capacitorcircuit analysis
I have a problem that I would like to have help with.
In the image above I need to calculate the capacitance. The values I was given in the beginning was:
So I calculated U3, which was 17.3 V and I have tried using Uc = (1/w*C) * Ic, but with no luck…I've been stuck on this problem way to long. I've been trying to find information on the internet and on youtube, however I didn't quite find anything that could help me. I've tried, but I'm stuck.
Any kind of help would make me glad!:)
If by charges you mean electric charges, then no, a capacitor does not store charges. This is a common misconception, maybe due to the multiple meanings of the word charge. When some charge goes in one terminal of a capacitor, an equal amount of charge leaves the other. So, the total charge in the capacitor is constant.
What capacitors store is energy. Specifically, they store it in an electric field. All the electrons are attracted to all the protons. At equilibrium, there are equal numbers of protons and electrons on each plate of the capacitor, and there is no stored energy, and no voltage across the capacitor.
But, if you connect the capacitor to something like a battery, then some of the electrons will be pulled away from one plate, and an equal number of electrons will be pushed on to the other plate. Now one plate has a net negative charge, and the other has a net positive charge. This results in a difference in electrical potential between the plates, and an increasingly strong electric field as more charges are separated.
The electric field exerts a force on the charges which attempts to return the capacitor back to equilibrium, with balanced charges on each plate. As long as the capacitor remains connected to the battery, this force is balanced by force of the battery, and the imbalance remains.
If the battery is removed, and we leave the circuit open, the charges can't move, so the charge imbalance remains. The field is still applying a force to the charges, but they can't move, like a ball at the top of a hill, or a spring held under tension. The energy stored in the capacitor remains.
If the capacitor terminals are connected with a resistor, then the charges can move, so there is a current. The energy that was stored in the capacitor is converted to heat in the resistor, the voltage decreases, the charges become less imbalanced, and the field weakens.
Further reading: CAPACITOR COMPLAINTS (1996 William J. Beaty)
Best Answer
One way to look at this is as a R-C low pass filter. The rolloff frequency is where the output has half the power of the input. Since power is proportional to the square of the voltage, this means the output voltage is 1/sqrt(2) of the input at the rolloff frequency.
Unfortunately, your problem is at the half voltage point, which is the 1/4 power point. For frequencies well above the rolloff, you can approximate the filter as attenuating 10 dB per decade above the rolloff. However, you are not far from the rolloff point, so that approximation would have some error. Simple RC filters are easy to analyze right at the rolloff, and then a decade in frequency or more either side of that, but not otherwise near the rolloff. That means looking at this problem as a R-C low pass filter isn't going to help simplify things.
This means you have to do things out the long way using complex impedances. What you have is a voltage divider, and normal voltage divider math will work except that the capacitor will have a complex component to its impedance. In fact, the capacitor's impedance is all complex with no real component.
To solve this, do the voltage divider math to find the capacitance at which the magnitude of the output is half that of the input.
As a alternative to complex numbers, you could do this using phasers (magnitude and angle instead of real and imaginary). The result will be the same either way.