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Inductor

Inductor

by Shashank Todwal -
Number of replies: 13

When a coil carrying steady current is short-circuited,the current becomes 1/n times initial current in time to.The time constant of the circuit is?

Ans:to/ln n

My Doubt:Steady current initially==> a DC source must be present, lets say cell of emf E.Resistance is ofcourse present.Now only inductor is shorted.So here we can't use the concept of decay of current wherein we assume that E is removed suddenly and we see the nature of i.Here E is not removed.But still the ans comes only by using this concept wherin we develop i=ioe-(R/L)t.PLz reply

In reply to Shashank Todwal

Re: Inductor

by Chandni Bhatia -
Initially, a constant current was flowing through inductor; inductor won't mind this ,inductor would  simply act as  wire or a resistor[ if it has some resistance].Now, inductor is short circuited and   current produced by the cell will not flow through it .Inductor won't like this and will send current in opposite direction .This is as good as the process in which current is decaying when cell has been removed  .
In reply to Chandni Bhatia

Re: Inductor

by Shashank Todwal -

Dear Chandni,try to write the equation for the situation and you will see what I mean to say.Here I can't write L(di/dt)=iR and so I think i=i0e-(R/L)t is invalid.

In reply to Shashank Todwal

Re: Inductor

by Chandni Bhatia -
Sorry for late reply.When inductor is short circuited,it sends current in opposite direction and sets an emf equal in magnitude of the emf of the cell connected but the emf generated across the inductor is opposite in polarity  and hence we can conclude that current is decaying
In reply to Shashank Todwal

Re: Inductor

by Chandni Bhatia -
When inductor is short circuited then the current flowing due to the cell connected does not reach the inductor.Now inductor  does not like the change in current .Whatever current was flowing through the inductor ,in say,right direction will now flow in left direction . E.M.F of the cell = iR where i is the current that was flowing in the inductor when there was no short circuit .As,said earlier inductor does not like the change in current so now it will send the current of same magnitude but in opposite direction and therefore it will set up an EMF across it  of equal magnitude of the emf of the cell but opposite in polarity . In problems where we   see the decay of current when cell  is removed ,we use kirchoff's law.Here emf across inductor is Ldi/dt=iR IS STILL APPLICABLE.EMF across inductor is iR [OHM'S LAW] and EMF=Ldi/dt [from  electromagnetic induction chapter]
In reply to Chandni Bhatia

Re: Inductor

by Shashank Todwal -
Hey friends please clear my doubt.Sorry Chandni,still not convinced.
In reply to Shashank Todwal

Re: Inductor

by Chandni Bhatia -

Can you tell me where you have not got the explanation ? if my explanation seems to be wrong ,please point out.

In reply to Chandni Bhatia

Re: Inductor

by Manoj Tirukodi Radhakrishn -

Hey todwal i think you are worrying too much man . 

You see bro if you just connect the ends of the inductor by a conducting wire your job is done. What happens to the cell after that is immaterial to you Just see that all you want to do is short circuit it and thats it

But here comes a complication because if you do that your cell would still give you a current due to the resistor if it was not the inductor itself.So you assume that the inductor itself is the Resistor we are talking about.More over if your inductor itself is not the resistor you can never short it effectively i.e you cant have a decay initiated.

Moreover in the question i dont think he mentioned about any ideal inductor stuff so my assumption is obviously sesible.

Now writing the equation is childs play, you should be knowing that inductor is like an infinite resistor which resists any sudden change.You shold be knowing the definition of inductance and why we write Ldi/dt stuff right.

So writing the loop equation Ldi/dt-iR=0 is not a problem i think and rest is TOO OBVIOUS.The inductor doesnt practically send a current in the opposite direction Like how bhatia specified may be she was only imagining things.

Anyways man physics is full of approximations and assumptions thats why I SIMPLY LOOOOOOOVE PHYSICS.

And chandni you did speak something wrong may be you do know it .

And todwal i'll kill you if you say you didnt understand it ANYWAYS PLEASE REPLY AS FAST AS POSSIBLE 

In reply to Manoj Tirukodi Radhakrishn

Re: Inductor

by Chandni Bhatia -
Manoj, I did not understand why you have written that inductor is an infinite resistor .If this would have been the casethen see XL=OMEGA L. for dc. supply omega=0 and XL=0 .
In reply to Chandni Bhatia

Re: Inductor

by Manoj Tirukodi Radhakrishn -

Ya chandini thats a great way to close someones mouth who really analyses the whole situation

Its like this girl, The inductor is a one which resists sudden changes that is at the moment of sudden change i.e when current is suddenly switched on, all that a inductor does is that it develops apposite EMF.

So I only said that the inductor is like an infinite resistor but not really that it as an infinite resistor. If i have confused you i am sorry , But since you solved that integration problem in CTP i consider you to be a smart one anyways how on earth did you crack it

In reply to Manoj Tirukodi Radhakrishn

Re: Inductor

by Chandni Bhatia -

Manoj,there was nothing great in that problem .I had given you the solution for that problem when I asked a question on reverse biasing .Anyway, I am giving you the solution again.

INTEGRATION PROBLEM:

Differentiate the integral with respect to e.Inside the integral will be partial differentiation with respect to e .While integrating w.r.t to e ,treat x as a constant .After you have done the partial differentiation w.r.t to e treat x as a constant  .Then your equation would be dI/de =integral of [cos2x /1+ 4e(e-1)cos2x] dx{ within given limits} Now,divide by cos2x .The integral will be now in sec2x. Put tanx=t ,change the limits.dI/de=integral[dt/(t2+1){t2+1+4e(e-1)}] within proper limits.use partial fractions and the integral will convert in tan-1x .You will get dI/de= 3.14/e , integrate this.You will get get I= 3.14 log e +c where c= integration constant . NOTE  that here e has been chosen as a variable and therefore I can put any value of e to get the integration constant. Put e=1 in the obtained integral and in the given integral ,you will find that c=0 .   

In reply to Chandni Bhatia

Re: Inductor

by Manoj Tirukodi Radhakrishn -

Oh Yeah! parameter differentiation actually we werent aware of this type of integration

Anyways thanks friend.

In reply to Manoj Tirukodi Radhakrishn

Re: Inductor

by Chandni Bhatia -
Manoj , I COULD NOT UNDERSTAND WHAT YOU SAID IN YOUR PREVIOUS TO PREVIOUS  REPLY.ALSO, I CANNOT PONT OUT MY MISTAKE.PLEASE GIVE YOUR EXPLAINATION AGAIN.