Current in only inductive AC circuit 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?
 A: 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.
A: 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.
A: 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} $.  
