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Capacitor Combination Question

  1. Jan 16, 2007 #1
    Question:

    Each capacitor in the combination shown in Figure P26.49 has a breakdown voltage of 15.0 V. What is the breakdown voltage of the combination?

    [​IMG]

    Work Done:

    I figured that the voltage on the two parallel capacitors on either side must be the same, so both can handle 15V. So I assumed that, since C = Q/V, the amount of charge to the central capacitor would be 3 X 10^-4, but since the whole thing is serial, we have: V = V_1 + V_2 + V_3. How should I go about setting this one up?
     
    Last edited: Jan 16, 2007
  2. jcsd
  3. Jan 16, 2007 #2

    Kurdt

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    What you need to do is work out the capacitance of the combination. What equations do you know that will relate the capacitance to the voltage equation you have written?
     
  4. Jan 16, 2007 #3
    I can easily determine the entire capicitance of the whole system.
    Let C_1 = 20 mu-F, C_2 = 10 mu-F, then
    [tex]\frac{1}{C} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1}[/tex]
    or [tex]C = \frac{4 C_1 C_2}{4C_1 + 2}[/tex]

    We also know the charge of the center capacitor if V = 15, and a serial circuit has all the same charge, so using the highest charge on the center capicitor and the above equation for the entire circuit,

    [tex]V = \frac{Q}{C} = \frac{Q_c}{\frac{4 C_1 C_2}{4 C_1 + C_2}} = \frac{Q_c(4 C_1 + C2)}{4 C_1 C_2}[/tex]

    that's not it - I plugged in the numbers and it's not correct.
     
  5. Jan 16, 2007 #4

    Kurdt

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    Ok I think you have the right idea but you've made a couple of mistakes.

    Firstly [tex]\frac{1}{C} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1} = \frac{4C_1 + 2}{4 C_1 C_2}[/tex]

    is not correct.

    [tex]\frac{1}{C_{eff}} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1} = \frac{1}{C_1} + \frac{1}{C_2} [/tex] that should get you started on getting the right fraction.

    For the second part like I say you have the correct method. So for the middle capacitor which you have labelled C2, the charge it stores at maximum voltage is:

    [tex] \Delta V_{max}=\frac{Q}{C_2}\Rightarrow Q = \Delta V_{max}C_2 [/tex]

    now substitute Q into the [tex] \Delta V_{max}=\frac{Q}{C_{eff}} [/tex] equation and see how you go from there.
     
    Last edited: Jan 16, 2007
  6. Jan 16, 2007 #5
    Yes! thanks ever much! :-)
     
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