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Hoping someone can help. I have been studying RLC series circuits. The first set of questions gave the value of resistor, inductor and capacitor, which were calculated as follows:

To calculate resonant frequency:

$$ fr = \frac{1}{2 \pi \sqrt{LC}} $$

$$ fr = 1200Hz $$

To calculate Q:

$$ Q = \frac{2 \pi fL}{R} $$

$$ Q = 80 $$

To calculate Bandwidth:

$$ BW = \frac{fr}{Q} $$ $$ BW = 15Hz $$

The question set then goes on to show another series RLC. It gives the value of the resistor, the resonant frequency, and the value of \$Q\$.

I have tried re-arranging the equations but can only find the bandwidth and have no idea how to find the inductor and capacitor values.

I have searched Google for answers but not turned up any equation to find \$L\$ and \$C\$ values.

Any help much appreciated.

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  • 2
    \$\begingroup\$ You know R and Q, and f, so L drops out from the 2nd equation. Then C falls out of the first equation, no? \$\endgroup\$ – Spehro Pefhany Mar 19 '15 at 18:14
  • \$\begingroup\$ Two equations with two unknowns... Really, no google or electrical engineering knowledge is required here.. \$\endgroup\$ – Eugene Sh. Mar 19 '15 at 18:16
  • \$\begingroup\$ I solved L first L = QR / 2πf - Then was able to solve C = 1 / L2πR \$\endgroup\$ – zeeman Mar 19 '15 at 20:11
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Ok, these are the formulas that you provided. Props to @Ricardo for making them look pretty.

$$ \color{red}{fr} = \frac{1}{2 \pi \sqrt{\color{red}{L}\color{red}{C}}} $$

$$ \color{red}{Q} = \frac{2 \pi \color{red}{fr}\color{red}{L}}{\color{red}{R}} $$

$$ \color{red}{BW} = \frac{\color{red}{fr}}{\color{red}{Q}} $$

Too bad, pretty much everything in \$\color{red}{red}\$ is unknown. (admittedly, \$ \pi \$ is never fully known, but let's not get carried away by such details)

What you say? You actually know a few \$\color{green}{green}\$ things, namely \$\color{green}{R}\$, \$\color{green}{fr}\$ and \$\color{green}{Q}\$?

Go ahead and put them in the equations: $$ \color{green}{fr} = \frac{1}{2 \pi \sqrt{\color{red}{L}\color{red}{C}}} $$

$$ \color{green}{Q} = \frac{2 \pi \color{green}{fr}\color{red}{L}}{\color{green}{R}} $$

$$ \color{red}{BW} = \frac{\color{green}{fr}}{\color{green}{Q}} $$

The maths people say that you can play the math game now. The goal is to turn all the \$\color{red}{red}\$ things that you care about into \$\color{green}{green}\$ things.

The game has only one rule: Whenever there's only one \$\color{red}{red}\$ thing left in an equation, that becomes a \$\color{green}{green}\$ thing.

As you said and the third equation shows, \$\color{green}{BW}\$ is not an unknown anymore. So far so good, leaving you with:

$$ \color{green}{fr} = \frac{1}{2 \pi \sqrt{\color{red}{L}\color{red}{C}}} $$

$$ \color{green}{Q} = \frac{2 \pi \color{green}{fr}\color{red}{L}}{\color{green}{R}} $$

$$ \color{green}{BW} = \frac{\color{green}{fr}}{\color{green}{Q}} $$

The second equation turns \$\color{red}{L}\$ into \$\color{green}{L}\$, because it is the only \$\color{red}{red}\$ thing left, which results in:

$$ \color{green}{fr} = \frac{1}{2 \pi \sqrt{\color{green}{L}\color{red}{C}}} $$

$$ \color{green}{Q} = \frac{2 \pi \color{green}{fr}\color{green}{L}}{\color{green}{R}} $$

$$ \color{green}{BW} = \frac{\color{green}{fr}}{\color{green}{Q}} $$ What? \$\color{red}{C}\$ is evolving! * plays 8 bit music *

Congratulations! Your \$\color{red}{C}\$ evolved into \$\color{green}{C}\$!

$$ \color{green}{fr} = \frac{1}{2 \pi \sqrt{\color{green}{L}\color{green}{C}}} $$

$$ \color{green}{Q} = \frac{2 \pi \color{green}{fr}\color{green}{L}}{\color{green}{R}} $$

$$ \color{green}{BW} = \frac{\color{green}{fr}}{\color{green}{Q}} $$

Now you have all the \$\color{green}{green}\$ things. That's good. If you need further instructions on how to solve the equations, please comment on this answer.

I hope that helps.

But be quick, people will downvote this answer to oblivion because I did not go for the \$\color{red}{C}\$ evolving into \$\color{green}{C++}\$ pun.

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  • \$\begingroup\$ Thank you very much for this. Please can you give me further instruction on how to solve equations. \$\endgroup\$ – zeeman Mar 19 '15 at 22:11

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