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Capacitance and Resistor Watch

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    Part (d) is the one I'm having difficulty with.

    Initially I assumed a voltage of 12 V was being applied across the resistor and used my previous calculation of current to find resistance.

    This gave me a wrong answer.

    I then realised that the energy stored in the capacitor must all be dissipated in the resistor.

    Hence, I used E=VIt to find the voltage (6V). I then used V=IR to find R.

    My questions are :

    1.) Why did the first method fail ?

    2.) When I used E=VIt I found the voltage was 6V. What does this value signify ? Is the average voltage ?

    3.) What would be the reasoning behind part (e) ?
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    And how would I approach part (b) in this question ?

    If half the charge flows out shouldnt half the energy be dissipated as well ?
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    (Original post by Ari Ben Canaan)
    And how would I approach part (b) in this question ?

    If half the charge flows out shouldnt half the energy be dissipated as well ?
    If half the charge flows out then the p.d. across the capacitor halves as well.
    So the energy stored in the capacitor would be a quarter of what it was before the discharge process (?).

    E=1/2QV=1/2 x CV^2.
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    (Original post by Killjoy-)
    If half the charge flows out then the p.d. across the capacitor halves as well.
    So the energy stored in the capacitor would be a quarter of what it was before the discharge process (?).

    E=1/2QV=1/2 x CV^2.
    Sorry, that was a typo. I meant one quarter the energy.

    However, the given answer is 1.08 J :/
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    (Original post by Ari Ben Canaan)
    1.) Why did the first method fail ?

    2.) When I used E=VIt I found the voltage was 6V. What does this value signify ? Is the average voltage ?

    3.) What would be the reasoning behind part (e) ?
    1) &2) The voltage varies throughout the discharge process. Sketch a graph of V/Q and you will see that the average must be 6V. (diagonal for 0-12V if the capacitor is fully discharged.)

    3) I think work would be done to heat the wires as well.

    I don't know how good my answers are but as you haven't had responses I thought I'd try
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    (Original post by Ari Ben Canaan)
    Sorry, that was a typo. I meant one quarter the energy.

    However, the given answer is 1.08 J :/
    If you charge it to half its total charge the energy will increase to 0.36J.
    Then charge it fully and the energy will increase by 1.08J to 1.44J.

    When half the charge leaks 1.08J will be lost.
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    (Original post by Killjoy-)
    If you charge it to half its total charge the energy will increase to 0.36J.
    Then charge it fully and the energy will increase by 1.08J to 1.44J.

    When half the charge leaks 1.08J will be lost.
    Ah, yes, silly me. Thank you !! I understand it now !
 
 
 
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