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Capacitor and voltage source in one?

  1. Apr 26, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    Consider circuit below.

    (a)Use integral calculus to determine how much heat is generated through R when switch S is closed.

    (b)Derive the expressions for the charge q(t) and the current i(t) as functions of time, for the discharging capacitor in the circuit.

    [​IMG]
    2. Relevant equations



    3. The attempt at a solution

    First of all I'm not sure what the capacitor is or the voltage source is. Is it a battery or a capacitor??
    It seems that it is one and both, but how can this be?

    Please help.
     
  2. jcsd
  3. Apr 26, 2008 #2

    Hootenanny

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    The voltage source in the circuit is the capacitor. All a voltage source does is provide a potential difference across the circuit, in this case the potential difference is provided by the potential across the plates of the capacitor.
     
  4. Apr 26, 2008 #3

    ~christina~

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    for part a where I have to

    (a)Use integral calculus to determine how much heat is generated through R when switch S is closed.

    I know
    [tex]\Delta Vo= 100V [/tex]
    [tex]C= 20x10^{-6} F [/tex]
    [tex] R= ? [/tex]

    and that [tex]E_{int}= Q [/tex] in a resistor (heat produced), and thus the heat that is generated is in the form of the internal energy that builds up in the resistor.
    And also that the rate at which the energy is converted to heat in the resistor is
    [tex] P= I \Delta V [/tex]

    but as to the equation I'm not sure how to set it up.

    I also don't know R or is it not necessary to know R?

    Thanks
     
    Last edited: Apr 26, 2008
  5. Apr 26, 2008 #4
    Here's what happened:
    You connected your capacitor to a voltage source with 100 V ..
    After long time, when no current flows, the voltage across cap is 100 V
    (This is step response)

    and now, you take capacitor and connect to a resistor (second thevenian circuit)
    you u know
    i = C.dv/dt
    so, i does not change intially, but v makes a sudden jump
    use x(t) = .. equation
    (see Natural response section in your book .. you will see detailed steps)

    and you don't need to know R,
    where does the energy lost from the capacitor is going?
    Ans: to R
    and using conservation of energy
     
  6. Apr 26, 2008 #5

    ~christina~

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    I get that. and it's because there is no current flow at the time when the capacitor is fully charged.
    I don't get this. Isn't there only 1 switch there? so how do I take the capacitor and connect it to the resistor if the capacitor is already connected to the resistor since it produces the potential difference in the first place?

    yes
    how does v make a sudden jump if the current is decreasing? and the V is increasing at the same rate? (I think that's what happens when you charge a capcitor)
    I can't find that, it would be in which section exactly?(under what topic..under electricity and circuits)
    Thanks.
     
  7. Apr 26, 2008 #6
    ook, say makes a negative jump :rolleyes:

    If you are using physics book, then just forget about this.
    If you are electrical, and using electric circuits book then it shouldn't be hard

    See page 2 (Transients) for quick solution:
    http://www.ece.uwaterloo.ca/~ece100/Final_W08_formulas.pdf

    OR

    and see my prof notes (for finding x(t) ... ) for little 30 seconds explanation:
    http://www.ece.uwaterloo.ca/~ece100/
    ---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/
    ---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/SummaryRL_RC_AC.pdf
    (this is what you need: a 5 minutes read)
     
  8. Apr 27, 2008 #7

    ~christina~

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    I'm using a physics book.
    I'm even more confused now. I read the 2nd link and just got through 1/2.
    It helped but I got up to power and then capacitor later. Then the notes mention energy in a capacitor as being [tex]W_{c}(t)= 1/2 CV^2(t)= 1/2 \frac{q^2(t)} {c} [/tex] (pg 77 second link)and since you said that all the energy goes from the capacitor to the resistor does this happen with this situation and thus the energy can be calculated from the above equation??

    and I don't know what the notation of x(t) is supposed to be anyhow. even though I read the derivation which wasn't really a derivation in the first place.

    BUT I do have a derivation in my book of the enrgy that is delivered to a resistor during the time interval it is discharged after it is charged to full potential difference of a battery.
    The equation that is solved is [tex]E_r= 1/2 C \epsilon^2 [/tex]

    is this the same as what I'm asked above, where I have to find the heat but the heat is from the energy that the resistor gets from the capacitor?.

    Thanks.

    Note: if anyone would like to help me out on this problem it'd be appreciated.
     
    Last edited: Apr 27, 2008
  9. Apr 27, 2008 #8

    Hootenanny

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    You are correct, all the energy stored in the electric field by the capacitor will be dissipated by the resistor as heat.
     
  10. Apr 27, 2008 #9

    ~christina~

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    so would the ammount of heat be

    [tex]E_r= 1/2 C \epsilon^2 = 1/2(20x10^{-6})(100V)^2= 0.1J[/tex] ?
     
  11. Apr 27, 2008 #10

    Hootenanny

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    Looks good to me :approve:
     
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