I have a problem that I would like to have help with.

enter image description here

In the image above I need to calculate the capacitance. The values I was given in the beginning was:

enter image description here

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!:)

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    \$\begingroup\$ Rule one for science and math: stay away from Youtube. \$\endgroup\$ – JRE Apr 22 '16 at 10:08
  • \$\begingroup\$ Why? I have learned alot from people explaining on Youtube. Some people explain things I need to understand way better than my teacher does actually.. \$\endgroup\$ – Vetenskap Apr 22 '16 at 10:10
  • \$\begingroup\$ Wikipedia can help. Note the section on a low pass filter. It includes all the info you need. \$\endgroup\$ – JRE Apr 22 '16 at 10:11
  • \$\begingroup\$ It took me 1 minute to find the info you need on wikipedia. How much time have you spent trying to find the same info in a video? \$\endgroup\$ – JRE Apr 22 '16 at 10:12
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    \$\begingroup\$ @JRE: Its a modern sickness that even simplest things that could be explained with one image and two sentences are ending up in 5 minute amateurish Youtube videos. Oh such a waste of data storage... \$\endgroup\$ – Rev1.0 Apr 22 '16 at 11:18

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.

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  • \$\begingroup\$ Hi, thank you for answering and explaining. :) If I want to calculate C then I should get it by C = 1/ (f*2*Pi *R) and when putting my values in, I get 5 micro Farad. I want to think that this is right, but it's wrong when I put that as the answer in our datasystem that my teacher provided us. \$\endgroup\$ – Vetenskap Apr 22 '16 at 12:02
  • \$\begingroup\$ I mean 50 micro Farad \$\endgroup\$ – Vetenskap Apr 22 '16 at 12:56
  • \$\begingroup\$ @Rod: I just realized I made a mistake. Give me a few minutes to update the answer. \$\endgroup\$ – Olin Lathrop Apr 22 '16 at 14:02

You can analyze this using the voltage divider equation and using \$\frac{1}{j\omega C}\$ for the impedance of the capacitor.

We get an output voltage \$\frac{V_{out}}{V_{in}} = \frac{1}{1+ j\omega R C}\$

Solve for the magnitude of output voltage ratio, we get \$\frac{\sqrt{1+a^2}}{1 + a^2}\$ where \$ a = \omega R C \$, setting that equal to 0.5 and letting x = a^2, we can solve this- it reduces to a quadratic in x - there is only one positive solution and that is x = 3.

Substituting the numbers back in, you should get C = 91.9uF.

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  • \$\begingroup\$ Thank you so so so much!! I was seriously going insane! thank you! \$\endgroup\$ – Vetenskap Apr 22 '16 at 21:05

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