just go the electric field is going to be emf over l so that's going to be three volts of the three, two, one point five volt batteries divided by the length of four times ten to the negative third meters and so I get point seven five times ten to the third volts per meter and that's the electric field in here, okay, that's it. Now if I do the loop rule here I'm going to get the exact same thing, right? If I do this loop it doesn't really matter what loop I choose if I do that loop it's going to give me the exact same equation with the exact same filament, exact same l, I get the exact same electric field. What about e2? Well I could do, I could do two ways I could do this loop which is the exact same thing or I could do this loop. Let me set up that loop just so you can see what it would look like. If I go that way the electric field is that way so I get minus e2l and then I'm going to go against the electric field plus e1, I'm sorry that's e3 I call it e3l equals zero so here I get e3 equals e2 so it's the same also so they're all the same they're all the same value. I don't need to even use the electron mobility I don't need to use the radius of the thing because I'm getting a phone call. I don't need to use those because it just doesn't come into play they're the same electric field. Okay the last part says how would you find electron mobility if they give you the current of i equals 0.08 amps equals q n a u e find the electron mobility. Okay so this is one this is the charge of electron 1.6 times 10 to the negative of 19th this is the charge carrier density which they give you a you could find this is just going to be pi r squared and you just found e so that's it. Now I can answer my phone call.