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.
Best Answer
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.