Newbie RLC Question

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.

• You know R and Q, and f, so L drops out from the 2nd equation. Then C falls out of the first equation, no? – Spehro Pefhany Mar 19 '15 at 18:14
• Two equations with two unknowns... Really, no google or electrical engineering knowledge is required here.. – Eugene Sh. Mar 19 '15 at 18:16
• I solved L first L = QR / 2πf - Then was able to solve C = 1 / L2πR – zeeman Mar 19 '15 at 20:11

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.

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