Electrical – Impedance phase angle relative to

You can analyze phasor notation in a quasi-2D Cartesian fashion. The real part is the "x", and the complex part is the "y".

So given a phasor magnitude M with angle Theta,

Using trig: \begin{equation} R = M \cos(\theta)\\ X = M \sin(\theta) \end{equation}

We now have the complex impedance R + Xj

To invert, you can multiply by the complex conjugate (R - Xj) to both the numerator and denominator.

\begin{equation} Y = \frac{R - Xj}{(R + Xj)(R - Xj)} = \frac{R - Xj}{R^2 + X^2} \end{equation}

To compute the magnitude of the admittance, use the distance formula:

\begin{equation} M_Y = \sqrt{\left(\frac{R}{R^2 + X^2}\right)^2 + \left(\frac{-X}{R^2 + X^2}\right)^2} \end{equation}

And the phase of the admittance:

\begin{equation} \theta_Y = \tan^{-1}\left(\frac{-X}{R}\right) \end{equation}

Note that tangent is a bit finicky for computing the phasor angle as you have to be careful about the quadrant. If you're using a computer, they often times have an "atan2" function which takes the x and y coordinates directly and computes the CCW angle from the positive X axis.

A closer look at the phase angle mapping, and it looks like the admittance phase angle is just the reflection of the impedance phase angle about the real/X axis.

For example, an impedance phase angle of 45 degrees is equal to an admittance phase angle of -45 degrees.

And this makes sense if I had used some identities above:

\begin{equation} \theta_Y = -\tan^{-1}\left(\frac{X}{R}\right) = -\theta_X \end{equation}

The impedance Z = V/I changing by time, even goes to infinity sometimes; yet we associate the reactance with what??

The impedance \$Z\$ of a circuit element is the ratio of the phasor voltage across and phasor current through, not the ratio of the time domain voltage and current.

So, for example,

$$v_C(t) = A\sin(\omega t) = \Re\{-jAe^{j\omega t}\}$$ $$i_C(t) = \omega C\; A\cos(\omega t) = \Re\{\omega C\;Ae^{j\omega t}\}$$

The phasor representation of the voltage and current are just the (constant) complex numbers multiplying \$e^{j \omega t}\$ thus, the phasor voltage and currents are

$$V_c = -jA$$ $$I_c = \omega C A $$

The impedance of the capacitor is then

$$Z_c = \frac{V_c}{I_c} = \frac{-jA}{\omega C A} = \frac{1}{j\omega C}$$

and so the reactance, the imaginary part of the impedance, of the capacitor is

$$X_c = -\frac{1}{\omega C} $$