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Capacitor value -- series/parallel caps to increase voltage

  1. Sep 16, 2014 #1
    1. The problem statement, all variables and given/known data
    The number of 2 uF, 400 V capacitors is needed to obtain a capacitor value of 1.5 uF rated for 1600 V is..............


    2. Relevant equations


    3. The attempt at a solution

    equating the energy on both sides,
    1/2*2*400^2*n=1/2*1.5*1600*1600,
    n=12, here n = number of capacitors
    [tell me my procedure is right or not,and how]
     
  2. jcsd
  3. Sep 16, 2014 #2

    Rocket50

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    Gold Member

    12 is the right answer.
     
  4. Sep 16, 2014 #3

    CWatters

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    Science Advisor
    Homework Helper

    Suppose the problem was slightly different eg...

    The number of 2 uF, 400 V capacitors is needed to obtain a capacitor value of 1.5 uF rated for 1500 V.

    Your approach would give..
    1/2*2*4002*n=1/2*1.5*15002
    and that gives n=10.5

    Rounding up to 11 doesn't work.

    Going back to the original problem (1600V)...

    To increase the working voltage you have to add caps in series. Four caps in series increases the working voltage from 400 to 1600V but at the same time the capacitance of a string of four reduces from 2uF to 0.5uF. So you need three strings in parallel to make 1.5uF. Three strings of four make 12.

    If the target rating was 1500V you would still need 12 capacitors. The actual rating would be 1600V but you can use it at 1500V.

    PS In practice it's not wise to do this. You need to factor in things like the capacitor tolerance which might mean the voltage isn't shared equally... and there are other issues.
     
    Last edited: Sep 16, 2014
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