Jan 03 phy6 question help? Watch

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Shem
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am trying to do this question and am having a blank at the min, so any help wud be appreciated...

A 4700 micro farad capacitor is charged to 25 V and is discharged through a tightly would bundle of fine insulated wire.

calculate the energy dissipated in the wire? its prob really simple but am having a major blank....
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S1M
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(Original post by Shem)
am trying to do this question and am having a blank at the min, so any help wud be appreciated...

A 4700 micro farad capacitor is charged to 25 V and is discharged through a tightly would bundle of fine insulated wire.

calculate the energy dissipated in the wire? its prob really simple but am having a major blank....
Energy= 1/2 x C x V^2
i think that does it...
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Shem
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is this the energy stored on the capacitor or the energy dissipitated in the wires?
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S1M
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its the energy stored... but the same amount of energy is disapated through the wire when the capacitor discharges... conservation of energy
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Shem
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ahh....thanx...
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qqqwwweee
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Does anyone know the answer to the next part?
Explain why it would be difficult to use this arrangement to demonstrate that the energy stored is proportional to V^2 for a range of potential differences up to about 50V.
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Rose in Bloom
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(Original post by keisiuho)
Does anyone know the answer to the next part?
Explain why it would be difficult to use this arrangement to demonstrate that the <a href="http://www.ntsearch.com/search.php?q=energy&v=56">energy </a> stored is proportional to V^2 for a range of potential differences up to about 50V.
probably becuz the amount of energy stored would result to a high power due to high voltage- this would heat up the insulator and cause it to conduct...
I guess that was a made up answer, its probably wrong!
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Shem
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maybe something to do with ummm... lack of potentiometer to vary the voltage...? dunno am guessing... not good at these explaning questions... ur explanation sounds pretty thought out... anyone else have any ideas?
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S1M
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which question is this? (what year? what question no?)

it has to be something to do with the high voltage damaging the capacitor... i think
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Shem
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june 02 it is quest 2 b
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S1M
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(Original post by Shem)
june 02 it is quest 2 b
thanks... now i see the circuit even though it was quite obvious anyway...
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S1M
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scrap that last answer i gave... its bs...

the answer must be something to do with the fact that theres more energy loss through the bundled wire (heat) as the current increases with voltage (R is constant so I proportional to V)
does this sound reasonable to you guys?
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AmanBOB
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i think its as you pass larger amounts of current through the wire the wire heats up therefore resistance of wire increases so you cant really show a relationship here as resistance varies
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Shem
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something vaguely related.... transformers, step up transformers increase the voltage and decrease the current so less energy is lost along the lines? does the increase in voltage cause a decrease in the current?
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Rose in Bloom
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(Original post by AmanBOB)
i think its as you pass larger amounts of current through the wire the wire heats up therefore resistance of wire increases so you cant really show a relationship here as resistance varies
Yeah thats similar to wat I wrote earlier....unaccrate results will be obtained due to increase in R as result of the wire heating up...
Or like I mentioned earlier in this thread that the head might cause the insulator material to conduct again giving in accurate results!
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qqqwwweee
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But the main point is how we can measure the energy stored or dissipated when discharged?

If the energy can be measured, there is nothing do with change in resistance because the energy is conserved. A change in resistance only varies the current and the discharge curve, but the total energy dissipated is still the same.

So the problem comes to the voltage of power supply? I think it may account for a mark only. You can only write one or more sentences about the power supply of fixed voltage not being able to charge with a range up to 50V.

Does anyone have the mark scheme?
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Emma Louise
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The energy stored in the capacitor is very small.
The wire might melt.
Heat will be lost to the surroundings.
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qqqwwweee
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(Original post by Emma Louise)
The energy stored in the capacitor is very small.
The wire might melt.
Heat will be lost to the surroundings.
but the question doesnt give the resistance of the fine insulated wire. How do you know that the wire might melt?
The resistance may be quite large and so the current flow is small
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S1M
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(Original post by keisiuho)
but the question doesnt give the resistance of the fine insulated wire. How do you know that the wire might melt?
The resistance may be quite large and so the current flow is small
hmmm i see your point... maybe its got to do with the resistances before it is charged or whilst it is charging... that resistance would be higher which will screw up the results through heat loss... iirc there is a resistor there too... wouldn't exposing it to high voltages heat it up resulting in energy loss whilst the capacitor is charging?
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Emma Louise
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(Original post by keisiuho)
but the question doesnt give the resistance of the fine insulated wire. How do you know that the wire might melt?
The resistance may be quite large and so the current flow is small
Because I have the answers written down.
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