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This might be absurd question for many here but I'm trying to picture a bit in depth what is happening in an LR circuit and confusion arises.

Below is how a series LR circuit behaves when the switch is turned ON:

enter image description here

enter image description here

I can understand the basic explanation and the derivations.

But there is a particular part that confuses me here conceptually.

The texts mention that, the moment the switch is turned ON the current wants to increase but due to the self-inductance an opposing voltage induced as -L*di/dt which is equal to the battery/source voltage hence no current flows at time zero. The equations also support that.

But they say: The self-inductance is fighting the current's will to go through the circuit, and there comes a time that the self-inductance loses the fight if we wait long enough and eventually the current reaches to the maximum value.

All of these are clear if we look at the equations, but what really happens at the coil that there comes a time that the self-inductance loses the fight? I mean why doesn't its induced opposing voltage stay constant forever and stop any current flow? Is there a more phenomenal explanation rather than maths here?

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The self-inductance is fighting the current's will to go through the circuit, and there comes a time that the self-inductance loses the fight if we wait long enough ...

First, it's just a silly analogy to say the current has a "will" or that the inductor is "fighting" it. This might help you visualize or remember what's the behavior going to be, but it has nothing to do with the physics of the circuit.

... and eventually the current reaches to the maximum value.

A more rigorous way to say this is, "in the limit as time goes to infinity, the current approaches a limiting value"

$$\lim_{t\to\infty}i(t) = I_{max}$$

or "for any positive finite error \$\epsilon\$, if you wait long enough, the difference between the current and the limiting current \$I_{max}\$ will be less than \$\epsilon\$."

I mean why doesn't its induced opposing voltage stay constant forever and stop any current flow?

Because it's not the value of the current that induces the back-emf of the inductor, it's the rate of change of current.

If the current were fixed at 0 (for example), then the inductor would produce no back-emf. Thus the full source emf would be applied across the resistor. Which would produce a current through the resistor. Which must also flow through the inductor, contradicting our assumption that the current through the inductor is zero.

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  • \$\begingroup\$ Yes I see your argument that if the current were fixed at 0 L*di/dt would be zero and there would be no back-emf. But unfortunately that only proves that the current is not fixed. But I'm wondering what physically starts to reduce the back emf of the inductor and lets the current to flow and most importantly how. Lets say we observe the back emf at time t = +0 which is equal to the source voltage in magnitude and opposing it. It seems that the power somehow is dissipated at the resistor since the inductor power is reactive. Somehow around t = +0 the resistor and inductor acts in a way but how. \$\endgroup\$ – user1999 Nov 16 '17 at 17:51
  • \$\begingroup\$ Every circuit analysis involves figuring out how the characteristics of the different devices balance each other to determine an operating point. Even if you just have a battery and a resistor, the operating point is determined by the intersection of the I-V curves of the two devices. There's no fixed causality where "the battery imposes a voltage on the resistor" or "the resistor draws a current from the battery" (although sometimes simple cases might work out as if there were). There is always really a mutual interaction between the different elements rather than one thing causing another. \$\endgroup\$ – The Photon Nov 16 '17 at 18:24
  • \$\begingroup\$ In this case we know that current is changing, but at the moment of connection, the current is 0. Since current is 0 we know the voltage across the resistor is 0. So we know the full battery EMF is found across the inductor. And we know \$L\frac{di}{dt}=V\$. So we know the current state of the inductor and how it is changing. \$\endgroup\$ – The Photon Nov 16 '17 at 18:26
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Consider the case when you do not have any inductance in the circuit, and you close the switch at t = 0, the current profile should look like a step as shown below: enter image description here

Before the switch was on (t < 0) there was no current and as soon as we close the switch at t = 0 we have a current \$\frac{E}{R}\$ flowing through the circuit. The current rises at infinite rate and reaches the final value. So what limited the current in this circuit and why did the current not rise to infinite value?
The answer is the resistance of the wire. It stops the current to rise beyond a certain point and this is exactly the reason why in an LR circuit the emf of the inductor starts to reduce because ultimately the resistor starts to reduce the \$\frac{di}{dt}\$ of the circuit.
To clarify this further, consider another case where we have no resistance but just an inductor connected with the battery and switch. In this case, there is nothing (no resistance) to limit the current of the circuit and the emf in the inductor will be constant and equal to battery potential. The current in this case would be: $$E = L\frac{di}{dt}$$ $$i = \frac{E}{L}t$$ Thus current continues to rise indefinitely and the emf stays constant.
This should make sense since there is no resistance we expect to get infinite current in steady state. But because of the inductance the rate of change of current is limited and we have a finite slope.

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  • \$\begingroup\$ Have you considered that any conductive 'circuit' will by its nature form an inductor (1 turn) and that it also has capacitance. At t = 0 we don't suddenly get E/R amps, there is a (steep but) finite slope due to this inductance. It gets worse when you try to consider an LC circuit of zero resistance. As regards infinite current you need to consider that any supply to the circuit is simply not capable of supplying infinite current. \$\endgroup\$ – JIm Dearden Nov 17 '17 at 17:04
  • \$\begingroup\$ @JImDearden I understand that, I am talking about ideal conditions to keep things simple. \$\endgroup\$ – sarthak Nov 17 '17 at 17:45
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Current = 1/L * integral of Vl * dT.

Thus as T becomes larger than zero time, there will be some current.

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