What I understand from Faraday and Lenz laws is that(I might be wrong): a voltage appears across an inductor when the current through it changes. And the polarity of that voltage is such that it opposes the increase of flux. In other words voltage appears in a polarity such that it tries to maintain the current.

So if we look at the below illustration:

enter image description here

Above the current increases in the shown flow direction. So the flux increases. According to law of induction v = L*di/dt. For polarity I assume that the voltage induced should oppose the flux or try to stop the increase in current. So I marked the polarity of the voltage I guess.

But obviously Im wrong, after seeing some examples in texts.

And here is a simulation(V(A,B) is the voltage across A and B) result also disapproves me. When the current increasing the polarity does not oppose the current:

enter image description here

Where am I wrong here in my thinking? Can you explain intuitively what's happening?


Here is the figure and the argument from the text which confused me: enter image description here

The voltage developed by the change in flux linkages has a polarity such as to oppose the change in current producing it. A current i, increasing in time in Fig. 1.7(b) or (c) induces a voltage +v as shown, thereby opposing a source which tends to increase i:

Is the book correct?

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    \$\begingroup\$ Try increasing your resistor to, let's say, 10k ohm. \$\endgroup\$ – WhatRoughBeast Jun 7 '17 at 17:12
  • \$\begingroup\$ @WhatRoughBeast In my first illustration is the voltage polarity wrong? I saw in a text it shows polarity the other way around for the increasing current. \$\endgroup\$ – user1245 Jun 7 '17 at 17:14
  • \$\begingroup\$ @WhatRoughBeast Please see my edit. Is the claim in this text book correct? (I uploaded the figure the text use along with its argument) \$\endgroup\$ – user1245 Jun 7 '17 at 17:24
  • \$\begingroup\$ Try drawing a better diagram, with a core and the windings passing, clearly, over the top of the core then wrapping around and under and back up, before beginning that 2nd turn, and the 3rd turn. Apply the right-hand-rule to that, to predict the flux and the increase-of-flux. Then.....bring Lenz Law into the circuit's voltage loop. \$\endgroup\$ – analogsystemsrf Jun 7 '17 at 17:39
  • \$\begingroup\$ @analogsystemsrf In my first illustration I showed the magnetic flux direction. This reveals the how the wrapping is(right hand rule). \$\endgroup\$ – user1245 Jun 7 '17 at 17:42

enter image description here

The polarity of the induced emf you have shown is incorrect : it is not opposing the increasing current. Replace those "- +" in you figure with a battery and think in which direction that battery drives the current. Your polarity drives current like this. (Recall that outside the battery current flows from + to -. Inside, current flows from - to +)

enter image description here

  • \$\begingroup\$ You got where I am confused. I dont get why my polarity doesnt oppose the current. In this polarity the current flows from B to A opposing the original increasing current flowing from A to B isnt it? How can I make it easy to understand? \$\endgroup\$ – user1245 Jun 7 '17 at 17:56
  • \$\begingroup\$ Ahh I think I see where you're stuck at :) I'll add another picture, one sec \$\endgroup\$ – Hiiii Jun 7 '17 at 17:57
  • \$\begingroup\$ Oh I see my polarity actually reinforces the current like a series battery right? Thanks for the clarification. \$\endgroup\$ – user1245 Jun 7 '17 at 18:01
  • \$\begingroup\$ You've got it ! \$\endgroup\$ – Hiiii Jun 7 '17 at 18:02

The induced voltage always try to oppose the change in current. So, your polarity is wrong. The inductor at the beginning has no current. And the current source wants to increase the current in the inductor. So the induced EMF must oppose this change (stop the current to flow).

The voltage across an ideal inductor is \$ V_L = L*\frac{dI}{dt}\$

This equation indicates that inductance voltage depends not on the current which actually flows through the inductance, but on its rate of change. This means that to produce the voltage across an inductance, the applied current must change. If the current is kept constant, no voltage will be induced, no matter how large the current. Conversely, if it is found that the voltage across an inductance is zero this means that the current must be constant but not necessary zero.

enter image description here

In summary:

When the current is increasing di/dt > 0, so V must be positive because L times a positive number yields a positive voltage.

When the current is decreasing di/dt < 0, so V must be negative because L times a negative number yields a negative voltage.

When we have no change in current over time then we cannot have any induce voltage V = L*di/dt = L * 0 = 0.

In the circuit below V2, V4, V6 source represent the "induced" voltage. As you can see only "positive voltage" oppose the current.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Please see also my edit. I would be very glad if you also have comment about the book's claim. That started my confusion. \$\endgroup\$ – user1245 Jun 7 '17 at 17:32
  • \$\begingroup\$ But why the voltage with that polarity in my first illustration does not oppose the current? Isnt that voltage(with red polarity) yield a current from + to - which would be opposite to the increasing current flow? \$\endgroup\$ – user1245 Jun 7 '17 at 17:35
  • \$\begingroup\$ Your book is right. The current will stop to flow only if there we be no potential difference. So, if the Source cause the voltage at point A to be positive the inductor also must induced positive voltage at point A (to stop the current from flowing). learnabout-electronics.org/ac_theory/images/… \$\endgroup\$ – G36 Jun 7 '17 at 17:47
  • \$\begingroup\$ Im sure yo're right. But what I don't get is how +v is opposing a source. it is drawn in the book like it is reinforcing the source or current not opposing. What is meant by opposing here. Can you show it in a circuit or a small figure. Im stuck with this... \$\endgroup\$ – user1245 Jun 7 '17 at 17:51
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    \$\begingroup\$ I think the confusion may arise because current enters a component such as a resistor at a + sign; but current exits an emf (such as a battery or inductor) at a + sign. \$\endgroup\$ – Chu Jun 7 '17 at 19:20

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