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Can u help me design my project

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Rockylink

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Hello, Im Jason Fernandes, 18, and a student at college in london. Basically i have to complete a planning exercise for my physics exam practical paper, I was given this task, I will post the details below:

'A student is comparing materials for use as electrical insulation. The student decided to measure the resisticity of a number of different materials. The circuit below is used to determine the resistance of the sample

-Diagram of a circuit, 3v d.c power supply connected in series to a milliammeter, and then to a sampler of insulating matetial. A voltmeter is connected across the insulator.

The student finds that the milliammeter give a small reading. However when the voltmeter is removed from the circuit the milliameter reads zero. The student concludes that the current registered on the milliammeter is due to the voltmeter and not the insulating material. The student decides to make several modifivcations to the experiment in order to obtain a measurable current in the insulating material.

Design a lab experiment to measure the resistuvty on an insulating material such as paper, PVC, Glass. You may asume that standard colelge equipment is avaiblable.

yoour answer should contain details of the following:
a) The steps which would be taken to ensure the current measured is not due to the current passing through the voltmeter.

b) The modifications you would make to circuit in order to obtain a measurable current.

c) What size and dimensions of insulating material you would use and reasons for ur choice.

d) how the resistivity would be found and measured

e) safety precautions.

f) any design featured that would improve the accuracy of ur results.'


and thats it. It might look complicated, but dont worry im not looking for someone to write it out for me. We have to go sit in a hall one morning and write the whole thing up from scratch to make sure we understand it.

I just need help in overcoming 1 problem, the current going through the voltmeter:
Treating the voltmeter (digital, typical resistance of 1M ohm supposedly)
like a resistor it takes up a large proportion of the current because the resistance of a sample that im supposed to be measuring is pretty high. For my test values, i used a glass block of resistivity in the region of 10^12 Ohm m^-1, being 0.003m in length, and 0.01m^2 in cross sectional area. using the equation R = pl/A , got a resistance of 3 x 1-611 ohms.

any suggestions on what to do? I dont think Im supposed to make the resistance of the insulator sample comparintively smaller that the resistance of the voltmeter. I think the solution lies in the circuit. I thought about increasing the voltmeter resistance somehow, or something else.
The goal is to acheive current in the insulator of no small than 10^-12 amps, using a high tension supply supply.

Most other problems i have covered, but any other help would be great.

If u can help me, could u please email me too, I would appreciate it.
thanks, Jason Fernandes
 

Gene

New Member
Here's one solution. Remove the voltmeter from the circuit and calculate the voltage across the insulator.
1. Configure the circuit as a 3 volt supply feeding a mAmeter feeding your insulator and return to the supply voltage. So you have two resistances (the meter and the insulator) in series with your 3 volt supply.
2. Using an ohm meter from the lab, measure the internal resistance of the mA meter - some meters have this printed on the back. Lets say it's 1 ohm.
3. Using the circuit (from #1 above) read the mAmeter. Lets say it reads 0.5 A with everything hooked up.
4. Calculate the total resistance of the circuit as e/i or 3/0.5 = 6 ohms.
5. Subtract the internal resistance of the meter as 6 (from #4) - 1 (from #2). Your insulator is 5 ohms.

By the way, I have heard of resistance and conductivity but resisticity is a new one to me - and it's not in my dictionary. I'm still learning.
 

Gene

New Member
Did a professor write the problem? I hope not and that this is just a series of your typos. I am the worlds worst speller and do not claim to be a typist - but I am also not a college professor. If this is verbatum from your professor, he deserves an "F." Also, all students deserve an increase of one letter grade for having to deal with a teacher who can not express himself clearly.

". . . then to a sampler of insulating matetial." Shouldn't this be a sample instead of a sampler - is this a special way of spelling material?

". . . asume that standard colelge equipment is avaiblable." What is a 'colelge'?

". . . the milliammeter give a small reading." No, it gave a small reading.

". . . student decides to make several modifivcations." No, modifications

"Design a lab experiment to measure the resistuvty. . . " I haven't got the slightest idea what this means.
 

mechie

New Member
Wheatstone Bridge ?

Could the good old Wheatstone bridge help ...

see diagram ...

If 'B' is your sample, the relationship of 'A' to 'B' is the same as 'C' to 'D'
ie if 'B' is 1G (gig - ohms) and 'A' is 10M (a ratio of 100 to 1 ?)
. then 'D' could be 10M and 'C' could be 100K (also a ratio of 100 to 1)

'D' is adjusted to show zero on the meter (uA or volts) at this point the two tappings are the same potential and the meter cannot be affecting the reading as there is no current through it.

As you know the ratio of 'A' to 'C' - you chose them as fixed values; you know the ratio of 'B' to 'D'.
'D' will be a more convenient value to measure conventionally (removed from the circuit of course).

A high voltage supply will be required
The technique can be refined somewhat but this should start you off :?:
 

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Ricky

New Member
I agree with sir Menchie of using the wheatstone bridge as it can easily determine the one unknown element value.
 

mstechca

New Member
The student decided to measure the resisticity of a number of different materials.
No wonder why you have some trouble. the textbook is showing greek!
resisticity isn't even an electronic word (I dont think). I think what the book means is that you measure resistance of materials.
 

Rockylink

New Member
My project

Ok ok point taken, I have terrible grammer and spelling. Its probaebly the reason I dont take english-y subjects at college. So no more critisizing my bad spelling, theres likely to be more to come.

I meant resistivity. I hope everyone worked that out from what the aim of my project.

Funnily enough I was literally just reading about The wheatstone brdige, say 2 minutes ago. I havent used it before, but I can understand the principle. I but better start with Genes proposal first.

The circuit you are saying i can try wont give me an measurable current. I would have to replace the mAmmeter with a picoammeter, and the 3v supply with a much higer supply. To get a resistance of 5 ohms for example, of a glass sheet of 0.001m length(thickness) and asuming the resistivity of glass is in the region of 10^12 ohm meters:

R = (pl)/A

A= (pl)/R
= ((1 x 10^12) x 0.001)/ 5
=200,000,000 m^2

I cant possibly justify that in the report.
Maybe my understanding here isnt correct though, Im open to corrections.

Here, the 2 diagrams are:
fig 1. The diagram given to me in the project
fig 2. A diagram we discussed today



In figure 1, the problem was that the current measured was that of both the insulator and the voltmeter. In figure 2. its just the insulator. Hopefully i could have some feedback on the circuits.

We also discussed using a C.R.O (Cathode ray oscilloscope) in place of the voltmete in figure 1, because it has a much higher resistance. but there are issues with accuracy and such which im not familiar with. also it will only measure a.c current right.

Finally onto the wheatstone brdige. As I said, explaining how and why i would chose to use this could be tricky, so i need a bit more explaination. I wanted to know if the relationship between the a/b and c/d needs to be somewhat similar, in term of that i will have a huge insulator resistance of maybe 10^12, compared to normal resistor sizes....

thanks for your time, I appreciate all your help with this everyone.
Jason
 

Rockylink

New Member
wheatstone bridge

Using a wheatstone bridge can be useful here, because of the high resistances used. all i have to keep the same is the ratio between resistors. There might be a problem with it though. Its not a big problem.

The way that the problem is given to us, we need to deisgn an experiment that has a dependant variable, such as the voltage, and a dependant, such as the current. Here, i wouldnt get a set of readings.
Maybe I could change the cross sectional area, for 7 samples. Or the length....
 

Gene

New Member
For a technical discussion, I think you have a good grasp of a solution. As I stated, mine was "one solution" - there is surely nothing wrong with the Wheatstone bridge.

Take the voltmeter out of the circuit - it is not useful.

As a practical side note, I don't think it is possible to actually read/calculate the resistance of a piece of glass with 3 volts and a milliampmeter - I would think much more current and a more sensitive meter/scope would be required. It has been an interesting discussion.

Since the write-up was your's, the professor is forgiven :p

Hope you do well on the exam.
 
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