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Need help answering an electrical question

  • 30-11-2012 1:36pm
    #1
    Jude_2010
    Registered Users, Registered Users 2 Posts: 118 ✭✭


    can anyone help me do this question please?


Comments

  • #2
    FISMA
    Closed Accounts Posts: 2,616 ✭✭✭FISMA


    Jude_2010 wrote: »
    can anyone help me do this question please?

    You should post this on the Physics forum.

    The instant you complete a circuit a capacitor acts like a perfect wire (a good conductor/a short circuit). "A long time later" the capacitor acts like a broken wire (open switch) as it is fully charged.

    Start by breaking the circuit down into more manageables pieces - show the potential drops across each resistor and the currents.

    I think the easiest way to do this is to break the circuit down into the simplest possible and apply the Law of Conservation of Energy - Kirchoff's Loop Rule and maybe the Law of Conservation of Charge - Kirchoff's Junction Rule.


  • #3
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,641 Mod ✭✭✭✭2011


    FISMA wrote: »
    You should post this on the Physics forum.
    Why?
    It is an electrical question.


  • #4
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,641 Mod ✭✭✭✭2011


    OP, try this:


    Q =CV

    Clue: V is the voltage applied across the capacitor.


  • #5
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    Probably 286.7 uC or so? Or something not even close:D


  • #6
    Jude_2010
    Registered Users, Registered Users 2 Posts: 118 ✭✭Jude_2010


    thanks robbie .. that was the right answer


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  • #7
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    Jude_2010 wrote: »
    thanks robbie .. that was the right answer

    Do you know how to get it?


  • #8
    Jude_2010
    Registered Users, Registered Users 2 Posts: 118 ✭✭Jude_2010


    get the voltage drop across the 1.2K resistor and 'use the formula Q=CV?


  • #9
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    Jude_2010 wrote: »
    get the voltage drop across the 1.2K resistor and 'use the formula Q=CV?

    Its actually the voltage drop across the 26k resistor R, that you use, since the capacitor when fully charged through its 10k resistor will be at the same voltage as R, as the cap and 10k are in parallel with the 26k R

    Then multiply that voltage, which happens to be 5.7348v x 0.00005 F (50uf) = 0.0002867 C or 286.7 uC


  • #10
    .G.
    Registered Users, Registered Users 2 Posts: 5,420 ✭✭✭.G.


    Jesus christ I only finished in college in March 2011 and this bloody thing fried my brain!

    Got there in the end though

    god help me in a couple more years :D


  • #11
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    superg wrote: »
    Jesus christ I only finished in college in March 2011 and this bloody thing fried my brain!

    Got there in the end though

    god help me in a couple more years :D

    You can do it by ratios as well as by current x resistance to find the voltage across R.

    1200 ohm resistor, 26000 ohm resistor. Common denominator = 27200

    Ratios are 1200/27200 and 26000/27200 so as voltage drop across R is needed, its 6v x 26000/27200 = 0.9558 which is the fraction of the 6v which is across the 26000 ohm resistor R.

    So 6 x 0.9558 will give the voltage across R. And from that you can get the charge on the capacitor.


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