6 part 2 loop RLC circuit complex impedance

[froldy]

Homework Statement

I'm trying to determine this insane circuit's |Z| so I can use an AC voltage to find the current as a function of time.

This is my circuit (the switch is closed, and ignore the values): http://i.imgur.com/mA55I.png

Homework Equations

Impedances for each of the RLC components

The Attempt at a Solution

Here, I make an equation for the total impedance, then tried to separate real and imaginary parts. First is the stuff on the bottom, and then the separation is on the top. The stuff in the big parentheses is the imaginary part. http://i.imgur.com/KArxr.jpg

So have I done it right? Any tips to make multiplying that behemoth by its complex conjugate any easier? I'm quite worried I did some fundamental thing wrong. Thanks!

 

[gneill]

Your life may be made considerably easier if you happen to know the AC voltage source's frequency. Then you can manipulate the numerical values of the resistances and reactances and the algebra won't spiral out of control

 

[froldy]

Your life may be made considerably easier if you happen to know the AC voltage source's frequency. Then you can manipulate the numerical values of the resistances and reactances and the algebra won't spiral out of control

Arg, I would but I need it in terms of variables :(

 

[gneill]

Arg, I would but I need it in terms of variables :(

Argh. You have my sympathies.

Okay, I'll make a few suggestion if I may.

First, let c1 = 1/C1, c2 = 1/C2. Also let s = jω so that s² = -ω². This makes Z_C1 = -s*c1 and Z_C2 = -s*c2.

Next, find the impedance of the subcircuit that excludes the 100Ω resistor and the 330μH inductor; Since they are in series with the subcircuit, the final impedance will just have the additional real (100Ω) and imaginary (jω330H) components tacked on in the end. You may find that determining real and imaginary parts of the subcircuit to be a less daunting task than taking the whole thing in one gulp.

 

[froldy]

Argh. You have my sympathies.

Okay, I'll make a few suggestion if I may.

First, let c1 = 1/C1, c2 = 1/C2. Also let s = jω so that s² = -ω². This makes Z_C1 = -s*c1 and Z_C2 = -s*c2.

Next, find the impedance of the subcircuit that excludes the 100Ω resistor and the 330μH inductor; Since they are in series with the subcircuit, the final impedance will just have the additional real (100Ω) and imaginary (jω330H) components tacked on in the end. You may find that determining real and imaginary parts of the subcircuit to be a less daunting task than taking the whole thing in one gulp.

Ok, I guess that's my only option... I still have to calculate sqrt(ZZ*) though, and then use a varying V expression to solve for current, and then use the current to find the V across each component as a function of time. Is there any easier way to do this, like with phasors or something? I have no clue how I'm going get each of the components out of that square root without spending like a month on it...

Thanks!

 

[gneill]

You could let each component have a symbolic impedance, say z1, z2,...z6, then write the KVL equations for the two loops and solve for the loop currents in terms of an assumed input voltage v and the z's. That will put you on track to find the individual component voltages since you'll have both of the loop currents.

Once you plug in the component impedances for the z's you'll still have the task of separating out real and imaginary bits, but at least the denominators will consist of a sum of products of the z's taken pairwise, so they'll be easy to sort out in terms of real or imaginary.

I don't think you'll be facing a square root. When you clear the denominator of an imaginary, you get something like:

[tex] \frac{c + id}{a + ib}*\frac{a - ib}{a - ib} = \frac{ac - bd +i(cb + ad)}{a^2 + b^2} [/tex]

 

[froldy]

Ok, that makes sense. Thanks a lot, gneill!