# A question about the argument part of impedance

The definition of impedance is Z = V/I. I'm trying to understand the idea behind the argument part of the impedance concept in a capacitor.

First here is what I know briefly so far:

I will begin this by assuming the voltage across the capacitor is sinusoidal as:

Vcap = A* sin(w*t)


one can also write:

Vcap = Im{A* e.^(j*w*t)} * (1/j) //complex representation


Current through the capacitor Icap is the derivative of the voltage so that:

Icap = C*dV/dt

Icap = C*A* w*cos(w*t)

Icap = Re{A* e.^(j*w*t)} * C*w //complex representation


So we can therefore write the impedance Zcap as:

Zcap = (1/j*w*C) *  [Im{e.^(j*w*t)}] ./ [Re{e.^(j*w*t)}]


or one can also write the following:

Zcap = Vcap./Icap

Zcap = (1/w*C)*tan(w*t)

tan(w*t) = -j * Im{e.^(j*w*t)}] ./ [Re{e.^(j*w*t)}]


My question is about the impedance's argument part:

Zcap = (1/w*C)*tan(w*t)


or

Zcap = (1/j*w*C) *  [Im{e.^(j*w*t)}] ./ [Re{e.^(j*w*t)}]


Reactance is defined as Xcap = |Zcap| = 1/(w*C)

But to hold the above equation the following must be true:

|tan(w*t)| = 1


or

| -j * Im{e.^(j*w*t)}] ./ [Re{e.^(j*w*t)}] | = 1


How can they be equal to one where they are time dependent?

Is reactance the maximum impedance?

Is there a way to illustrate or explain the argument part of the impedance in this case?

EDIT:

My confusion comes from the following fact:

In the above plot red represents the voltage, blue is the current accross the capacitor. They have 90 degrees phase difference.

And since we defined Z =V/I:

Look at now the green encircled region where voltage becomes max and current becomes zero at the same time. If we freeze the time at that moment, as you see V/I goes to infinity. What is going on here?

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

• A guide to increase readability: math.stackexchange.com/help/notation Jun 14, 2016 at 9:31
• What is Re ? You introduce this term with no expaination. Jun 14, 2016 at 9:31
• @JImDearden Re means the real part of the complex exponential in this context. Im is the imaginary part. Jun 14, 2016 at 9:42
• But for a capacitor (or inductor) there is no real part as the current is 90 degrees out of phase with the voltage. Jun 14, 2016 at 9:46
• I used complex representation to introduce j Jun 14, 2016 at 9:52

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

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

• How can we define the impedance Zcap in time domain? Jun 14, 2016 at 11:09
• 'Impedance' has no meaning in the time domain
– Chu
Jun 14, 2016 at 14:33