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    Question and answer from the A2 CGP book, I dont underrstand the reasoning!

    Question: For a capacitor of capacitance x, what will be the effect on the charging time of increasing the charging voltage?

    Answer: Nothing.

    Could someone please explain this? Thanks
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    Hmm, thinking about it, charging a capacitor has the equation Q=Qoe^(-t/rc)

    It is clear that the only factors that influence that charge on the capacitor are the time it has been charging for and the time constant. The voltage actually doesnt come into it at all. Strange really, as you would kinda expect it to.

    Good luck tomorrow!
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    Thanks bslforever, but I have a small problem with that becuase T=RC and C=V/Q so a greater voltage would change the time constant!
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    (Original post by lefneosan)
    Thanks bslforever, but I have a small problem with that becuase T=RC and C=V/Q so a greater voltage would change the time constant!
    :eek3: Well thats me out then If that is the case then im not sure. I only have the nelson thornes book and it doesnt mention this.
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    (Original post by lefneosan)
    Thanks bslforever, but I have a small problem with that becuase T=RC and C=V/Q so a greater voltage would change the time constant!
    Hmm it's always a tricky one...

    but instead of thinking of the capacitance of a capacitor as the formula you gave (which is upside down, btw ), it's better to think of capacitance being dependent on the geometry of a capacitor. I don't know if you study any equations for capacitance as a function of geometric variables, but, for example, the formulae for a spherical or parallel plate capacitor are only dependent on areas, radii, distances, etc...

    Also, seeing how V is proportional to Q, that means that C = Q/V isn't going to change much as you increase V.

    Back to the time constant: yes, it's that and that only which determines how rapidly a capacitor will charge or discharge. Keeping in mind that C in this case is actually the equivalent capacitance of the circuit, you change the time constant by changing the circuit arrangement or geometry of the capacitors in question. Alternatively, you could change the (equivalent) resistance of the circuit.
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    Think about it this way:

    You know rate of charge and discharge is always exponential. Half-life=ln2*RC. A constant half-life is a property of an exponential curve(Q against t), you can see from that equation that the only thing effecting half-life is R and C, if these are the only factors effecting half-life then they are also the only factors effecting dQ/dt.

    I see what you mean about C=Q/V, but it's the fact that charge and discharge is exponential that is important. Increasing V just means you can store less charge per volt, but all charge is effected equally so it has no effect on the pattern of how charge is distributed around the circuit.

    I suspect i've used affect and effect in the wrong context, i just dont care! :p:
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    Thanks folks got it now. All legends the lot of you
 
 
 
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