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How much energy in microjoules is stored in the capacitor

  1. Nov 18, 2014 #1
    1. The problem statement, all variables and given/known data
    a) How much energy in microjoules is stored in the capacitor at t=500 micro sec
    b) Repeat (a) for t=infinite

    2. Relevant equations
    Will be on attachment

    3. The attempt at a solution
    I am not sure how to approach the problem.. The graph shown is a graph of amps respect to time. So the tangent line to the function curve is joules?
     

    Attached Files:

  2. jcsd
  3. Nov 18, 2014 #2

    gneill

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    Staff: Mentor

    The tangent to a current versus time curve would be ΔI/Δt, or di/dt. That's not energy.

    What's a formula for the energy stored on a capacitor? You should know one in terms of voltage and another in terms of charge (Coulombs). How might you find the total charge or voltage on the capacitor from the given curve? Note that you also have the formula for the curve, so you can work the problem symbolically.
     
  4. Nov 18, 2014 #3
    Energy stored on a capacitor is given by C=(epsilon*A)/(d)

    Epsilon is a constat
    A is the area of the plate
    d is the distance between the plates

    the equation with voltage is Q=C*dv/dt
     
  5. Nov 18, 2014 #4
    I am currently working on the problem. I am trying to understand what the tangent to the graph or the total area underneath a graph represents when you have a graph that may have voltage respect to resistance, power respect to resistance, current respect to time, etc.
     
  6. Nov 19, 2014 #5
    I said Q=integral of 50e^-2000t with limits of integration of 0 to 500microsec. I get 0.015803mA then I used C=Q/v to get volate. After that I use w=1/2(CV^2)

    But I don't get the correct answer.
     
    Last edited: Nov 19, 2014
  7. Nov 19, 2014 #6

    gneill

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    Staff: Mentor

    You seem to be having some difficulty distinguishing the units of various electrical quantities. The above formula yields C, the Capacitance of the capacitor, not the energy. Energy has the units Joules.
    And that's an expression for the current (I in amps) , or rate of change of change charge (Coulombs) on the capacitor.
     
  8. Nov 19, 2014 #7

    gneill

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    Staff: Mentor

    If you integrate current over time what do you get? Remember that current is the rate of flow of charge, or dQ/dt. That means the area under the curve represents....?
     
  9. Nov 19, 2014 #8

    gneill

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    Staff: Mentor

    Okay, that method is correct, but your value for charge Q is not correct and the units will not be current (mA) but Coulombs (C). Can you show more detail for your integration steps?
     
  10. Nov 20, 2014 #9
    I got it figured out... I used V(t)=1/C integral i dT+V(0) I got 11.06v and then used w=1/2*CV^2 to find energy
     
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