Battery connected to a pure inductor with a Zero-resistance wire

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I know the mechanism of a circuit containing just a capacitor connected to battery in resistance less wire. The particles are just tending towards making same emf on capacitor.
But, I am not able to think clearly for the case of inductor. As, we connect battery and as resistance is 0.So,current will be infinite in 0 seconds. So, when the inductor will have chance for the induction of emf(my view: not possible)? Also, there an expression comes out of result of kirchoff's voltage law i. e. " i=-(E/L) *t, i=current, t=time". Please, help me.
 

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  • #2
jbriggs444
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But, I am not able to think clearly for the case of inductor. As, we connect battery and as resistance is 0.So,current will be infinite in 0 seconds.
An inductor resists changes in current. A change from zero to infinity in zero seconds is rather a high rate of change. An inductor would resist that.
 
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Thank you for the logic.
 
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CWatters
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The equation for the voltage on an inductor is...

V=Ldi/dt

di/dt is the slope or rate of change of current.

So if you apply a voltage of say 1V to an inductor of 1H then the current increases at a rate of 1A/s. So after 10 seconds its 10A. After 20 seconds it's 20A etc. Soon the smoke escapes.
 
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CWatters
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PS... I'll let you work out how fast the current increases for a 1mH (1*10^-3 H) inductor. Something else in the circuit might effect the result.
 
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PS... I'll let you work out how fast the current increases for a 1mH (1*10^-3 H) inductor. Something else in the circuit might effect the result.
As a current even starts to gain a slight value in the circuit, for even such a very short value of current like 0.000001mA, the slope (di/dt) would still be very large or same for the current rising afterwards. And thus, voltage across the inductor will be infinity or very large. Doesn't it violate the kirchoff's voltage law? Or, if it does not, how the battery is in action?
 
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jbriggs444
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As a current even starts to gain a slight value in the circuit, for even such a very short value of current like 0.000001mA, the slope (di/dt) would still be very large
Can you explain why you think this?

Surely, if the current attains a value of 0.000001 mA after a time of 0.000001 seconds then the applied voltage to produce this result would be 1 millivolt.
 
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Surely, if the current attains a value of 0.000001 mA after a time of 0.000001 seconds then the applied voltage to produce this result would be 1 millivolt.
You may be right.
So, is there anyone and you who knows the mechanism of this circuit?
 
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jbriggs444
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You may be right.
So, is there anyone and you who knows the mechanism of this circuit?
The behavior is specified in the title of this thread: "pure inductor". The mechanism is irrelevant if the behavior is specified.
 
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CWatters
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As a current even starts to gain a slight value in the circuit, for even such a very short value of current like 0.000001mA, the slope (di/dt) would still be very large or same for the current rising afterwards. And thus, voltage across the inductor will be infinity or very large. Doesn't it violate the kirchoff's voltage law? Or, if it does not, how the battery is in action?
Going back to the equation I posted...

V=Ldi/dt

I was thinking of the situation where you apply a fixed voltage say 1V across the inductor, what happens to the current...

So rearrange the equation to..

di/dt = V/L

The smaller the inductance L the faster the current I increases.

Plug in numbers I suggested...

di/dt = 1V/1mH = 1000A/S

So in 1 second the current would reach 1000A (Assuming the voltage source and wires could supply that much and the inductor didn't melt :-)
 
  • #11
Merlin3189
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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 possess inductance, even if they have zero resistance.)

... 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.
=================================================
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.
 

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