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Can You Answer this Question?

  • 10-04-2006 5:05pm
    #1
    Ra1ph
    Closed Accounts Posts: 85 ✭✭


    question5.JPG Can someone answer this question for me? I know it's pretty easy but have a mental block at the moment.

    Formula needed are:

    I=Q/t and C=Q/V

    I know you don't like doing other peoples work so some hints will do fine also. Thanks


Comments

  • #2
    dudara
    Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    Calculate the combined capacitance of the parallel capacitors.

    Find an expression that relates current flowing to time ie I(t) = ...

    When the capacitors are charged, there will be no current flowing.


  • #3
    Ra1ph
    Closed Accounts Posts: 85 ✭✭Ra1ph


    Cheers,
    Am I on the right path?

    C1 + C2 = 300uF + 500uF = 800uF
    The Total Charge: Qt = Ct x V
    = 800 uF x 240
    =192000 uCoulombs

    The Charge on each capacitor:
    Charge on 300 is: 72000 uCoulombs
    Charge on 500 is: 120000 uCoulombs
    I = V/R
    = 240/68000 = 0.00353Amps
    I = Q/t therefore
    t = Q/I...


  • #4
    Ra1ph
    Closed Accounts Posts: 85 ✭✭Ra1ph


    apologies I think I worked it out:

    C1 + C2 = 300 + 500 = 800uF
    RC = t
    68 x 10-3 * 800 x 10-6 = 54.4secs

    Capacitor will become fully charged in five time constants (5RC).

    272secs


  • #5
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    That's it exactly! I had big long post written but it didn't post for some reason, but you have the answer anyway so it's grand. Anywhere around 272 seconds is acceptable really as its arbitrary how many time constants you take really. I mean technically, the capacitors will never be fully charged. The question should really be worded better to avoid the smart student who just writes down that...


  • #6
    Ra1ph
    Closed Accounts Posts: 85 ✭✭Ra1ph


    Thanks Michael. I eventually found the part of the notes which the question was based on.
    "The time constant is an indication of the length of time it takes to charge a capacitor. The Capacitor charges to about 63.2% of the supply voltage in one time constant. In each successive time constant the capacitor will charge to 63% of the remaining voltage. In this way we can see that the capacitor will never reach the supply voltage. For all practical purposes, however, we can say that the capacitor will become fully charged in five time constants (5RC)."


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  • #7
    Pal
    Closed Accounts Posts: 4,115 ✭✭✭Pal


    Don't know how I stumbled in here, but I did.

    suffice to say I haven't a facking clue what you're talking about.


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