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Homework Help: Charging up capacitors

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A 50 [tex]\mu[/tex]F capacitor is being charged from a 6V battery via a 100k[tex]\Omega[/tex] resistor . what is the initial charging current? after a period of time, the charging current is 30[tex]\mu[/tex]A. what are the pd's across the resistor and the capacitor at that moment? how much charge has been stored on the capacitor up to that time?

    2. Relevant equations

    Q = CV

    Initial Charging Current = emf/resistance

    Q = EC (1 - e-t/RC)

    3. The attempt at a solution

    the initial charging current wasnt a problem, using V=IR.
    I0 = 60[tex]\mu[/tex]A

    from there on, i struggled to get the pd's but i realised that one half time had gone seeing as the current had halved so the pd must be 3V on each.

    i tried using Q = CV to find the charge on the capacitor at that point, but it gave me 1.5 x 10-4C which is not the answer given. i also used the third equation given to get the charge, given the time- but the e term was negative, giving a negative amount of charge. the answer given in the book is 6[tex]\mu[/tex]C, but i can't see how to get this.
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2


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    Homework Helper

    Double-check your math. 6V, 50kΩ, ___μA?
  4. Sep 22, 2009 #3
    sorry i copied the question wrong, it was meant to be 100kilo ohms.
    i still dont see how you can get the answer given though.
  5. Sep 22, 2009 #4


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    What are "pd's"?

    What is a "half time"? Is this a "half-life"? If you meant "time constant" (i.e. RC), then that would be incorrect. The quantity (say current) reduces to one-half in a bit less than one time constant of decay. (Remember, the current is the time derivative of the charge, which is an exponential function of time.) Hmm... I guess your result for 3V on the cap and res should be correct anyway.
  6. Sep 22, 2009 #5
    because the current has halved, the potential difference across the resistor must also have
    halved because of ohms low. So it is indeed 3V. The rest of the 6V must be across the capacitor.
    if your statement of the problem is correct, 1.5*10^-4 C is the right answer.

    the e term can't go negative BTW. if t>0 and R>0 and C>0 then -t/RC < 0 so [itex] e^{-t/RC} < 1 [/itex] and [itex] 1- e^{-t/RC} > 0 [/itex]
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