# Two questions about an AC electric circuit: Have I interpreted the dots correctly and where does the negative sign come from?

I asked this question over at the Physics Stack Exchange (https://physics.stackexchange.com/q/549814/263454) and I thought that it might stand a better chance of being answered here.

So I am looking at two mutually coupled circuits that are not at resonance. In particular I am looking at a pair of electric circuits discussed on pgs: 259 - 261 of the book "Electricity and Magnetism", 4th edition by W.J. Duffin. I focus on pg 260 and a sketch of the two circuits is shown below (Fig 10.14a in Duffin). Apologies for the lob sided drawing.

My first question pertains to my understanding of how to assign the potential differences across the coils when applying Kirchhoff's voltage law to each mesh. Before showing what I have done, I will show Fig 9.13 and Fig 10.1 from Duffin which use and discuss the dot notation and quote the explanation of how to assign the potential difference across each coil given by Duffin in the paragraph preceding Fig 10.1.

In the paragraph preceding Fig 10.1 Duffin explains how to assign the potential difference across each coil using the dot notation and the direction of the current in each coil as follows:

" For a mutual inductor, details of the directions of the windings are avoided by using the dot notation of Fig. 9.13. A current flowing in at the dotted end of one coil produces a voltage $V = M\, dI/dt$ across the other which is higher at its dotted end. The $sign$ attached to this voltage will depend on the assumed direction of current in the second coil. In any given circuit with directions for all currents decided, taking one dot to the other end of its coil will reverse the sign of $M$."

So with this in mind I think the potential differences across each coil are as shown in

where plus and minus signs are in reference to the potential difference for the mutual inductance effects.

This leads me to the following for Kirchhoff's equations for the physical voltage and currents in each mesh:

$$v_1 - i_1R_{\rm p} - L_{\rm p}\frac{di_{1}}{dt} + M\frac{di_2}{dt} = 0\\ \,\\ M\frac{di_1}{dt} - i_2R_{\rm s} - L_{\rm s}\frac{di_2}{dt} - v_2 = 0\,.$$ In terms of phasors, where $$i_1 = {\rm Re}({\rm {\bf I}}_1)\, {\rm etc.}$$ $${\rm {\bf I}}_1 = I_1e^{j\omega t}\,, {\rm{\bf I}}_2=I_2e^{j\omega t}\,, j=\sqrt{-1}\,,$$

where $$\omega$$ is the driving frequency of the AC source of emf, $${\rm {\bf Z}}_{1} = R_{\rm p} + j\omega L_{\rm p}$$, and $${\rm {\bf Z}}_{\rm s} = R_{\rm s} + j\omega L_{\rm s}$$ the above equations can be written as $${\rm {\bf V}}_1 - {\rm {\bf I}}_1{\rm {\bf Z}}_1 + M\frac{d{\rm {\bf I}}_2}{dt} = 0\\ \,\\ M\frac{d{\rm {\bf I}}_1}{dt} - {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm s} - {\rm {\bf V}}_2 = 0\,.$$

Letting $${\rm {\bf V}}_2 = {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm L}$$ and defining $${\rm {\bf Z}}_{2} = {\rm {\bf Z}}_{\rm s} + {\rm {\bf Z}}_{\rm L}\,$$ this can be written as

$${\rm {\bf V}}_1 - {\rm {\bf I}}_{1}{\rm {\bf Z}}_{1} + j \omega M{\rm {\bf I}}_2 = 0\\ \,\\ j\omega M{\rm {\bf I}}_{1} - {\rm {\bf I}}_{2}{\rm {\bf Z}}_{\rm 2} = 0\,.$$

Now to me this identical to what Duffin writes, which is

$${\rm {\bf V}}_1 = {\rm {\bf I}}_{1}{\rm {\bf Z}}_{1} - j \omega M{\rm {\bf I}}_2\\ \,\\ 0 = {\rm {\bf I}}_{2}{\rm {\bf Z}}_{\rm 2}- j\omega M{\rm {\bf I}}_{1}\,.$$

I understand that I might be being silly, but this brings me to my first question: have I understood the meaning of the dots correctly and assigned the potential differences associated with mutual inductance properly? I ask this because when Duffin calculates the ratio $${\rm {\bf V}}_2/{\rm {\bf V}}_1$$ he states that $${\rm {\bf V}}_2/{\rm {\bf V}}_1 = - {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm L}/{\rm {\bf V}}_1\,,$$ which implies $${\rm {\bf V}}_2 = -{\rm {\bf I}}_2{\rm {\bf Z}}_{\rm L}$$ whereas I think $${\rm {\bf V}}_2 = {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm L}\,.$$

So my questions are: (1) Is my understand the meaning of the dots correct and hence have I properly determined the p.d. due to mutual inductance across each coil?, and (2) If so, why does Duffin write $${\rm {\bf V}}_{2} = - {\rm {\bf I}}_{2}{\rm {\bf Z}}_{\rm L}$$ instead of $${\rm {\bf V}}_{2} = {\rm {\bf I}}_{2}{\rm {\bf Z}}_{\rm L}$$ ?

Lastly, a screenshot of the relevant piece of text from Duffin is attached.

I should note that Duffin writes elsewhere that the impedance for mutual inductance can be $\pm j\omega M$ so maybe I have messed up the signs in the second mesh but still obtained the circuit equations that he did.

The point of my asking is that when he applies this circuit to an ideal transformer he gets

$$\frac{{\rm {\bf V}}_{2}}{{\rm {\bf V}}_{1}}= -\frac{N_{\rm s}}{N_{\rm p}}$$ where as I would get $$\dfrac{N_{\rm s}}{N_{\rm p}}$$. Duffin states that the negative sign means that potential differences $${\rm {\bf V}}_{1}$$ and $${\rm {\bf V}}_{2}$$ are out of phase by $$\pi$$ radians.

So am I going wrong somewhere?

• Why? They were originally not showing and then @ppmbb formated the original text of the question so that the images show. In fact, I don't know what you mean by inline. Commented May 8, 2020 at 21:26

You are not wrong. However, the red signs in your own drawing is not correct. The signs of the instantaneous voltages follow the dot convention, as does the ampere-turns law where the dots indicate the instant of current flowing into the transformer.

If you use Faraday’s law for the transformer winding voltages $$\u_p\$$ and $$\u_s\$$ $$u_p(t) = - n_pA\frac{dB}{dt}$$ and $$u_s(t) = - n_sA\frac{dB}{dt}$$ then $$\frac{u_p}{u_s} = \frac{n_p}{n_s}$$ for an ideal transformer. Therefore I suspect that the "-" sign is a relic from that relationship.

(I also see some other bad editing errors so rather find a better source or a version of the book that has been corrected.)

Also note that for a transformer with k windings, $$\Sigma n_ki_k(t) = Hl$$

and the magnetic relationship of the core material $$B = \mu H.$$

Note that in many practical examples the ampere-turn law assumes the magnetising current, $$\Hl = 0\$$.

If you draw the two windings on the same side of the core, wound in the same direction the dots will be indicated at the top as well. Maybe that clarifies the issue issue for you.

The dot convention is correctly explained in this transformer phasing article.

• Thank you for your answer to my question and for the link but I don't understand why you write the signs I've given for the mutually induced voltages are incorrect. I have tried to follow the dot convention as I understand it and after writing my original post and posting here I looked at youtube.com/watch?v=VHbqrs1gkAc and at chapter 16 in eng.harran.edu.tr/~msuzer/files/edt/edt.pdf. These both seem to confirm what I wrote. As far as I can tell my choice $${\rm {\bf V}}_2 = {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm L}$$ is correct. Commented May 7, 2020 at 10:19
• Look at the diagram posted by @ppmbb. At $t=t_1$, $u_p(t_1) = n u_s_(t_1)$ with the dots indicating the polarities. Carfully review electricaltechnology.org/2013/12/… if this is not clear. Your choice is correct, your signs not. Commented May 7, 2020 at 10:46
• I've also looked at images.app.goo.gl/rHbx3LZiw2f8vVhX6 and images.app.goo.gl/rHbx3LZiw2f8vVhX6 Commented May 7, 2020 at 10:56
• Those are very confusing! There is in fact no dot relationship between the voltage and the current. The relationship is between the voltages or between the currents. The relationship between voltage and current is due to the rest of the circuit. Commented May 7, 2020 at 11:16
• Thanks for your help. Commented May 7, 2020 at 17:58

Just a side note, when an inductor is acting as a load in a circuit, its polarization is such its positive terminal is located where current 'enters' the inductor.

While an inductor acting as a source (secondary side in this case), it's positive terminal is located where current 'leaves' the inductor.

This behavior is indicated by the blue polarization marks in the image.

So, in short, what dot marks means is that when a current $$\I_1\$$ enters by the dotted terminal of $$\L_p\$$, the inducted current $$\I_2\$$ in the coupled inductor $$\L_s\$$ will leave by its dotted terminal.

• Effectively it stands for 'Load' and in generic terms, it can be of any nature. Commented May 7, 2020 at 9:13
• Sorry, I decided to delete and repost my original question because I was trying to add my answer after my original question and tag you but I wasn't allowed to because the five minutes for editing had elapsed. It was only after I deleted my original question that I saw your answer. Thanks for your answer. Your answer confirmed what I thought. Commented May 7, 2020 at 9:29
• Thanks for your help. Commented May 7, 2020 at 17:58

Okay so I am going to attempt to answer my own question. In order to understand how to employ the dot notation I have purchased and consulted "Electric Circuits" by James W. Nillson and Susan Reidel (11th edition, Global Edition, published by Pearson). In Section 6.4 on mutual induction the dot convention for magnetically coupled coils is stated in two alternative ways on pages 227 and 228:

1."When the reference direction for a current enters the dotted terminal of a coil, the reference polarity of the voltage that it induces in the other coil is positive at its dotted terminal."

or

2." When the reference direction for a current leaves the dotted terminal of a coil, the reference polarity of the voltage that it induces in the other coil is negative at its dotted terminal."

So applying these to the mutually coupled circuits in question

the polarity of the voltages due mutual induction across each coil are shown in blue in the figure below.

Next I will include the polarity of the voltages due to self-induction. These are shown in green in the figure below.

This leads me to the following for Kirchhoff's equations for the physical voltage and currents in each mesh:

$$v_1 - i_1R_{\rm p} - L_{\rm p}\frac{di_{1}}{dt} + M\frac{di_2}{dt} = 0\\ \,\\ M\frac{di_1}{dt} - i_2R_{\rm s} - L_{\rm s}\frac{di_2}{dt} - v_2 = 0\,.$$

Looking at the overall behaviour of both circuits, the inductor in the left circuit functions as a load whereas the inductor in the right circuit serves as voltage source. Thus the net voltage polarities on the coils is that given in https://electronics.stackexchange.com/a/498093/251725

In terms of phasors, Kirchhoff's voltage law applied to each mesh reads

$${\rm {\bf V}}_1 - {\rm {\bf I}}_1{\rm {\bf Z}}_1 +j\omega M{\rm {\bf I}}_2 = 0\\ \,\\ j\omega M{\rm {\bf I}}_1 - {\rm {\bf I}}_2{\rm {\bf Z}}_{\rm s} - {\rm {\bf V}}_2 = 0\,,$$

where $$\{\rm {\bf V}}_{2} = {\rm{\bf I}}_{2}{\rm{\bf Z}}_{L}\,.\$$

So in conclusion my initial application of the dot convention is correct and the definition of $$\{\rm {\bf V}}_{2} = -{\rm{\bf I}}_{2}{\rm{\bf Z}}_{L}\$$ by Duffin as mentioned above in the opening post is incorrect.

Edit two things to add: Firstly, I applied the dot convention as explained to another circuit in Nillson's book on page 375 and got the answer given on page 376 so I think I understand what is going on.

Secondly, to address the comment below and an earlier comment made skvery in https://electronics.stackexchange.com/a/498087/251725

Yes, it is possible to have the dots in the same location as above and have the current in the right circuit in the anti-clockwise direction. This is shown Fig 9.46 from Nillson's book. So there is no inherent relationship between the voltage and the currents. It is all about magnetic fluxes reinforcing and Faraday's law.

simulate this circuit – Schematic created using CircuitLab

What the circuit above shows is that the current in a transformer is totally independent of the voltage. Therefore the dots work either for current or for voltage.

• I can only say it again! The dot notation does not describe the voltage-current relationship at all. You can put a voltage source on the primary and a current source on the secondary and do whatever you want with the voltage-current relationship! Commented May 7, 2020 at 19:32
• Okay, but you have just down voted an answer than has referenced the 11th edition of a book that was first authored by a man who has nearly 40 years experience teaching electric circuits. I applied what was written in section 6.4 to a circuit appearing in Section 9.10 that discusses transformers and I got what was printed therein. The discrepancy about signs arises in the next section. It may be so that the dot convention has nothing to do with the voltage-current relationship but what I've referenced and used is the way it is taught to undergraduates so what I am to do? My answer suffices. Commented May 7, 2020 at 19:41
• It is still wrong! If you want to I can add a sample circuit to your question. Commented May 7, 2020 at 21:47
• Thanks for offering. How about doing an example of applying the dot notation to a sample circuit? Commented May 8, 2020 at 9:05
• Thanks. However, I don't know how to run that simulation and all I wanted to know was had I applied the dot notation correctly to construct Kirchhoff's voltage equation to each circuit. Based on the book by Nillson I quoted above and the 3rd edition of the "Fundamentals of Electric Circuits" by Alexander and Sadiku that I found online and @ppmbb suggested I consult, I have done so because when I follow the rules for the dot convention they give I get what I got initially. After that my question was about why Duffin writes $${\rm {\bf V}}}_{2} = -{\rm {\bf I}}}_{2}{\rm {\bf Z}}}_{\rm L}$$? Commented May 8, 2020 at 21:34