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    http://qualifications.pearson.com/co...e_20130613.pdf
    For the paper above for question 15aii, is the answer that the time for discharge is so much greater than the time constant so it can fully discharge
    http://qualifications.pearson.com/co...c_20130815.pdf
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    (Original post by runny4)
    http://qualifications.pearson.com/co...e_20130613.pdf
    For the paper above for question 15aii, is the answer that the time for discharge is so much greater than the time constant so it can fully discharge
    http://qualifications.pearson.com/co...c_20130815.pdf
    The question states that the battery has negligible internal resistance, that's important because the resistor isn't in series with the capacitor during charging (when the switch is closed) - charging will happen instantly because RC is zero.

    during discharge (switch open) the time constant will be some non zero value because the current is flowing through the resistor.

    you've just got to pick the graph with the closest shape to what you'd expect

    I'd expect to see exponential decay shape to start with, followed by a step back up to the full value of V... which is close to B
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    (Original post by Joinedup)
    The question states that the battery has negligible internal resistance, that's important because the resistor isn't in series with the capacitor during charging (when the switch is closed) - charging will happen instantly because RC is zero.

    during discharge (switch open) the time constant will be some non zero value because the current is flowing through the resistor.

    you've just got to pick the graph with the closest shape to what you'd expect

    I'd expect to see exponential decay shape to start with, followed by a step back up to the full value of V... which is close to B
    thanks for the reply but i think you have misread my question. my question was 15aii not 5
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    oh sorry,

    Well it's approximately correct. you never reach full discharge, it just keeps getting closer and closer to zero potential difference... forever.
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    (Original post by Joinedup)
    oh sorry,

    Well it's approximately correct. you never reach full discharge, it just keeps getting closer and closer to zero potential difference... forever.
    but then the capacitor could supply charge for infinite time but that doesn't really answer the question which is to do with the capacitor supplying charge for 4.7 time constants
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    (Original post by runny4)
    but then the capacitor could supply charge for infinite time but that doesn't really answer the question which is to do with the capacitor supplying charge for 4.7 time constants
    The capacitor will keep charging and discharging, at however minuscule a level, indefinitely. That's the nature of a tank circuit.
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    (Original post by Tootles)
    The capacitor will keep charging and discharging, at however minuscule a level, indefinitely. That's the nature of a tank circuit.
    why does it talk about the time being 4.7 time constants and what would get the mark for this point 'Relates time constant to the time for which current is required'
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    (Original post by runny4)
    why does it talk about the time being 4.7 time constants and what would get the mark for this point 'Relates time constant to the time for which current is required'
    No idea. I'm a self-taught amateur radio operator, all I know is from what I've picked up when making my own caps. As radio transmitters and tuners are based on tank circuits, I know about how the tank circuit's resonant frequency is exactly how those things are tuned. I always "played by ear" rather than mucked around with mathematics though - the limit of my maths knowledge is sixth-form electronics ten years ago.
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    (Original post by runny4)
    why does it talk about the time being 4.7 time constants and what would get the mark for this point 'Relates time constant to the time for which current is required'
    if the TC was much less than the time it's required to provide current for you'd have a problem because there would be very little charge left in the capacitor while you still wanted the current to be flowing, afaict they just want you to calculate the TC and show that you recognise it's roughly in the 'right ballpark' for the application. right ballpark just means the duration of the required pulse is a few times the TC.
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    (Original post by Joinedup)
    if the TC was much less than the time it's required to provide current for you'd have a problem because there would be very little charge left in the capacitor while you still wanted the current to be flowing, afaict they just want you to calculate the TC and show that you recognise it's roughly in the 'right ballpark' for the application. right ballpark just means the duration of the required pulse is a few times the TC.
    thank you for the reply. So the answer that they would be looking for is that since the time constant is much 4.7 times less than the time for which the current is supplied, since the initial current is the same due to the supply voltage and resistance being constant, the current won't decrease too rapidly so the capacitor can supply the average current
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    (Original post by runny4)
    thank you for the reply. So the answer that they would be looking for is that since the time constant is much 4.7 times less than the time for which the current is supplied, since the initial current is the same due to the supply voltage and resistance being constant, the current won't decrease too rapidly so the capacitor can supply the average current
    seems to be what the markscheme is looking for... the question appears to be asking you to show by calculation that the average current over the required duration is similar to the value specified earlier but the markscheme doesn't seem to want to give you marks for that... I'm only trying to interpret the markscheme - I don't write them.

    I also note that these pdf's have watermarks saying they're draft examples so maybe they weren't checked very well.
 
 
 
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