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Offline harryneild

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electrics exam question
« on: 21/04/2007 17:47:50 »
Heres a question i had in an exam which i didnt have a clue how to answer...

The attached picture is the situation.

The first few questions are -

1.What is the potential at X? Explain your reasoning

2.What is the potential difference between A and X?

3.What is the potential gradient along the cable from A to X?

Could anybody answer these with explanations?

Thankyou


 

another_someone

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electrics exam question
« Reply #1 on: 21/04/2007 21:20:28 »
If I read the question correctly, then it seems to be that there is no voltage or current source at point B, and there is thus no current flowinf between X and point B.  If that is the correct reading of the question, then there can (in the absence of any current flow between X and B) be no voltage drop between X and B, so the voltage at X must be the same as the voltage at B.

If the above is correct then the potential difference between A and X must be 200 - 40, and thus is 160V.

Then again, I may have fundamentally misunderstood the question (and in any case it is a good few decades since I took this kind of thing at college).
« Last Edit: 21/04/2007 21:56:46 by another_someone »
 

Offline harryneild

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electrics exam question
« Reply #2 on: 21/04/2007 22:16:24 »
Hmmm yeh thats the answer i put, but i didnt get the mark...

Its a bit tricky and i cant find a website or anything that the papers solutions may be on.

It was an english physics olympiad.

Thanks!
 

lyner

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electrics exam question
« Reply #3 on: 21/04/2007 22:32:39 »
I agree with the preceding posts.
I feel there must be something missing from the question or a misprint.
Having seen many GCSE and A level questions in the past I can say that it isn't that rare for questions to be, at least a bit, dodgy.
If you are to use just the information, supplied, the problem is trivial, I think.
1. 40V
2. 160V
3. 160/d V/km
If there is an implied leakage per km on the cable (i.e. a practical problem) they haven't included it in the question. Schematic diagrams are like maths - they  should show the whole of the problem so the question is flawed.

There must be a mark scheme, somewhere - if those aren't the answers in the scheme, challenge them !
« Last Edit: 21/04/2007 22:37:19 by sophiecentaur »
 

Offline harryneild

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electrics exam question
« Reply #4 on: 21/04/2007 22:43:12 »
Well thats the whole of the first part of the question. I thought that if u answered that, then i could do the second part myself. But, ow well...

"The potential at A is removed and it is insulated from the ground instead. Potential at B is raised to 300V, at which point the potential at A is measured to be 40V."

1) What is the potential at X now?

2) Having measured 40V at end B initially, why is it that 40V has also been required at end A for the second measurement?

3) What is the potential gradient from B to X?

4) Why is the potential gradient from A to X the same as B to X

5) From the two potential gradients obtained earlier, deduce the value of d.
 

Offline that mad man

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electrics exam question
« Reply #5 on: 23/04/2007 00:17:07 »
I'm embarrassed as I should know this but its one of those things you learn to pass exams and then forget.

Its taken me a while to try and understand the question with no textbooks so Its reasoning and guesswork!

In the length of cable the voltage drop is A - B = 160v
The total length of the cable is 50k so I think the gradient would be 160/50 = 3.2v per kilometre?

Then the potential difference at X should be 200 - 3.2 x D
and the potential difference between A and X should be A - X

Maybe, just maybe.
 ;D
 

lyner

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electrics exam question
« Reply #6 on: 24/04/2007 00:16:50 »
I reckon you can find the position of the short because you get 160V drop for one length and 260V drop for the other length.
The Current is the same in both cases, because the same voltage is measured   at X.
If the resistance of the 'short' is R, then the resistance of AX must be R(160/40) = 4R and the resistance of BX must be R(260/40) = 6.5R. This tells you that the ratio of the lengths AX: BX is 4:6.5. This gives you d.
The voltage per ohm (resistance) of each length of cable must be enough to drop the voltage to 40V in each case so that means that the voltage gradients must be the same. i.e same number of  volts per ohm (which is current) corresponds to same number of volts per metre.
 

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electrics exam question
« Reply #6 on: 24/04/2007 00:16:50 »

 

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