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As you can see there are electrical circuits of power supply, resistance, inductive reactance combination in one case and power supply resistance and capacitive reactance combination in another case are there. Their respective phasor diagrams are also drawn.

But why is the inductive reactance or capacitive reactance phasor on imaginary axis while the resistance phasor is taken on the real axis? What will happen if we take resistance as the imaginary component and reactance as the real component?

Phasor diagram

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    \$\begingroup\$ In a series circuit, current is common to all components, so current is normally selected as the reference \$0^o\$ phasor. In a parallel circuit, voltage would be selected as reference phasor since voltage is common to all components. For more complicated circuits involving series and parallel elements, the selection of reference phasor is not quite so straightforward. \$\endgroup\$ – Chu Jun 26 '16 at 15:58
  • \$\begingroup\$ @pandu: Note: Although reactances are complex quantities they are not phasors. I think your question is automatically answered when you go back and ask yourself "What actually is a phasor"? (and "When and why are they used?"). See my answer below for details. \$\endgroup\$ – Curd Jun 26 '16 at 17:48
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    \$\begingroup\$ Well why nobody used the simple answer, do the fourrier transform of a resistor, an inductor and a capaticance and you will end up with inductor and capacitance on the imaginary axis. \$\endgroup\$ – MathieuL Jun 27 '16 at 14:40
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    \$\begingroup\$ @MathieuL: ...because it is not "the simple answer". There is no need to even mention Fourier Transform for explaining phasors or for deriving complex impedance of L or C (See my answer below or - for more details - chapter 10 "Sinusoidal Steady State Anlysis" in "Engineering Circuit Analysis" (8th ed.) by Hayt et al. They introduce phasors 8 chapters before they introduce Fourier Transform) \$\endgroup\$ – Curd Jun 27 '16 at 21:18
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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:

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

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  • \$\begingroup\$ Having changed the axes you could also make the change by making the voltage across the resistor $jV_r$ and all would be well? \$\endgroup\$ – Farcher Jun 26 '16 at 15:05
  • \$\begingroup\$ @Farcher If you declare the voltage as \$jV\$, then the equation becomes mathematically correct again. To rotate it you actually have \$jV=jIR - IX\$ (note the minus sign) but the resistive part must still be real - its the current that becomes out of phase to match the out of phase voltage. In other words the alternate impedance equation is still wrong. \$\endgroup\$ – Tom Carpenter Jun 26 '16 at 15:40
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    \$\begingroup\$ The choice of which convention which is made is arbitrary but with in practice with a view to make the processing of the data as simple as possible. Changing the axes round can be done by multiplying every term by $j$ or $-j$ and modifying the convention as appropriate. This would add extra complexity as compared with the normally used convention? \$\endgroup\$ – Farcher Jun 26 '16 at 19:37
  • \$\begingroup\$ I don't see how it explain why the inductor or the capacitor end up on the imaginary axis... You seem to just say it is a convention while it is the result of a fourrier transform. \$\endgroup\$ – MathieuL Jun 27 '16 at 14:45
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  • 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).

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  • \$\begingroup\$ Phasor analysis? It is by applying the fourrier transform to the V-I expression of a inductor or a capacitor that you end up with these expressions. Phasor analysis isn't a fourrier transform. \$\endgroup\$ – MathieuL Jun 27 '16 at 14:43
  • \$\begingroup\$ @MathieuL: Who claimed that it is a Fourier Transform? Just because there happens to be a \$e^{j\omega t}\$ term somewhere does not mean we are doing a Fourier transform (BTW F.T. would contain an integral in any case; the only integral we have here is the V-I-relationship of the capacitor). The \$e^{j\omega t}\$ comes solely from the fact that the source is a sinusoidal source (and only then you can apply Phasor analysis). I think you are mixing up some things just because of some superficial similarities in formulas with F.T.. \$\endgroup\$ – Curd Jun 27 '16 at 19:29
  • \$\begingroup\$ In case my short explanation is not clear enough I recommend reading Chapter 10 "Sinusoidal Steady-State Analysis" (8th ed.) in "Engineering Circuit Analysis" by Hayt, Kemmerly and Durbin. \$\endgroup\$ – Curd Jun 27 '16 at 19:31
  • \$\begingroup\$ Apply a fourrier transform to Ldi/dt, you end up with jwL... same apply to the V-I equation of a capacitor which you end up with -jwC. We didn't shove out of a magic hat the complex impedance of the inductor or the capacitor.. It is because we can transform the V-I with fourrier that there is complex number in electronics. \$\endgroup\$ – MathieuL Jun 27 '16 at 20:34
  • \$\begingroup\$ Therefore because we can use fourrier on those V-I equation, we can therefore transform our sin source into phasor because we have demonstrate that fourrier is linear, so we can resolve the response of a sin with 2 phasors. Not the other way around, you reverse the reasonning. \$\endgroup\$ – MathieuL Jun 27 '16 at 20:36
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In a resistor voltage and current are in phase so to imply that a resistor's impedance is totally imaginery is missing the point. In capacitors and inductors, the voltage and current are 90 degrees apart so this naturally means that the impedance expressed in polar co-ordinates is totally imaginery.

I can't begin to see a counter argument to this straightforward way of looking at things.

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    \$\begingroup\$ The choice is arbitrary but your statement with the phrase "this straightforward way of looking at things" in it explains that what is perceived to be the easier route is taken. \$\endgroup\$ – Farcher Jun 26 '16 at 15:11
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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.

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  • \$\begingroup\$ Well, it is a complete arbitrary decision that we put real power on the real axis, we could put reactive power on the real axis and put real power on the imaginary axis. Nothing can prevent us from that. It like using the generator or machine convention. As long as you stay coherent in your convention, everything gonna be alright. \$\endgroup\$ – MathieuL Jun 27 '16 at 14:39
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I don't know your level of understanding of the concept of impedance. In any case, the properties of capacitors and inductors are such that, for both of them, the current through them and the voltage across them are 90 degrees out of phase. Since the real and imaginary axes represent quantities that are 90 degrees out of phase, it is natural to express reactance on the imaginary axis and resistance on the real axis.

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  • \$\begingroup\$ we always write Z= R +j X . but why? Also, why Z=X+j R is wrong ? \$\endgroup\$ – pandu Jun 26 '16 at 12:38
  • \$\begingroup\$ The basic reason is that if the current and voltage have the same phase angle, their phasors point in the same direction. Since V = IR (Ohm's law), Z = R is right, and Z = jR is wrong. Of course you could invent a different system of "Pandus" instead of "phasors" where current and voltage "Pandus" are drawn at right angles when the current and voltage are in phase. In "Pandu notation" Z = X + jR would be correct, but you would probably be the only person who used that notation. \$\endgroup\$ – alephzero Jun 26 '16 at 20:59
  • \$\begingroup\$ @alephzero, in pandus, how could voltage and current be in phase if they are at right angles to each other? \$\endgroup\$ – Chu Jun 27 '16 at 0:01
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    \$\begingroup\$ @Chu You can draw a diagram any way you like. If somebody wants to draw voltages as blue arrows where zero phase is horizontal, and currents as red arrows where zero phase is vertical, they won't get arrested for doing it. Then, adjust all the standard textbook formulas by adding extra factors of j and -j as necessary. I'm not suggesting "Pandus" are a sensible idea, of course. \$\endgroup\$ – alephzero Jun 27 '16 at 0:09
  • \$\begingroup\$ @Chu: Phase diagram notation can be understood as definition. If we introduce Pandu notation as we want and there will be no doubts. But we will have to keep every single equation to fulfill Pandu notation and totally forget about Phase notation. But as alephzero wrote, it will be working but it won't be sensible. In that notation, P=VI sin phi etc. \$\endgroup\$ – Crowley Jun 27 '16 at 7:25
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You need to consider what the phasor represents. For a general AC circuit, the phasor is a way of representing a sinusoidally (sine wave) varying input signal. The phasor has fixed length in phasor space, and rotates counterclockwise at a set frequency. The actual voltage or current is the real component of the phasor, or the projection of the phasor onto the real axis. We use the phasor as a tool to tell us how the RLC circuit responds to the specific input voltage/current and frequency.

So when you ask why is the resistive component on the real axis and the capacitive/inductive component on the imaginary axis, that's just because we choose an instant in time when they are aligned that way. Look at the circuit a moment later, and the voltages and currents that we measure will change as the phasor rotates.

Consider the comparison between circular motion and simple harmonic motion. Take the projection of the circular motion onto a line, and the result is simple harmonic motion. The phasor is like the circular motion, and the output voltage/current is like the simple harmonic motion.

Now, for a resistor, the voltage and current are in phase, so V = I R, as Ohm's law tells us. For the capacitor, the current leads the voltage by 90 degrees, and for an inductor, the voltage leads the current by 90 degrees.

All that said, the answer to your question is the reason we take the resistance on the real axis is convenience. The phasor rotates over time, and we're looking at a snapshot when we assign the resistance to the real part.

A classic physics question is to take a RLC circuit with a fixed AC voltage (or current) applied, and then ask what the current (voltage) is when the instantaneous voltage (current) is some given value. The first real step in the problem is to determine the instantaneous angle of the phasor to get the given voltage (current) value. I've assigned this kind of problem regularly on homeworks and tests (and the students just LOVE me for it).

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In addition to the more mathematically rigorous comments regarding the phase angles between the voltage and current in reactive and non-reactive components, it helps to remember that an ideal inductor and an ideal capacitor do not dissipate energy (they accumulate and release energy), while an ideal resistor does dissipate energy.

In my own mind, I like think of the imaginary values on the Y-axis as representing the energy that's just "sloshing around" in the circuit's reactive components and is not being consumed—i.e., electric field energy or magnetic field energy that is accumulated and released by capacitors and inductors, respectively—while the real values on the X-axis represents energy that's actually being consumed (e.g., a resistor that converts electrical energy into heat). Obviously, this is not a mathematically rigorous definition; it's more a just a memory aid that people sometimes find useful.

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The best answer is actually the comment by Chu.

Remember a reference is only an arbitrary decision by someone, usually to help understanding. So we select references for AC circuits to make it easier to understand.

Current is constant in a series circuit. So current is used as a reference. This is where students start. We choose current as the reference to simplify calculations.

By selecting current (\$I\$) as the reference, \$V_R\$ will be along the x-axis (no y-component). and \$V_C\$ and \$V_L\$ will be along the y-axis (no x-component). This simplifies circuits and calculations to vector addition (no need to break vectors into components) and allows circuits to be solved using right triangles and trigonometry or complex numbers. The alignment allows complex numbers to be used and complex numbers simplifies calculations as circuits become more complex.

$$ \overrightarrow {V_S} = \overrightarrow {V_R}\ +\ \overrightarrow {V_L}\ +\ \overrightarrow {V_C}$$

$$ V_S = \sqrt{(V_R)^2\ +\ (V_L - V_C)^2}\ \measuredangle\ arctan\ \left ({\frac {V_L - V_C} {V_R}} \right ) $$ $$ V_S = V_R\ +\ j (V_L - V_C)$$

Phasor Diagram for Series Circuit

For a parallel circuit, voltage is constant, so the source voltage \$V_S\$ is chosen as reference, and vector addition now becomes addition of currents.

$$ \overrightarrow {I_S} = \overrightarrow {I_R}\ +\ \overrightarrow {I_L}\ +\ \overrightarrow {I_C}$$

By changing the reference, circuit can be represented by right triangles and calculations can be simplified. The reference is chosen to make understanding simpler.

Phasor Diagram for Parallel Circuit

For a series/parallel circuit, the source voltage \$V_S\$ is typically chosen as reference, but the vectors typically do not align with the axis. Hopefully the student has some knowledge by that point.

As for the question. \$V_R\$ (or \$I_R\$) along the x-axis represents Real Power, something we get out of the circuit (heat, mechanical power, etc.). This (in my mind) has always corresponded to real numbers.

For a circuit with a capacitor and an inductor, leading capacitive reactive power supplies lagging inductive reactive, decreasing the reactive power the source must supply [vertical leg of right triangle is less than \$V_L\$ (or \$V_C\$)]. Power flows back and forth between capacitor and inductor. Reactive power is required to make the circuit work (create a magnetic field in a motor), but does no useful work. It appears we get something for nothing, which corresponds as imaginary.

So reactances (reactive power) along imaginary and resistance (real power) along real.

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