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Series Combo Of Capacitors, Smaller Than Single Capacitor

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data

    "A series combination of capacitors must always be smaller than the smallest individual capacitor."

    This is the statement that I am trying to validate, but I am stuck.

    2. Relevant equations

    3. The attempt at a solution

    I know that if you have a circuit that consists of a cell and a number of capacitors in parallel, the overall effect due to the 'inside' plates neutralising each other, is that of one capacitor with a large plate separation. This large plate separation is what causes the equivalent capacitance to be lower than the sum of it's parts, but I don't understand how I can show that this value must be smaller than that of the smallest capacitor in the circuit.

    I can show that it is the case for specific examples, but not a general case.

    Thanks for any help you can give,

  2. jcsd
  3. Jan 27, 2015 #2


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    Perhaps start by showing that 1/Ceq = 1/C1 + 1/C2 + 1/C3 etc
  4. Jan 27, 2015 #3


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    Have you obtained the formula for the equivalent capacitance of two capacitors in series? (You can always deal with more than two by combining two at a time).
  5. Jan 27, 2015 #4

    [itex]V = \frac{Q}{C}[/itex]

    So the voltages on the individual capacitors is given by;

    [itex]V_{1} = \frac{Q}{C_{1}}[/itex],

    [itex]V_{2} = \frac{Q}{C_{2}}[/itex],

    [itex]V_{3} = \frac{Q}{C_{3}}[/itex] etc...

    The total voltage is [itex]V_{tot} = V_{1} + V_{2} + V_{3} + ...[/itex]

    [itex]V = \frac{Q}{C_{eq}} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} + ...[/itex]

    Dividing through by [itex]Q[/itex] yields

    [itex]\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + ...[/itex]
  6. Jan 27, 2015 #5


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    Okay, now using what you've done, write and simplify the equation for the equivalent capacitance of just two capacitors in series.
  7. Jan 27, 2015 #6

    [itex]\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}[/itex]

    [itex]C_{1}C_{2} = C_{eq}C_{2} + C_{eq}C_{1}[/itex]

    [itex]C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}[/itex]
  8. Jan 27, 2015 #7


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    Excellent. Now does that last expression suggest anything to you about showing that ##C_{eq} < C_1## and ##C_{eq} < C_2##?
  9. Jan 27, 2015 #8
    Not really...
  10. Jan 27, 2015 #9


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    What if I write the expression this way:
    ##C_{eq} = C_1 \frac{C_2}{C_2 + C_1}##
  11. Jan 27, 2015 #10
    Ah ha!

    ##C_{eq} \propto C_{1}##

    ##C_{eq} = k C_{1}##

    ##k = \frac{C_{2}}{C_{2} + C_{1}}##

    ##C_{2} < C_{2} + C_{1}## (Capacitors can't have negative values)

    Therefore ##k < 1##


    ##C_{eq} < C_{1}##

    The same argument applies to ##C_{2}##

    That makes a lot of sense to me, but is the argument general enough to apply to any number of capacitors?.
  12. Jan 27, 2015 #11


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    See what you can do with what you have now and the comment I made in post #3.
  13. Jan 27, 2015 #12
    For that, wouldn't it be necessary that C1+C2>1 ?
  14. Jan 27, 2015 #13
    Ah, I can see where this is going. With three capacitors I can use my previous result to reduce it to an equivalent of two capacitors, and then repeat the argument.

    Thanks for the help gneill!
  15. Jan 27, 2015 #14


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    Why? For any positive values of ##C_1## and ##C_2##, can ##C_2## ever be greater than ##C_2 + C_1##?
  16. Jan 27, 2015 #15


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    @siddharth23: In addition to that, how would you compare capacitances to real numbers?
  17. Jan 27, 2015 #16


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    You have a capacitor, C, connected to a voltage source, V. Were you to add a second capacitor in series with C, some voltage will appear across that second capacitor, leaving less than V volts across C. Less voltage on C means there is less charge on C. Less common charge means the circuit's capacitance has been lowered, so must now be less than C.
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