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Capacitance - combining multiple capacitors for equivalent C

  1. Feb 12, 2014 #1
    Capacitance -- combining multiple capacitors for equivalent C

    In the given problem, Ceq = C1C2C3C4/(C1 + C2 + C3 + C4)

    The answer says the equivalent capacitance is always less than C1 but I can't come up with thy. When I do this, I can't seem to prove that equivalent capacitance is always less than C1. For example, if I let:

    C1 = 998 F, C2 = 999 F, C3 = 1000 F and C4 = 10001 F, then I get an answer (767119326 F) and this is much larger than C1. Any suggestions?

    (I realize the capacitances used here are very large, but they are just used to make a point.)
     

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  3. Feb 12, 2014 #2

    BvU

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    Use Q=CV and discover that now four capacitors have to share the voltage that C1 would have had all on its own if C2,3,4 were absent!
     
  4. Feb 12, 2014 #3

    BvU

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    Perhaps your formula is questionable ?
    When I do
    $$ {1\over C_{eq}} = {1\over C_1} + {1\over C_2} + {1\over C_3} + {1\over C_4} $$
    I get something completely different ! Which has the right dimension. Your ##C_{eq}## does not have the dimension of Farads ....

    Hint: The ##C_1 C_2 C_3 C_4## in the numerator is correct.
     
  5. Feb 12, 2014 #4
    Hmmm....
    Well I know:

    Ceq = Q(V1 + V2 + V3 + V4)/(V1V2V3V4)

    I just don't quite see how this shows Ceq < C1...do you mind explaining further?
     
  6. Feb 12, 2014 #5

    BvU

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    You know this for two capacitors, because then it comes out right.
    What you wrote here has the dimension Coulombs * Volts / Volts4 and that is not Farads ...
    So: from where did you get this ?
     
  7. Feb 12, 2014 #6

    BvU

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    You mean you know that
    $$ {1\over C_{eq}} = {V\over Q} = {V_1\over Q}+{V_2\over Q}+{V_3\over Q}+{V_4\over Q} $$
     
  8. Feb 12, 2014 #7
    Sorry, yes.

    Ceq = Q/(V1 + V2 + V3 + V4)

    Also, Veq = V1 + V2 + V3 + V4

    So since C1 = Q/V1 then Ceq must be smaller (i.e. it is divided by more positive numbers).

    Thank you!
     
  9. Feb 13, 2014 #8

    BvU

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    Just in case you ever need to calculate the ##C_{eq}## for 4 capacitors in series: do you know how to work out
    $$ {1\over C_{eq}} = {1\over C_1} + {1\over C_2} + {1\over C_3} + {1\over C_4} $$
    to ## { C_{eq}} = {C_1 C_2 C_3 C_4 \over ???} ## ? And: can you guess why anyone would ever use capacitors in series in a circuit ?
     
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