rahaverhma said:
I know the mechanism of a circuit containing just a capacitor connected to battery (with) resistance less wire. ... for the case of inductor. As, we connect battery and as resistance is 0. So,current will be infinite in 0 seconds. ...
This is your error. The current will be zero up to 0 seconds. It can only change after that.
You seem to be thinking of applying the battery simply to the zero resistance wire,
without any inductance* , so you would calculate an infinite current, immediately
after connection.
But zero resistance wire, simply means that all the battery voltage appears across the inductor, rather than being split between the inductor and the resistance.
Applying voltage to an inductor causes the current to start changing: So, as has been said before, the current starts at zero and changes at a rate ## \frac {V} {L} ##
If there is really no resistance in the circuit and the battery maintains its voltage, then the current keeps on increasing - but it takes for ever to become infinite.
And if there's no resistance, it shouldn't get hot. ( I chuck that in as naive suggestion, in the hope that the experts will entertain me with all sorts of other ways that heat/smoke will be generated.)
You should have been more worried about your capacitor, because if there had truly been no resistance and no inductance in that circuit, that's where you'd have gotten infinite current. (I think inductance is what always saves the day in these "zero resistance" problems ! )
* ( I think a circuit without inductance would be impossible, since all conductors possesses inductance, even if they have zero resistance.)
rahaverhma said:
... So, when the inductor will have chance for the induction of emf(my view: not possible)? ...
As soon as current starts to flow, the magnetic field in the inductor will change. As soon as the field starts to change, an emf is generated.
So immediately you connect the battery, there is an emf.
That emf is equal to the inductance times the rate at which the current starts to increase (from its zero starting point.)
And the current, which is zero up until you make the connection, then increases steadily with time from that zero starting point, at the rate ##\frac {V} {L} ##
I expect this is the expression you mention, ## i = \left( \frac {E} {L} \right) *t ## because I've been using V for E.
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Just as a couple of BTW points, which might interest you;
It would not matter how much resistance the wire had (or the battery, or the inductor itself) the current in this circuit must start from zero and it will always start at the rate ## \frac {V} {L} ##
Although it is easy to start the current flowing, you cannot stop it instantly. If you try, the current must continue for a while, either by charging up stray capacitance, or if that's not enough, by continuing to flow through a spark. This is a consequence of your expression above, which transforms to
## V= L \frac { i} {t} ## so if ## t ## is very small, ## V ## becomes very large.