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

In summary, the conversation revolved around finding the current in a circuit with three capacitors and a resistor in series, given their values and an initial charge. The equations t = -RC ln(q/Q0) and I = -Qo/RC e^(-t/RC) were discussed, as well as the relationship between charge and energy in a capacitor. The question of whether losing 80% of charge equals losing 80% of energy was raised, and it was determined that there is not a 1 to 1 linear correspondence between the two. The equation for energy in a capacitor, U = 1/2 Q^2/C, was also mentioned. The conversation concluded with a discussion on how to find the current
  • #1
smashyash
28
0

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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  • #2
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
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 ...
 
  • #4
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
"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
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
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
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
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?
 

What is the formula for finding the current at the instant when capacitors have lost 80% of initial energy?

The formula for finding the current at the instant when capacitors have lost 80% of initial energy is I = (Q/Q0) * (E0/R) * e-t/RC, where I is the current, Q is the charge, Q0 is the initial charge, E0 is the initial energy, R is the resistance, t is the time, and C is the capacitance.

How do you calculate the initial energy of a capacitor?

The initial energy of a capacitor can be calculated using the formula E0 = (1/2) * Q0 * V0, where E0 is the initial energy, Q0 is the initial charge, and V0 is the initial voltage.

What is the significance of capacitors losing 80% of their initial energy?

Capacitors losing 80% of their initial energy means that they have reached a state of approximately 80% discharge. This can be an important consideration in circuits and systems that rely on capacitors to store and release energy.

How do you determine the time constant in the formula for finding current?

The time constant (RC) can be determined by dividing the resistance (R) by the capacitance (C). It represents the rate at which a capacitor discharges.

What factors can affect the accuracy of calculating the current at the instant when capacitors have lost 80% of initial energy?

The accuracy of calculating the current at the instant when capacitors have lost 80% of initial energy can be affected by factors such as the accuracy of the initial energy and charge measurements, variations in the resistance and capacitance values, and external factors such as temperature and humidity.

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