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

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Homework Help Overview

The discussion revolves around a circuit containing three capacitors in series and a resistor, where participants are tasked with finding the current at the moment when the capacitors have lost 80% of their initial stored energy. The problem involves understanding the relationship between charge and energy in capacitors.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between charge loss and energy loss, questioning whether losing 80% of charge equates to losing 80% of energy. They discuss the equations for energy in capacitors and the implications of their calculations.

Discussion Status

The conversation is active, with participants examining the calculations related to energy loss and its connection to charge loss. There is a recognition of the need to clarify the timing of energy loss versus charge loss, and some guidance is offered on how to interpret the results of their calculations.

Contextual Notes

Participants are navigating the complexities of circuit analysis, particularly in the context of energy and charge in capacitors. There is an acknowledgment of the challenge posed by the problem's requirements and the potential for confusion between different loss percentages.

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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??
 
Last edited:
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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?
 
Matterwave said:
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 let's 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 ...
 
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?
 
"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...
 
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.
 
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??
 
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?
 
We are actually looking for 0.2E_0, right??
 
  • #10
Yes. So what must Q be, in order to end up with 0.2 Eo?
 

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