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- Homework Statement
- For the circuit shown, say ##V(t)=V_0sin\omega t##. Calculate the current in the resistor

- Relevant Equations
- V=ZI

Complex impedances: ##Z_R=R,\ Z_L=j\omega L,\ Z_C=\frac{1}{j\omega C}##

Hello guys,

LRC circuits with an AC source are having the best over me... had some confusion in class with respect to which method is best using(phasors diagram, reactances or complex impedances) which I am trying to desperately sort out before my exam; here I will show you my best attempt on this one, which although is missing on something!

Applying Ohm's law ##I(t)=\frac{V(t)}{Z_{eq}}=\frac{e^{j(90-\omega t)}}{Z_{eq}}##, and so the need to compute ##Z##.

Considering the schematics of the circuit, we can just sum up the various impedances accordingly to get to the equivalent one:

$$Z_{eq}=Z_R+(\frac{1}{Z_L}+\frac{1}{Z_L})^{-1}=R+(j(\omega C-\frac{1}{\omega L}))^{-1}=

R+j\cdot \frac{1}{\frac{1}{\omega L}-\omega C}$$

Substituting the result above and manipulating the equation:

$$I(t)=\frac{V_0}{\sqrt{R^2+\frac{1}{\frac{1}{\omega L}-\omega C}^2}}\cdot \frac{e^{j(90-\omega t)}}{e^{j\phi}}=\frac{V_0}{\sqrt{R^2+\frac{1}{\frac{1}{\omega L}-\omega C}^2}}\cdot e^{j(90-\omega t-\phi)}=\frac{V_0}{\sqrt{R^2+\frac{1}{\frac{1}{\omega L}-\omega C}^2}}\cdot sin(\omega t +\phi)$$

All we are left with is, at this point, to compute the phase angle ##\phi## which originated from taking the module of Z; for this, what I would do is: ##tan\phi = \frac{Im(Z)}{Re(Z)}=\frac{L}{RC(\frac{1}{\omega C} - \omega L)}## and then write the phase angle as the arctangent of such expression.

The correct result for ##\phi## is, though, ##\phi=atan\frac{L}{RC(-\frac{1}{\omega C} + \omega L)}##, which is what I get but multiplied by '-1'.

This really bothers me, because I am pretty sure to have not made any calculations mistakes and signs are a problem I've been getting quite often when dealing with such problems, which are still somewhat confused in my mind... this makes me think I am missing out on something more important wrt to AC circuits