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Homework Help: Question about Capacitor

  1. Feb 21, 2005 #1
    A 7.7 uF capacitor is charged by a 125-V battery and then is disconnected from the battery. When this Capacitor [tex]C_1[/tex] is then connected to a second (initially uncharged) capacitor, [tex] C_2[/tex], is the final voltage on each capacitor is 15V. What is the value of [tex]C_2[/tex]
    (The hint from the book is that, charge is conserved.)

    Here is how I have approached the problem:
    Q= CV, so the charge on [tex]C_1[/tex] is 7.7 * 10^-6 * 125 = 9.625 * 10^-4
    And this is also the total charge before removing the battery. After inserting [tex]C_2[/tex], the total voltage is 30V, so we have:
    [tex] 9.625*10^{-4} = (C_1 + C_2)(30)[/tex]
    and solve for [tex] C_2[/tex], which is 2.4 * 10^-5

    Am i right?
  2. jcsd
  3. Feb 21, 2005 #2
    The total Voltage after the charging is done is 30V however the voltage is then split into 15V for both

    [tex] C_{1} (15) + C_{2} (15) = C_{1} (125) [/tex]

    because the charge is conserved the sum of charges on both the capacitors is the charge on the capacitor 1 on how it started. Also Read teh question ti says the FINAL VOLTAGE ON EACH CAPACITOR is 15V.
  4. Feb 21, 2005 #3
    Thx, I similar problem like the previous one:

    A 2.7 uF capacitor is charged by a 12V battery. It is disconnected from the battery and then connected to an uncharged 4 uF capacitor. Determine the total stored energy (a) before the two capacitors are connected, and (b) after they are connected. (c) what is the change in energy?

    (a) it is just [tex]\frac{1}{2}\frac{Q^2}{C}[/tex], so it is 1.944* 10^-4J
    (b) The total capacitance in the circut is 6.7uF, so we can find the total voltage by using Q= CV where Q is the total charge from [tex]C_2[/tex] or 2.7*10-6 * 12 = 3.24 * 10 ^-5, so the total voltage is [tex]3.24 * 10^{-5} / 6.7 * 10^-6 = 4.8358 V[/tex]
    Then we setup the equation :
    [tex](2.7{\mu}F)(V_1)+(4{\mu}F)(V_2) = 3.24*10^-5[/tex] and [tex]V_1+V_2 = 4.8358[/tex],
    then we calculate the electric energy on each capacitance individually by using [tex]\frac{1}{2}CV^2[/tex], and add them together

    (c) it is just the difference of (a)-(b)

    I am not quite sure if i am right on part (b)
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