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Current, Power and Energy in a Capacitance

  1. Sep 18, 2011 #1
    The problem statement, all variables and given/known data
    The graph below shows the voltage across a 100 μF capacitor vs. time.
    Find the current (Ix), the power delivered (Px) and the energy stored (Wx) at time tx = 8 s.


    The attempt at a solution

    I found the current, Ix using the capacitance multiplied by the slope between 7s and 13s. However, I got stuck when I tried to integrate the voltage in order to find power because I don't know what the limits t and t0 should be.
     

    Attached Files:

  2. jcsd
  3. Sep 18, 2011 #2
    Hi,

    You seem to have calculated the current correctly, so you are on the right track. But to calculate the power, I think you are over-complicating the problem; no integral is necessary. There is a very simple relationship between the power delivered to, and the voltage across a capacitor in a DC circuit.

    However, the energy stored in the capacitor will require an integral (well, strictly speaking anyway). But perhaps once you've worked out the power delivered to the capacitor, you may be able to work out this part.

    Let me know how you go.
     
  4. Sep 18, 2011 #3
    well the method i was using to calculate the power was first finding v=1/c multiplied with the integral of current already found out and then using using the relationship p=vi with the valuse we already have.
     
  5. Sep 18, 2011 #4

    uart

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    That's correct. Note that both the voltage and the current are negative over interval [0..8ms], so the energy absorbed by the capacitor is positive. You can calculate it as the integral of "vi dt" over that interval, but it's much easier to just use [itex]W= \frac{1}{2} C V^2[/itex].
     
  6. Sep 18, 2011 #5
    I know that to calculate energy is W=1/2CV^2, I'm just still stuck on how to get voltage and what limits to put for the integral.
     
  7. Sep 18, 2011 #6

    uart

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    What integral? You've already got [itex]i = c \frac{dv}{dt}[/itex], which you've done graphically. And then you've got P=vi, which is just multiplication. Then you've got W = 1/2 cV^2.

    You've already got the voltage, it was given in the question.

    I'm confused about exactly what you're asking.
     
  8. Sep 18, 2011 #7
    ok i get it now i just didn't realise that you have 2 use the point gradient formula . Thanks alot.
     
  9. Apr 9, 2012 #8
    Lol I'm doing a very similar question on webtutor I'm guessing ur question is from that as well just wanted to say it was really helpful to me thnx :)
     
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