How can we say whether the applied emf and the inductor in the the image are in parallel or in series? In some discussions I have seen people taking the applied emf and inductor to be in parallel and concluding that way that the applied emf = back emf, but I believe they are in series and this reasoning will not hold good.

enter image description here

  • $\begingroup$ The circular symbol in your diagram is not "an EMF." Its proper name is, "AC voltage source," and it applies electromotive force (a.k.a., "voltage") to the inductor in your circuit. $\endgroup$ – Solomon Slow Jul 16 '18 at 16:45
  • $\begingroup$ yes but as the voltage is generated,its called emf. $\endgroup$ – sachin Jul 16 '18 at 16:46
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    $\begingroup$ Yes, but calling the component in the diagram "an EMF" is like saying "a water," when the thing that you are talking about actually is a pump. $\endgroup$ – Solomon Slow Jul 16 '18 at 16:48
  • $\begingroup$ hope m getting u,still we say the emf of a battery is the work done in taking a charge thru a complete circuit,even though i dnt understand it fully i believe here a complete circuit basically means from the negative terminal of the cell to the positive one ie only in the internal circuit of the battery not the external one,still we say its as a complete circuit,can u help,thanks. $\endgroup$ – sachin Jul 16 '18 at 16:59
  • $\begingroup$ @sachin, going from one terminal of a battery to another is not a complete cycle. Complete cycle means the charge has to return to its starting point. However, battery's emf is localized to vicinity of the battery, so both paths results in about the same work - when the charge is out of the battery, forces due to battery are almost zero and so work done by the battery (against the electric forces in the circuit) is zero. $\endgroup$ – Ján Lalinský Jul 16 '18 at 17:37

They are both.

They are in series because the current through one must go through the other.

They are in parallel because the potential difference across one is the same as the potential difference across the other.

More importantly, there is a single loop there, and you can use KVL around that loop to show that the potential drop across the inductor must be equal to the potential generated by the source.

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  • $\begingroup$ "They are in parallel because the potential difference across one is the same as the potential difference across the other" basically means we take the potential difference to be granted as the same and conclude as parallel,but the discussion goes with reverse ie it says its parallel so potential differences are the same in youtube.com/watch?v=Uc6GYqRhjwA $\endgroup$ – sachin Jul 16 '18 at 17:02
  • $\begingroup$ @sachin, another way to say it is, label the two nodes in your circuit A and B. No matter whether you consider the path through the voltage source or the path through the inductor, or even some path through the air between the components, the voltage between A and B must be $V_A-V_B$. All of this of course requiring we satisfy the assumptions of the lumped circuit approximation. $\endgroup$ – The Photon Jul 16 '18 at 17:06
  • $\begingroup$ hope m getting u,it has to be VA−VB but may be + ve or − ve,can u intimate when we are giving a sinusoidal ac supply,in the beginning we get a negative current in the inductor,what it means is the back emf is able to arrest the flow of the current in the forward direction so also reverses it in the inductor with the emf of the same magnitude as that of the supply emf,if the emfs are equal and opposite,the current should go on moving in the same direction without increasing,but here it reverses and also decreases. $\endgroup$ – sachin Jul 17 '18 at 1:13
  • $\begingroup$ @sachin, no, at any instant in time $V_A-V_B$ is either positve or negative. It is not uncertain. It isn't one sign by one path and opposite sign by a different path. The AC source means this value is changing with time, it doesn't mean the sign is uncertain at any particular instant in time. $\endgroup$ – The Photon Jul 17 '18 at 2:58
  • $\begingroup$ l request to give in a bit clearer and simpler way,the uses of not uncertain,doesn't mean uncertain does not let me know much. $\endgroup$ – sachin Jul 17 '18 at 11:13

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