Electronic – Solving for the impedance and phase shift in a series-parallel AC RCL Circuit with Complex Numbers

accircuit analysisimpedance

Having a lot of trouble working through an AC circuit, asking for some help.

The circuit is relatively simple – I have a capacitor of capacitance \$C\$ and resistor of resistance \$R\$ hooked in parallel, which is hooked in series with a coil loop of impedance \$L\$ to an alternating voltage source of \$e_0\sin(\omega t)\$.

I'll work through the steps to show you where I'm having my problems.

So, first, solving the parallel combination. $$ \frac{1}{Z_t} = \frac{1}{Z_c} + \frac{1}{Z_r} $$

and we know that \$Z_c = \frac{-i}{\omega C} \$ and \$Z_r = R \$

So $$ \frac{1}{Z_t} = \frac{1}{R} – \frac{\omega C}{i} = \frac{1}{R} + \omega Ci $$

and thus $$ Z_t = \frac{R}{1+i \omega CR} $$

Now, to find the total impedance, \$ Z = Z_t + Z_l \$

$$ Z = \frac{R}{1+i \omega cR} + i \omega L = \frac{R(1- \omega^2 LCR) + i \omega L}{1+i \omega CR}$$

Is that right?

Additionally, the next problem I then have is the phase shift. By definition, the phase shift of a complex circuit is $$ \phi = \arctan\left(\frac{\text{Imaginary}}{\text{Real}}\right)$$

How the heck am I supposed to get an imaginary and a real out of that equation? What parts are imaginary and what are real?

Best Answer

Assuming this equation is correct: $$Z = \frac{R}{1+i \omega cR} + i \omega L$$

We multiply the first term's nominator and the denominator by the complex conjugate of the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+i \omega cR)(1-i\omega c R)} + i \omega L$$ and thus getting rid of imaginary stuff in the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+ \omega^2 c^2R^2)} + i \omega L$$ And this thing is easily separated to real and imaginary parts: $$.. = \frac{R}{1+ \omega^2 c^2R^2} + i \left(\omega L - \frac{\omega c R^2}{1+ \omega^2 c^2R^2} \right)$$

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