RC-circuit -- Need help understanding why I get the wrong answer

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  • Thread starter Jonas E
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In summary, the correct formula for finding the time required to dissipate a certain percentage of initially stored energy in an RC circuit is to set the dissipated energy in the resistor equal to that percentage multiplied by the initial energy stored in the capacitor. It is important to use the correct formula for energy dissipated in a resistor (V^2/R) and to set the equation equal to the desired percentage of initial energy stored in the capacitor, not the remaining energy in the capacitor.
  • #1
Jonas E
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Homework Statement


I have an RC circuit where C = 0.4 * 10^-6, R = 50k Ohms. The problem says: "Find the length of time required to dissipate 75% of the initially stored energy."

Homework Equations

energy 1: 1/2 * C * v(t)
energy 2) w = ∫p(t)dx, from 0 to t

The Attempt at a Solution


This is the last part of the problem. I solved everything else up to this point correctly. v(0) = 200 V, Tau = 0.02s. The initial energy stored in the capacitor is 8mJ. Now I set p(t) = v(t) / R, and integrate from 0 to t. That gives me the expression for the energy, which I then set equal to 0.25 * w(0) = 2 * 10^-3 J. However this gives me that t = 2.877ms, which is wrong according to the solutions manual. It uses the first energy formula and gets t = 13.86ms. Can someone explain why these two formulas give two different answers, or point out any mistake I have done.
 
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  • #2
The energy dissipated in a resistor is V^2/R, not V/R.
 
  • #3
Your equation for energy stored in a capacitor doesn't look right.
 
  • #4
Jonas E said:
energy 1: 1/2 * C * v(t)
This is incorrect.
 
  • #5
Sorry, it should say v^2 in the energy formula, but that doesn't change anything since I used the other formula anyway. My problem is that the 1st formula (which I mistyped) gives a different answer than the second one (which is the one I used).
 
  • #6
If you have dissipated 75% of the energy in the capacitor, then the dissipation in the resistor should be that 75% energy.
 
  • #7
Jonas E said:
Now I set p(t) = v(t) / R, and integrate from 0 to t.
That formula is also wrong. Did you read and understand post #2?
 
  • #8
JonasE:

We're trying to help, but I think we are all lost as to what you did, which formulas you used and which ones you mis-typed. Can you correct the two errors in your original post:

P = V^2 / R, not V/R
E = C*V^2/2, not C*V/2

and tell us if this fixes your issue or not? If not, what are you getting now?
 
  • #9
Don't know how to edit the original post, so I'll post the edit here:

I set p(t) = V(t)^2 / R and integrate from 0 to t. This gives me the following expression: w(t) = -1/125 * (e^(-100t) - 1), which I set equal to 0.25 * 8 * 10^-3 in order to find the time. (0.25 because I need to find how long it takes to dissipate 75% of the initial energy). Solving that gives me t = ln(3/4) / (-100) = 2.877 ms. According to the solutions manual, this is wrong.
 
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  • #10
OK, good. What you are doing makes sense. The problem is what NascentOxygen said in Post #6. w(t) is the energy dissipated in the resistor, not the energy stored in the capacitor. So you want to set this equal to 0.75*E0, not 0.25*E0. Then both methods give the 13.86 ms. Do you see?

Also, these problems are much easier if you do the algebra with symbols, and only put in the numbers at the very end. You would then see that it doesn't matter what V(0) is, for example.
 
  • #11
phyzguy said:
OK, good. What you are doing makes sense. The problem is what NascentOxygen said in Post #6. w(t) is the energy dissipated in the resistor, not the energy stored in the capacitor. So you want to set this equal to 0.75*E0, not 0.25*E0. Then both methods give the 13.86 ms. Do you see?

Also, these problems are much easier if you do the algebra with symbols, and only put in the numbers at the very end. You would then see that it doesn't matter what V(0) is, for example.
I understand now, thanks a lot for clearing it up!
 

1. How does an RC circuit work?

An RC circuit, also known as a resistor-capacitor circuit, is a type of electronic circuit that consists of a resistor and a capacitor connected in series. The resistor limits the flow of electric current, while the capacitor stores and releases electrical energy. When a voltage is applied to the circuit, the capacitor charges up until it reaches its maximum capacity, at which point the current stops flowing.

2. What is the formula for calculating the time constant in an RC circuit?

The time constant, denoted by the symbol τ (tau), is calculated by multiplying the resistance (R) in ohms by the capacitance (C) in farads. This can be represented by the formula τ = RC. The time constant represents the time it takes for the capacitor to charge to approximately 63% of its maximum capacity.

3. Why am I getting the wrong answer in my RC circuit calculations?

There could be several reasons for getting the wrong answer in your RC circuit calculations. Some possible reasons include using incorrect values for the resistor or capacitor, not considering the tolerance of the components, or not accounting for factors such as parasitic capacitance or resistance. It is important to double-check your calculations and make sure all factors are taken into account.

4. How does the time constant affect the behavior of an RC circuit?

The time constant determines the speed at which the capacitor charges and discharges in an RC circuit. A smaller time constant means the capacitor will charge and discharge faster, while a larger time constant means it will charge and discharge slower. This affects the overall behavior of the circuit, including the voltage and current levels.

5. Can I use an RC circuit for filtering purposes?

Yes, an RC circuit can be used as a low-pass filter to remove high-frequency components from a signal. The cutoff frequency of the filter can be controlled by adjusting the values of the resistor and capacitor. This makes RC circuits useful in applications such as audio amplifiers, power supplies, and electronic filters.

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