Capacitors and finding Voltage

Discussion in 'Homework Help' started by mer584, Feb 24, 2008.

  1. mer584

    mer584 Thread Starter New Member

    Joined:
    Feb 23, 2008
    Messages:
    6
    Problem: A 3.00uF and a 4.0uF capacitor are connected in series and this combination is connected in parallel with a 2.0uF capacitor. If 26V is applied across the whole network, calculate the voltage across each capacitor. (See attached figure)

    My Attempt/Questions:
    So my thought was to find the total charge and the total capacitance and use V= Q/C to find the voltage across each

    so the total capacitance will be 1/(.33 +.25) +2 = 3.71uF
    then the total charge by Q=CV = 3.71 x 26 = 96.5 uC
    so the voltage at 2.0uF capacitor will be V=Q/C = 96.5/2 = 48.2V

    What I don't get now is how I relate what I know to the remainder of the parallel series at the top of the box. I know that the voltage wont be split evenly between them, but that they will share the same amount as the one in parallel. That is to say capacitor 3.0 and 4.0 should have a cumulative voltage of 48.2V but that it wont be split evenly between them.

    Is this correct, and if so how do I go about calculating that? Thanks

    Attached Files:

  2. spar59

    spar59 Active Member

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    Aug 4, 2007
    Messages:
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    Location:
    Doncaster, England.
    If the only answers required are the voltages across the capacitors then from the circuit diagram the voltage across the 2.0µF capacitor is already shown as 26V - no calculation required! Ignore the 2.0µF capacitor and concentrate on the voltages across the remaining 2 capacitors.

    P.S. is the voltage AC or DC - if it is the latter then in a practical circuit after the initial charging period the voltages will be governed by the relative leakage currents rather than relative capacitance. If it is AC then consider it as a potential divider using the reactance of the capacitors ( 1/(2.Pi.f.C) )- the frequency doesn't matter, just use a variable f - since it will cancel out in the calculation anyway.

    Steve.
  3. mer584

    mer584 Thread Starter New Member

    Joined:
    Feb 23, 2008
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    I've never seen that formula before :( I don't think we've learned it yet. Why, or how can you tell that this will be an alternating current??

    I can understand now why it is 26V across the middle capacitor at least. Which would make me correct in assuming that the other branch is cumulatively going to be the same aka 26V?? But i havent made any progress towards the latter half
  4. spar59

    spar59 Active Member

    Joined:
    Aug 4, 2007
    Messages:
    51
    Location:
    Doncaster, England.
    The reactance (similar to resistance) of a capacitor is as given by the formula stated - if you haven't met it before then I shall point you to the site's E-book page for more information since I was struggling to find the appropriate symbol for Pi.

    http://www.allaboutcircuits.com/vol_2/chpt_4/2.html

    Unfortunately the diagram you provided does not specify A.C. or D.C. which is a rather large omission.

    Steve.
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