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Finding current at instant when capacitors have lost 80% of initial energy?

  • Thread starter smashyash
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  • #1
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Homework Statement



I have three capacitors and a resistor in a closed circuit. I'm given the values of C: 15,20,10pF. They are all in series. The magnitude of each is 3.5nC.

Homework Equations



t = -RC ln(q/Q0)
I = -Qo/RC e^(-t/RC)

The Attempt at a Solution



i've tried finding time first and it's not working...

I've used a q value of Q0 * 0.2.

Everything else is just plug and chug essentially. Is there a conversion that I'm missing maybe??
 
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Answers and Replies

  • #2
Matterwave
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So, does losing 80% of charge equal losing 80% of energy? Phrased another way, is there a 1 to 1 linear correspondence between charge and energy? What is the equation for energy of a capacitor?
 
  • #3
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So, does losing 80% of charge equal losing 80% of energy? Phrased another way, is there a 1 to 1 linear correspondence between charge and energy? What is the equation for energy of a capacitor?
Well lets see...

I know that C = Q/V..

and the energy of a capacitor can be written as U = 1/2 QDV = 1/2 C(DV)2 = 1/2 Q2/C ...
 
  • #4
Matterwave
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Yes so [tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex].

So notice:

[tex]E_0=\frac{1}{2}\frac{Q_0^2}{C} \rightarrow E(Q=.2Q_0)=\frac{1}{2}\frac{(.2Q_0)^2}{C}=.04(\frac{1}{2}\frac{(Q_0)^2}{C})=.04E_0[/tex]

So reducing the charge by 80% has actually reduced my energy by 96%...Do we want the current after 96% of the energy has been lost or after 80% of the energy has been lost?
 
  • #5
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"what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy? " -homework

so you would need the time at which the capacitors has lost 80% of initial energy...
 
  • #6
Matterwave
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Ok...

So...did you find the time for that?

The whole point of my previous 2 posts was just to show you that the time till 80% of energy is lost, as was asked by the question, is NOT what you found. What you found was the time for 80% of the charge lost.
 
  • #7
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I know. I've gotten that point since your first response! But this whole circuit thing is very new to me and I have no idea how to incorporate energy. I'm not looking for you to tell me exactly how to do this.. I'm just trying to understand the idea! Is there a way to use that number that you found, 80% lost charge-96% lost energy, to figure out the percent charge loss to get the 80% energy lost??
 
  • #8
Redbelly98
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Look at the calculation in Post #4 again. It ended up with 0.04Eo. Instead, what do we want it to end up with, if 80% of the initial energy is lost?
 
  • #9
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We are actually looking for 0.2E_0, right??
 
  • #10
Redbelly98
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Yes. So what must Q be, in order to end up with 0.2 Eo?
 

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