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I'm kind of confused about the principles of self induction; Does change in current represent the change in the net current (battery's current + induced current) or just battery's current? among the EMF and change in current; who causes who? When we increase the inductance, the rate of change of current decreases by the same rate in a way that would keep the induced EMF constant, does that mean that it doesn't depend on the inductance?

By the way, I'm dealing here with a DC circuit that consists of a power source (battery), and an inductor

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  • $\begingroup$ What do you mean by "battery's current + induced current"? If you just connect a battery to an inductor, then the battery current and the inductor current are the same (by KCL). Please include the schematic of the circuit you are considering, even if it's just two components it helps make your question more clear. $\endgroup$ – The Photon Apr 8 '19 at 21:55
  • $\begingroup$ edited. I mean, there would be the current that the battery produces, and the induced current as a result of self induction (which could be in the opposite direction of the other current) $\endgroup$ – Just_Cause Apr 9 '19 at 2:53
  • $\begingroup$ First point: Changes in current induce EMF, not the other way around (see Faraday's Law of Induction. Second: If we just connect the inductor across the battery terminals, then the battery current and inductor current must be equal, whether we say this is "battery current" or "inductor current". $\endgroup$ – The Photon Apr 9 '19 at 3:09
  • $\begingroup$ I don't really like the way inductors are taught, but if we talk about an ideal inductor, then the "back EMF" is exactly the same thing as the "voltage across the inductor". They aren't two different things, no matter how some texts try to make the back EMF seem like something separate from an ordinary circuit voltage. $\endgroup$ – The Photon Apr 9 '19 at 3:12
  • $\begingroup$ (To be pedantic, I should have said, changes in flux induce voltage. But if you're talking about a single inductor, current is proportional to flux) $\endgroup$ – The Photon Apr 9 '19 at 3:22
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Does change in current represent the change in the net current (battery's current + induced current) or just battery's current?

First, we shouldn't talk about an "induced current". According to Faraday's Law of Induction, a changing magnetic flux through a wire loop "induces" an EMF around the loop. In the case of self inductance, this magnetic flux is caused by current through the loop itself.

This view sees the EMF as being caused by the current. This can also lead to confusion because the cause and effect isn't inherent in the operation of the inductor. Whether the EMF causes the current or the current causes the EMF depends really on the kind of sources being applied to the inductor rather than an inherent process going on in the inductor itself. If we apply a current source, we can say the voltage across the inductor is caused by the current through it. If we apply a voltage source we can say the current is caused by the voltage. If we apply something in between (say, a 50-ohm sourced) then we can only say that the voltage and current are caused by the source, but there isn't any way to really distinguish whether the voltage causes the current or vice versa.

In any case, the simple answer to this question is that the flux in the inductor is proportional to the current through the inductor. Not the current through the inductor due to one source or another, the total or net current through the inductor. And the potential difference across the inductor (the sum of the EMFs around all the turns of the inductor) is proportional to the time derivative of the flux. So we get the usual circuit-level equation for the ideal inductor

$$V_L = L\frac{dI_L}{dt}$$

When we increase the inductance, the rate of change of current decreases by the same rate in a way that would keep the induced EMF constant, does that mean that it doesn't depend on the inductance?

First, we need to know exactly what circuit we're talking about. I'm assuming you're talking about this circuit:

enter image description here

In this circuit, by KVL, the voltage across the inductor (VL) must be equal to the voltage across the batter (VB). Also, the current through the inductor must be equal to the current through the battery (I labelled the loop current "I"). (If we use the passive current convention, we would say the battery current is negative because the current enters the battery's negative terminal and exits its top terminal, that is $I_B=-I$)

In this case, the inductor voltage is fixed by being connected to an ideal voltage source. The EMF around each of the $N$ turns of the inductor (assuming everything is uniform) must be $V_L/N$. We don't calculate the induced voltage from the current, we calculate the current from the induced voltage.

And we get the result that the current (I) will be constantly increasing

$$\frac{dI}{dt} = \frac{V_B}{L}.$$

The current does depend on the inductance. The higher the inductance, the more slowly the current increases.

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  • $\begingroup$ you mean that the induced EMF is forced to be equal to battery's voltage? $\endgroup$ – Just_Cause Apr 10 '19 at 3:18
  • $\begingroup$ Yes. The external constraint is that the potential difference across the inductor is the same as the battery voltage. In some other circuit, it might be different. $\endgroup$ – The Photon Apr 10 '19 at 4:17
  • $\begingroup$ Like, more components or what? $\endgroup$ – Just_Cause Apr 10 '19 at 15:48
  • $\begingroup$ @Just_Cause, I addressed this in my answer already. For example, if you used a current source instead of a voltage source, you'd get an EMF induced by the (changing) current from the current source. $\endgroup$ – The Photon Apr 10 '19 at 15:54
  • $\begingroup$ Right, so we would have a different way to calculate the emf. But anyways, when we connect 5 volts battery to an inductor of 5 henries, current's avarage rate of change would be 1 amp/sec, right? that's why EMF isn't dependent on the inductance.. because if we increase the inductance by some rate, avg rate of change of current would decrease by 2 as well? $\endgroup$ – Just_Cause Apr 10 '19 at 16:32

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