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Current through a LR circuit is given by $$i=i_0 (1-e^{\frac{-tR}{L}})..1$$From here I can observe that as I tend R towards 0 no current flows through the circuit , and current will be 0 when the resistance in circuit actually becomes 0.
Though for an A.C. LR circuit I have seen my book presenting a proof where there is a sinusoidally varying emf source and an inductor in a circuit, and nothing else, still the expression of current is derived which is of finite magnitude.
Now I might be wrong but I am taking the liberty of focusing my observation on the AC curcuit at a particular moment of time ,I assume here that my 'moment' is small enough to avoid any appreciable change in magnitude of current or its direction, so I am assuming it to be a DC source for this moment of time(which I don't know is valid or not).Given these assumptions and now comparing my momentary AC circuit to the one which is actually DC,how is the momentary AC circuit working when it has no resistance with reference to equation 1?

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  • $\begingroup$ "I can observe that as I tend R towards 0 no current flows through the circuit , and current will be 0 when the resistance in circuit actually becomes 0." You didn't say how you reached that conclusion, but it is wrong. If you connect a voltage source V to an ideal inductor with no internal resistance, the current increases linearly at a rate where the back EMF equals the applied voltage - i.e. $V = L\dfrac{dI}{dt}$. $\endgroup$ – alephzero Aug 4 at 16:20
  • $\begingroup$ How? Ok I'll tell you how I reached my conclusion.First let the DC circuit have inductor ,resistance and battery.Then I derive equation 1, now as R tends to 0 tau tends to infinity and as tau tends to infinity , e is raised to power 0, therefore i=i0(1-1)=0.This is the process through which I came to my conclusion that "if an ideal battery and ideal inductor are the sole elements in circuit , there can never be a current in it" , correct me if i am wrong. $\endgroup$ – ADITYA PRAKASH Aug 4 at 16:30
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The time constant of an resistor, $R$ and inductor, $L$, circuit is $\tau= \frac LR$.
As $R$ gets smaller and smaller the time constant gets larger and larger.

The initial charging current is $I_0 = \frac {\mathcal E}{R}$ where $\mathcal E$ is the emf of the voltage source.

The term $e^{- \frac {t}{\tau}}= e^{- \frac {Rt}{L}} $ can be expanded as far as the second term $1-\frac {Rt}{L}$ to a better and better approximation as $R$ becomes smaller and smaller.

So now your equation for the current is $I(t) = \frac {\mathcal E}{R}\left ( 1-e^{- \frac {t}{\tau}}\right )\approx \frac {\mathcal E\,t}{L} $ and the approximation gets better and better as the resistance gets smaller and smaller.

Indeed if $R=0$ then $I(t) = \frac {\mathcal E\,t}{L} $.

This expression can be found directly by having the voltage source connected an inductor with no resistance.
Then Kirchhoff’s voltage law for such a circuit is $\mathcal E - L \frac{dI}{dt}$ and integration gives the same equation for the current, $I(t) = \frac {\mathcal E\,t}{L} $.

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  • $\begingroup$ Ok I was evaluating the limit incorrectly ,the current increases indefinitely $\endgroup$ – ADITYA PRAKASH Aug 4 at 16:57
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I don't know what is meant by "...1" at the end of the equation, but otherwise the equation is for a series LR circuit connected to a dc (e.g., battery) source when a switch is first closed to connect the circuit to the battery and thereafter as time progresses.

The equation describes the dc transient behavior of the circuit between $t=0$ and $t=∞$ and assumes no initial current flowing in the circuit. The equation does not apply to the series LR behavior in an ac circuit.

The boundary conditions are $i=0$ at time $t=0$ the instant the switch is closed because you can't change the current through an ideal inductor instantaneously. At time $t=∞$ $i=i_0$ where $i_0$ is the final current and equals $\frac{V}{R}$ where $V$ is the battery voltage. This is because an ideal inductor looks like a short circuit to dc after transients have died out.

I should add that if you want to analyze the response of a series LR circuit to an ac source you will have to solve a first order differential equation. The following link may be of help to you in this regard:

https://www.mcvts.net/cms/lib07/NJ01911694/Centricity/Domain/134/1st%20order%20SystemsTransient%20Analysis.pdf

Hope this helps.

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  • $\begingroup$ Hey Bob you did not adress my question the way I wanted , I'm not confused regarding these topics I'm confused when I approach these topics from the process I described in my question. Help $\endgroup$ – ADITYA PRAKASH Aug 4 at 16:34
  • $\begingroup$ @ADITYAPRAKASH Not sure why you didn't since the answer you accepted was basically the same as mine except that it discussed varying R, which, as far as I can tell, was not part of your question.. But that's always your prerogative. $\endgroup$ – Bob D Aug 4 at 17:12
  • $\begingroup$ I'm sorry , but @Farcher corrected my mistake in a way , the point where I was wrong was that current is 0 at all times in a DC circuit of inductor and battery because R is 0 , which ofcourse is not the case , and neither is the current equal to i0 at t=infinity , interestingly current will increase indefinitely, without an upper limit which I found in Farcher's answer , I'm sorry for not accepting your answer , though I'm genuinely indebted for your contribution towards clearing my doubt , but the framework of PSE is such that multiple answers cannot be accepted.Please understand $\endgroup$ – ADITYA PRAKASH Aug 4 at 17:21
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In the DC case powered by constant voltage source, the smaller the R, the greater is the final current. This is because in the end, all voltage is on the resistor, so it takes greater current to make RI equal to voltage of the source.

In both AC and DC case the current will flow even if resistance R is zero.

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  • $\begingroup$ How can current flow in DC with no resistor $\endgroup$ – ADITYA PRAKASH Aug 4 at 16:37
  • $\begingroup$ There is no resistance , where do I drop the potential at t=infinity , if there is no resistor $\endgroup$ – ADITYA PRAKASH Aug 4 at 16:38
  • $\begingroup$ There is inductor in the circuit. All voltage drop is on that inductor. That also means current never stops increasing. It is idealization, in reality resistance of wires and inductor will be nonzero for high enough current. $\endgroup$ – Ján Lalinský Aug 4 at 16:48
  • $\begingroup$ @ADITYAPRAKASH wrote "How can current flow in DC with no resistor" - please understand that setting the resistance to zero doesn't mean remove the resistor. Removing the resistor is equivalent to setting the resistance to 'infinity'. $\endgroup$ – Alfred Centauri Aug 4 at 19:57

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