# AC RCL circuit

1. Apr 9, 2014

### TheEvenfall

Hello there, I'm not sure if my solution is correct for $\hat{I}_{C}$

1. The problem statement, all variables and given/known data

In the given circuit, calculate the current in each circuit element given that V = V$_{o}$sin(ωt)
R, L and C are given.
http://imgur.com/yO3flg8

2. Relevant equations
Z = R + jX (j$^{2}$ = -1)
X$_{L}$ = jωL
X$_{C}$ = $\frac{-j}{ωC}$
$\hat{V}$ = V$_{o}$e$^{jωt}$
$\hat{I}$ = I$_{o}$e$^{j(ωt-ø)}$
I$_{o}$ = $\frac{V_{}}{|Z|}$
tan(ø) = $\frac{\Im(Z)}{\Re(Z)}$

3. The attempt at a solution
First for the impedance, $1/Z_{XL} = 1/X_{L} + 1/X_{C}$
$Z_{XL} = j\frac{ωL}{1- ω^{2}CL}$
$Z = R + Z_{XL} = R + j\frac{ωL}{1- ω^{2}CL}$
The current phasor in the resistor R: $\hat{I_{R}}$= $\hat{I}$= $\hat{I_{C}} + \hat{I_{L}}$
$\hat{V_{C}} = \hat{V} - \hat{V_{R}}$
$\hat{I_{C}} = (\hat{V} - \hat{V_{R}} )/X_{C}$
$\hat{I_{C}} = (V_{o}e^{jωt} - I_{o}e^{j(ωt-ø)})(jωc) = V_{o}ωCe^{jwt}(1-\frac{R}{|Z|}e^{-jø})(j)$
$\hat{I_{C}} = V_{o}ωCe^{j(ωt+\pi/2)}(1-\frac{R}{|Z|}e^{-jø})$ ø and |Z| are known.
And$I_{L}$ can be found the same way.
I'm not entirely sure my solution is correct. Also, since $I_{R}$ is always in phase with V, does that mean that ø is 0? If so, then tan(ø) is also 0 but that would mean that either ω or L are 0...

Note: sorry if it seems slobby and for the skipped steps, my exam is in less than 3 hours and I'm really nervous and running out of time.

Last edited: Apr 9, 2014
2. Apr 9, 2014

### BvU

Good luck with your exam! Don't worry too much about this last exercise. The assumption Ir is in phase with V seems incorrect to me.

3. Apr 9, 2014

### BiGyElLoWhAt

well i'm still working through it, but one thing I want to point out is your source voltage is wrong. e^iwc =cos(wc) +isin(wc)
if you want to use eulers id. you want (e^iwc - e^-iwc)/2 , but that seems unnecessary, i'd just keep it v_s or something until you have your final answer, then sub in V(not)sin(wt). also it looks like complex analysis only? no time dependent ? just wondering.