g485 physics question

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nmjasdk
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http://www.ocr.org.uk/Images/62267-q...of-physics.pdf

http://www.ocr.org.uk/Images/58522-m...ysics-june.pdf

in question 2c im wodering what the infinite resistance of the voltmeter do because i calculated things normally and my answer turned out OK i dont get whats the point in saying that

in q2c iv i just dont get that answer at all a better explanation of the mark scheme would be helpful

thanks
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Doctor_Einstein
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(Original post by nmjasdk)
http://www.ocr.org.uk/Images/62267-q...of-physics.pdf

http://www.ocr.org.uk/Images/58522-m...ysics-june.pdf

in question 2c im wodering what the infinite resistance of the voltmeter do because i calculated things normally and my answer turned out OK i dont get whats the point in saying that

in q2c iv i just dont get that answer at all a better explanation of the mark scheme would be helpful

thanks
The voltmeter being infinite resistance means you can ignore it and do calculations normally.

Because the voltmeter has infinite resistance, no current will pass through it and it is thus not part of the circuit. Thus if you perform calculations as if the voltmeter isn't there, you will get the exact same result.

The only reason they say it has infinite resistance is in case some silly students want to include the voltmeter circuit in their calculations.
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uberteknik
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(Original post by nmjasdk)
http://www.ocr.org.uk/Images/62267-q...of-physics.pdf

http://www.ocr.org.uk/Images/58522-m...ysics-june.pdf

in question 2c im wodering what the infinite resistance of the voltmeter do because i calculated things normally and my answer turned out OK i dont get whats the point in saying that

in q2c iv i just dont get that answer at all a better explanation of the mark scheme would be helpful

thanks
If the voltmeter has a finite resistance, then the parallel resistance between it and the resistor would need to be taken into account. Hence the time constant for the circuit would be affected.

Infinite resistance means you can assume no current will flow via the voltmeter and hence all of the discharge current flows through the resistor alone. In practice there is no such thing as an infinite resistance voltmeter - although internal resistance is designed to be very high and can be safely ignored for most practical applications.

part iv: think about the charge stored on each capacitor.

HINT: concentrate on how the charge between the isolated plates (i.e. the central plates) exchange charge.

Next, you need to work out the energy stored on each capacitor independently and then find the ratio of those energies. Use E = CV[SUP]2[/SUP / 2

As t increases from 0, the charge will flow from the capacitors through the resistor which will convert the energy to heat. Does the energy equation involve time? Will that then affect the ratio?
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nmjasdk
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(Original post by uberteknik)
If the voltmeter has a finite resistance, then the parallel resistance between it and the resistor would need to be taken into account. Hence the time constant for the circuit would be affected.

Infinite resistance means you can assume no current will flow via the voltmeter and hence all of the discharge current flows through the resistor alone. In practice there is no such thing as an infinite resistance voltmeter - although internal resistance is designed to be very high and can be safely ignored for most practical applications.

part iv: think about the charge stored on each capacitor.

HINT: concentrate on how the charge between the isolated plates (i.e. the central plates) exchange charge.

Next, you need to work out the energy stored on each capacitor independently and then find the ratio of those energies. Use E = CV[SUP]2[/SUP / 2

As t increases from 0, the charge will flow from the capacitors through the resistor which will convert the energy to heat. Does the energy equation involve time? Will that then affect the ratio?
energy equation doesnt involve t yes but it does involve the voltage so wouldnt the results be affected because voltage decreases at a different rate at each capacitor (since CR value different)?
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uberteknik
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(Original post by nmjasdk)
energy equation doesnt involve t yes but it does involve the voltage so wouldnt the results be affected because voltage decreases at a different rate at each capacitor (since CR value different)?
The charge is shared equally between the capacitors. i.e. there is a finite amount of charge to be shared between the plates connected to each other. No more charge can be introduced or taken away because these plates are isolated from the supply.

That means, once one of the capacitors is saturated with charge, the other capacitor can no longer increase it's charge because it can then no longer be influenced by charge migrating between the isolated plates. i.e. charge migration between the isolated plates is controlling the maximum charge both capacitors can hold.

Charge sharing between the capacitors (and hence equality of charge) will be maintained throughout the discharge period and the hence the p.d. developed across each capacitor is a function of that charge together with capacitance.
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