Why is the inductive reactance or capacitive reactance phasor on the imaginary axis?

In a resistive element, current and voltage are in phase with each other. However for an inductive element the voltage leads the current by \$90^\circ\$, and for a capacitive element voltage lags current by \$90^\circ\$.

So lets look at how we define impedance and why. We define impedance as:

$$Z = R + jX$$

Now an impedance respects Ohms law, so what we are saying is:

$$V = ZI=RI+jXI$$

When the reactance is zero, you can see we are left happily with the Ohms law we all know and love:

$$\begin{align} V_r&=RI+j0I\\\\ V_r&=IR\\ \end{align}$$

So that works. Now what about when resistance is zero. We get:

$$\begin{align} V_x&=0I+jXI=jXI\\\\ V_x&=|X|\angle90^\circ\times I\\ \end{align}$$

We can see now that the current and voltage must be \$90^\circ\$ out of phase in order to satisfy this equation. Great, that is what we needed too. So basically this formation of impedance matches what we require.

So lets look at what you said in a comment to @Barry. Why not define impedance as:

$$Z = X + jR$$

Well, lets go through the derivations again. From Ohms law:

$$V= ZI = XI + jRI$$

So, lets first look at what happens when reactance is zero:

$$\begin{align} V_r = 0I + jRI = jRI\\\\ V_r = R\angle90^\circ\times I \ne IR\\ \end{align}$$

Now we have a big problem. We have just said that current and voltage must be out of phase by \$90^\circ\$. But as we well know this is not the case. So clearly the impedance equation cannot correctly be expressed in this form.

If you want to put the resistive part on the imaginary axis, you simply rotate both the voltage and current by 90 degrees. You don't however change the impedance equation.

Ohms law in effect becomes:

$$jV = jIZ$$

Substituting in the correct impedance equation we get:

$$jV = jI(R + jX) = jIR - IX$$

This is now perfectly valid. The resistance remains a real number meaning that voltage and current remain in phase - we see this by again setting the reactance to 0, resulting in:

$$jV=jIR \rightarrow V=IR$$

In fact this shift doesn't have to be by 90 degrees - you can shift the Ohms law equation by any arbitrary angle and it still holds true:

$$V\angle35^\circ=(I\angle35^\circ\times R) \rightarrow V=IR$$


  • First of all I suggest that you review what a phasor actually is. Reactances (or impedances in general) are never phasors. Just because they are complex quantities doesn't make them phasors.

    Now what is a phasor? A phasor is a complex quantity that has a \$e^{j\omega t}\$ term cancelled as a result of the phasor transformation, e.g. a voltage \$V(t) = V e^{j(\phi + \omega t)} = V e^{j\phi} e^{j\omega t}\$ → \$V e^{j\phi} = V_{re} + j V_{im}\$.
    The "→" is the phasor transformation that drops the \$e^{j\omega t}\$ resulting in a complex quantity that is time independent (which makes further handling easier).

    Although impedances \$Z\$ may be complex quatities they are not the result of a phasor transformation and therfore are not phasors.

  • Now back to your original question that I assume should just be
    "Why are the inductive impedances or capacitive impedances imaginary?".

    Answer: That is just the result when you expose
    an inductor (with voltage/current relationship \$V(t) = L \frac{d}{dt}I(t)\$) or
    a capacitor (with voltage/current relationship \$V(t) = \frac{1}{C}\int I(t) dt\$)
    to a sinusoidal source (i.e. voltage source of the form \$V(t) = V_{src} e^{j(\phi_{src} + \omega t)}\$ or similar current source) and apply KCL and/or KVL.

    Only then you can apply phasor analysis and miraculously all the \$e^{j\omega t}\$ terms can be canceled and the terms containing \$L\$ and \$C\$ "automatically" become pure imaginary constants \$Z_L = jX_L = j\omega L\$ because \$L\frac{d}{dt}Ie^{j\omega t} = j\omega LIe^{j\omega t}\$ i.e. the operator \$L\frac{d}{dt}\$ is the same as multiplication by \$j\omega L\$ (and similarly for capacitances).

    That way you get a simple time independend equation for voltage and current \$V = Z I\$ (where \$V\$ and \$I\$ are phasors, and \$Z\$ is a complex quantity) that looks the same as the equation for a circuit with a resistor and a DC source: Ohm's Law \$V = RI\$ (where all three \$V\$, \$I\$ and \$R\$ are real quantities).


You might be able to construct such a system and make it work.

But the fact that the physically real power dissipated as heat in a resistance appears on the real axis, makes the system we currently use, vastly more convenient and simpler to use.

Andy's point is more fundamental : the phase between voltage and current is zero in a resistance, so V=I*R (Ohm's Law) works and is useful in situations where you can neglect reactances altogether. Then, P = V*I = V^2/R = I^2*R describes this real power directly.

Then, taking reactance on the imaginary axis allows it to be described and calculated in a manner completely consistent and backwards compatible with basic Ohm's Law.

So there is nothing to be gained from changing axes when moving from resistive to reactive calculations, and a lot of simplicity to be lost.

Tags:

Phasor

Recent Posts