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Voltage would a 737. microfarad capacitor

  1. Sep 23, 2009 #1
    To what voltage would a 737. microfarad capacitor be charged by a 5.0. mA current lasting 2.0 seconds.



    t=R/C
    T=Time
    R=Resistance
    C=Capacitance

    Q=CV




    R=TC
    2x.000737=.001474
    Then I'm stuck at this point


    Can anyone explain this please?
     
  2. jcsd
  3. Sep 24, 2009 #2

    tiny-tim

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    Hi xpack! :smile:

    From the PF Library on capacitor (put V0 = 0 :wink:) …

    Inverse exponential rate of charging:

    A capacitor does not charge or discharge instantly.

    When a steady voltage [itex]V_1[/itex] is first applied, through a circuit of resistance [itex]R[/itex], to a capacitor across which there is already a voltage [itex]V_0[/itex], both the charging current [itex]I[/itex] in the circuit and the voltage difference [itex]V_1\,-\,V[/itex] change exponentially, with a parameter [itex]-1/CR[/itex]:

    [tex]I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}[/tex]

    [tex]V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}[/tex]

    So the current becomes effectively zero, and the voltage across the capacitor becomes effectively [itex]V_1[/itex], after a time proportional to [itex]CR[/itex].
     
  4. Sep 24, 2009 #3
    Re: Voltage

    I'm not sure I understand
     
  5. Sep 24, 2009 #4

    MATLABdude

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    Re: Voltage

    That's because tiny-tim was answering a different question (assumption of voltage source and equalization of capacitor voltage).

    Instead, use the capacitor equation: [itex]Q=CV[/itex]

    (Yes, you'll need to do some rearranging, and use a formula for current--c'mon, we can't do all the work for you!)
     
  6. Sep 24, 2009 #5
    Re: Voltage

    Well I've never seen a problem like this. Our teacher gives very poor lectures and then assigns homework another teacher created so I really dont know anything. I'd rather yall not give me the answer and instead explain it, so that way for future reference I know how to work it myself.
     
  7. Sep 24, 2009 #6

    MATLABdude

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    Re: Voltage

    I suspect that this is THE standard capacitor question (since I'd seen it in both my intro electromagnetics course, as well as my intro circuit analysis course). It's designed to show you that a capacitor is a device that stores charge, and that current is just the flow of charge.

    First off, the fundamental capacitor equation [itex]Q=CV[/itex] means that the voltage across a capacitor is related to the charge accumulated by the capacitor, by the (inverse of) the capacitance.

    I'll use an analogy here. Imagine you have some flowing water, 0.1 L/s. Now you have a beaker with a cross-section of 100 cm^2. In 2 seconds, how high is the water in the beaker?

    Now, current is just a flow of electrons [itex]i=\frac{Q}{t}[/itex]. With a flow of 5 mA (or 5 mC/s), and 2 seconds of flow, how many coloumbs of electrons would accumulate in the capacitor, and consequently, how much voltage would you measure across this capacitor?
     
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