Capacitor and inductor in series - Page 1
Forum section: electronics / beginners

author: jharig23
author_rank: Newbie
timestamp: February 08, 2014, 04:31:35 pm
content: I'm working on a homework problem. If I have a 1F capacitor in series with a 1H inductor, how can I compute the transfer function across the capacitor? I tried voltage divider, but get divide by zero. Is there an algebraic trick I can use?Thanks!

author: Rudane
author_rank: Regular Contributor
timestamp: February 08, 2014, 07:49:26 pm
content: Is this the circuit?

author: denelec
author_rank: Regular Contributor
timestamp: February 08, 2014, 09:04:33 pm
content: When the frequency reach the resonant frequency, the impedance of the capacitor and the inductor cancel each other.That where the divide by zero come from.http://en.wikipedia.org/wiki/LC_circuit

author: Rudane
author_rank: Regular Contributor
timestamp: February 08, 2014, 09:29:47 pm
content: For a series LC circuit, at the resonant frequency the imaginary part of the impedances cancels, and the circuit appears real only. The parallel circuit would appear as an infinite impedance.

author: Rerouter
author_rank: Super Contributor
timestamp: February 08, 2014, 09:37:59 pm
content: add a few hundred micro ohms to the circuit as a parasitic and try it again,

author: Marco
author_rank: Super Contributor
timestamp: February 08, 2014, 10:11:38 pm
content: Quote from: Rudane on February 08, 2014, 07:49:26 pmI kinda doubt he got far enough to learn Laplace yet ...Jharig, what you have to do is go back to your previous lessons and dig out what you need to solve it. It's almost the same arithmetic as a resistor divider, you just need to use the complex representation of the L&C impedances and some basic simplification.

author: jharig23
author_rank: Newbie
timestamp: February 08, 2014, 10:30:49 pm
content: Here's all of the information. It's a two part question.Part 1 is about solving an unknown circuit for Thevinen equivelance. Open Circuit measurement is 'sin(t)'. Circuit shunted with a 1ohm resistor gives '1/sqrt(2) * sin(t + pi/4)'I believe that VTH = sin(t), since that is the open-circuited voltage.Finding ZTh, I converted 1/sqrt(2) * sin(t + pi/4) to cos form: 1/sqrt(2) * cos(t - pi/2)Then to complex imaginary: VOUT = 1/sqrt(2) * e^(J*(t - pi/2))Then wrote the equation:VOUT == VIN*(ZTh/(Zth+R)Pluging everything in, and solving for ZTh, I got ZTh = JPart 2, replace the resistor with a Capacitor, C = 1 F.I solved for the frequency as: 2*pi*f = 1 (since Vth = sin(t)), so f = 1/(2*pi)ZC = 1/(I*2*pi*f*C) simplifies to ZC = 1/(I*C)Now using voltage divider, I getVOUT = VIN * (ZC / (ZTh + ZC))Simlifies to VIN * ( 1/(I*C) / (I + 1/(I*C)))Which is where I got division by 0, since ZTh and ZC cancel eachother out.

author: jharig23
author_rank: Newbie
timestamp: February 08, 2014, 10:33:57 pm
content: I parameterized ZC(C) = 1/(J * C), and plotted ZC / (Zth + ZC) over [0, 1) U (1, 2), and got a graph of 1 for all C.

author: jharig23
author_rank: Newbie
timestamp: February 08, 2014, 10:34:50 pm
content: BTW, what's the best way to post equations on here?

author: Rudane
author_rank: Regular Contributor
timestamp: February 09, 2014, 01:27:28 am
content: What is the capital I in your equations? Current? The imaginary number little i?

author: Rudane
author_rank: Regular Contributor
timestamp: February 09, 2014, 01:31:26 am
content: The only thing I can see is maybe you are plugging a negative into the capacitive reactance there and you shouldn't be. I may also be reading this wrong, but wouldn't Zth change with the addition of the capacitor?