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Electric Circuit Question 
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#1
Jan2713, 12:49 PM

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1. The problem statement, all variables and given/known data
I have a square that is 5 cm by 5 cm. I have filled it in with graphite from a pencil. The top left corner I have labeled A. The top right corner B. The bottom right hand corner C. The bottom left hand corner D. I take a DMM probe and set it to measure resistance, I put the red probe at A, the black probe at C. I then take a second DMM and set it to measure DC voltage. I place the black probe at C. At this point I take the red probe and place it around the corners of the squares within the square that are each a one centimeter by one centimeter and measure the voltage. I collect 25 data points. I observed the following data through measurement. The units are Volts. Please note I labeled the corners in my picture. My measured data is obviously not completely accurate, but according to my measured data, I would say that the further away you get from corner C the less the voltage is. This however isn't true along the lines DC and CB. The voltages seem to increase along these lines. I'm asked to explain my collected data for a lab report. 2. Relevant equations R = [itex]\frac{ρl}{A}[/itex] Where: R = resistance ρ = resistivity (in the case of graphite ρ≈1*[itex]10^{5} Ωm[/itex] l = length A = cross sectional area A = wh Where: w = width h = height (thickness of the graphite layer) V = IR Where: V = voltage I = current R = resistance 3. The attempt at a solution I started off by putting in the formula for resistance into Ohm's law in the hopes that I would be able to explain the observed data. V = I[itex]\frac{ρl}{A}[/itex] Now, I guess I just don't understand this. Sense the square is completely filled in, isn't the surface area and length remain constant. I'm not looking at a wire but a completely filled in square. So no matter where I put the red probe within the square to measure the voltage between the red probe and black probe the same resistance exists. I guess I just don't understand why I'm getting different measurements. Thanks for any help you can provide! 


#2
Jan2713, 07:33 PM

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PF Gold
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In the first place I think it is likely that your graphite layer is (1) not uniform in thickness, and (2) uneven in texture and therefore conductivity. Also, your measurement setup looks bad. For example, the voltage in the extreme bottom righthand corner should be zero regardless of the graphite layer error sources.



#3
Jan2713, 09:04 PM

P: 1,184

I agree. The collected data is probably not accurate. There is no way that the graphite layer is of uniform thickness or that I actually measured the voltages at the correct points. Yes, it's supposed to be zero. But, what exactly was my data supposed to show, and why? this is where I'm confused.
Thanks for any help =) 


#4
Jan2713, 09:16 PM

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Electric Circuit Question
Careful, many multimeters in resistance mode will place the positive measurement voltage on the black lead, contrary to expectations. Did you measure the voltage at the probes with the other DMM?



#5
Jan2713, 09:21 PM

P: 1,184

So you mean I was supposed to get negative voltages?
So the red probe at A and the black probe at C should be measuring resistance. This is done by giving of a voltage and finding the difference. The black probe with the second DMM set to measure voltage at point C as well. I used the red probe from second DMM to measure the voltages So no I did not measure the voltages at point A (the red probe set to measure resistance). I did measure the voltage at the two other black probes at C. I should have gotten zero but I got close to it. Not sure though if it helps. 


#6
Jan2713, 09:42 PM

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#7
Jan2713, 09:45 PM

P: 1,184

Ok so then in that case my data is really screwed up. I'm however not going to lie about the data I received. It's probably because the thickness of the graphite layer wasn't uniform and I didn't actually measure the data at the right spots.
In an ideal case though of uniform thickness and correct measuring instruments, what should I have seen? Shouldn't the resistance been the same throughout the whole entire square and there the voltage readings as well? 


#8
Jan2713, 10:05 PM

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Just from intuition I'd guess that there would be a general gradient of potential from the + lead down to zero at the  lead of the source along the diagonal connecting the source leads, and some symmetrical distribution of values surrounding the diagonal. Offhand I don't know a quick way to compute the distribution for a continuous conducting slab of material.



#9
Jan2713, 10:12 PM

P: 1,184

So this is because why? So between A and C, the shortest path would be the diagonal of the square. So voltage should be the highest on this line. The voltage should than decrease uni formally as you move away from the line.
So basically current flows through the square from point A to point C as if it where one giant wire and as you move away from the shortest path (the diagonal) the current has to go through more graphite. Surface area increases so resistance decreases so voltage does as well? The whole square experiences the same current just how one end of a node must experience the same current at the other node. Is this the logic your suppose to use? Also do you know how to create a color map on a computer? I'm supposed to create a color map with this data and don't know of software to do it =( 


#10
Jan2713, 10:36 PM

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I'm reminded of heat transfer problems with heat sources and sinks, where differential equations with boundary values need to be solved over the slab to determine the temperature profile. Laplace and Fourier spring to mind... Not my cup of tea, really. Sorry I can't be more helpful. 


#11
Jan2713, 10:49 PM

P: 1,184

See that's the thing. I don't know what to think of this thing. It was easier I was just dealing with wires and components. This however confuses me. Ideally I'm suppose to have a uniform layer of graphite. I'm sure if I should think of it as one giant variable resistor perhaps where R = (rho*l)/A. But see... in that case it's not a variable resistor just one giant resistor with a defined A and l that dose not change no matter where put the red probe. But see like in this case every voltage should be the same. Like I don't know if current travels the shortest distance or it just travels through the whole square or what. I just don't know what to think of this square in terms of one resistor or many, and how placing your probe in a certain location has anything to do with it.
Like in your logic the current goes through the diagonal resistor first. But why not just through the whole square equally? This is very confusing... especially when I was never told how to think of this... I'm just left to guess =(... I don't know what think of the square??? 


#12
Jan2713, 10:59 PM

P: 1,184

Like I don't even know what voltage difference I'm measure when I place the red probe in spots. Like in the picture below pretend the red circle is the red probe. Am I measure the voltage difference that's the red line despite the whole square being covered in graphite and the whole square is conductive of electricity with the red probe and not just the red line?
Or something like this or perhaps the whole entire square sense it's all connected together. 


#13
Jan2713, 11:11 PM

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#14
Jan2713, 11:19 PM

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#15
Jan2713, 11:32 PM

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Interesting. So the .485 volts that makes no sense. This means that this point is .485 volts higher than point C. This is all this means. It's a scalar. Now what exactly is the length or cross sectional area that you are suppose to in vision for the resistance at this point, as opposed to other points?
I have no idea. 


#16
Jan2713, 11:44 PM

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If there's a trick to making the calculation simple for this geometry I haven't spotted it yet. 


#17
Jan2713, 11:46 PM

P: 1,184

Perhaps I'm over complicating things. But lets say we take the red probe and place it as close as it can be to the black probe without touching it . Lets make corner C the origin of a coordinate system. To the left is positive. Upwards vertically in the same plane is positive. Vertically upwards out of the plane is positive. Assuming uniform thickness
So the probe would be at point (dx,dy) I'll just call the width x and the length y. So the voltage would be [itex]dV = I\frac{ρ dy}{dx*h}[/itex] Now as you move away from the corner C... idk but I think maybe perhaps I'm on the right path??? 


#18
Jan2813, 12:04 AM

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There's no one current value you can plug in because the current is spread out over a 3D volume and we don't know offhand how its distributed.



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