LC/Resonant circuit - the maximum capacitor energy

[fawk3s]

Homework Statement

The effective voltage in an LC circuit is 200V while the capacitance of the capacitor in the circuit is 20pF. What is the maximum capacitor energy?

Homework Equations

We = CU^2/2

Uefective = U / sqrt(2)

The Attempt at a Solution

Well it seems a fairly simple problem. It solves by just calculating the energy by putting the given information in the energy equation above, and the answer is supposed to be 4*10^-7 J.
However, this is not the way I solved it. As it states in the problem, the effective voltage of the system is supposed to be 200V, as there is AC going through it. So shouldn't we calculate the peak voltage first, which is 200*sqrt(2) and use it to calculate the maximum energy of the capacitor, which would be when the capacitor is fully charged and there is no current goingh through the system? By doing this we get the answer to be 8*10^-7 J.

So where does my logic fail?

Thanks in advance,
fawk3s

 

[gneill]

I suspect that the author of the problem made a poor choice in using the word "effective" here, and didn't realize that it could be taken to have the connotation of RMS value.

 

[fawk3s]

English is not my native language, but I am pretty certain he intended to use these words, as in the effective value of the AC voltage. Now does my logic fail somewhere or did the author make a mistake (which is highly unlikely)?
Because by my logic, the maximum voltage on the charged capacitor (when there is currently no current in the system) should be equal to the peak voltage of the AC voltage, not the effective value of the AC voltage.

Thanks in advance,
fawk3s

 

[gneill]

The energy stored in a capacitor is

$$\xi = \frac{1}{2} C\; V^2$$

Where V is the voltage on the capacitor (Joules/Coulomb). There's no quibbling about that. Since there's no concept of "effective energy", the answer should use the actual instantaneous voltage on the capacitor to calculate the stored energy.

So either the author's choice of word was poor, or his answer is wrong. Your choice!

Edit: Maybe I spoke to soon. After thinking about it, it might not be unreasonable to speak of the RMS energy handled by the capacitor in some context. But still, this would not be the maximum capacitor energy.

 

[rwisz]

I would say that your logic is not wrong, but it is not the most efficient way to solve this problem. The effective voltage, as you mentioned, is the peak voltage divided by the square root of 2. This is because in an AC circuit, the voltage is constantly changing and the effective voltage is the equivalent DC voltage that would produce the same power. So, using the effective voltage in the energy equation is the correct approach.

However, in this problem, the effective voltage is already given to be 200V. This means that the peak voltage is 200V multiplied by the square root of 2, which is approximately 283V. This is the maximum voltage that the capacitor will reach in the circuit. So, using this value in the energy equation will give you the correct answer of 4*10^-7 J.

Your approach of first finding the peak voltage and then using it to calculate the maximum energy is also correct, but it involves an extra step and may lead to more chances of error. It is always better to use the given information directly in the equation.

In conclusion, both approaches are correct, but using the given effective voltage directly in the energy equation is the more efficient and accurate way to solve this problem.