How an inductor acts as an open circuit to DC at starting?

[abdulhafeez]

hi

Can any one explain how an inductor acts as an open circuit to dc at the starting ?



thanks

 

[yjkwon57]

Re: inductor starting

Since the reactance of an inductor is given as X(L) = jwL, where w = 2(pi)f, when A DC voltage is applied to the inductor, the f approaches to infinite at the starting time, the reactance becomes infinite, thus, the inductor seems to be opened component.

 

[abdulhafeez]

Re: inductor starting

Actualy i want to know what physically is happening over there in the inductuctor that makes it to show such a behavior.

 

[sinatra]

Re: inductor starting

Hello All.
If you remember your basic lecons in the University you will probably understand.

1-The relation between current and voltage in an inductor is:

Eq.1
V = L * dI/dt

or

Eq.2
I = (1/L) * Integral ( V dt ) + I0

I= current crossing the inductor
V= Voltage across the inductor
I0= initial condition, continuous current crossing the inductor at t<0, t=0 and t>0.


2-Looking at the Eq.2 you can see that at the time t=0, the moment when the voltage V was applied to the inductor the current crossing the inductor will be the integral of the applied voltage plus a continuos current (stead state).

3-Assuming for simplicity that second term I0 is equal to zero (there was no continuous current crossing the inductor wen the voltage was applied).

4-Now, looking at the first term you can see that for any value of the applied voltage V the integral will be zero at t=0 (when the voltage was applied).

5-The current will be the summ of the 2 terms (first and second of Eq.2), that means, the current is zero at t=0.


6-The impedance offered by the inductor is:

Z=dV/dI

Z= impedance of the inductor in function of the time.
V= applied voltage, V=0 for t<0, V=V for t>=0.
I= current crossing the inductor given by Eq.2

7-According to item 5, the current is equal to zero at t<0 and at t=0. So dI, at t=0, i.e., the difference of the current just before t=0 and at t=0 is zero. And will grow linearly with the time. That means that the impedance will be infinity at t=0 (since dI=0 at t=0), and will decrease following 1/x relation with time.

Hope it helps.
S.

 

[ender84567]

inductor starting

look to maxwells equations if you need more of an explination than above.

 

[yjkwon57]

Re: inductor starting

Right!, after solving the Maxwell's equation, you will get v = L * di / dt, which becomes v = j * w * t in the steady-state.

 

[aryajur]

Re: inductor starting

Physically this can be explained as follows - the basic law of electomagnetics says that when you have a potential difference then charge will flow from higer potential to lower potential if it is not stopped. So now if you have an inductor and place a potential difference across it, ie a DC voltage current will start flowing from positive to negative terminal through the inductor wire since thats the path of least resistance. From Nature we see that any moving charge (current) produces a magnitic field. Now again from Nature we know that a changing Magnetic field produces a potential difference and hence a current if there is a path.
The only time when the magnetic field produced by a DC source is changing is when the source switches on and the voltage climbs up to the DC value, so we have a very fast changing magnetic field, in fact if you see from right hand thumb rule it will have a proper direction and it will be expanding out of the wire. Now from the Right hand rule you can determine the direction of the induced current by this magnetic field on the conductor itself and that will turn out to be opposite to the direction of current producing it(this is in fact Lenz law or the Law of conservation of energy in disguise). Since the turn On time is so short and plus the inductor is a conductor made in such a way that the induced current is maximized, it turns out that the magnitude of induced current is almost the same as that of the current that tries to produce it and hence the net current flowing in the conductor is nil. Thus at turn ON it seems that the inductor is a open circuit with no current flowing in it.
This is how you can imagine it physically without looking into equations and stuff.

 

[rigdoctor99]

inductor starting

Your all wrong, its all about the angle of the dangle and the vector of the sector........

 

[pmonon]

Re: inductor starting

Your all wrong, its all about the angle of the dangle and the vector of the sector........

You are right you are just short of b for bright...

 

[Electronica]

inductor starting

XL=2(pi)fL ,therefore at dc f=0 and XL=0 thus it acts like short circuit

 

[pmonon]

Re: inductor starting

XL=2(pi)fL ,therefore at dc f=0 and XL=0 thus it acts like short circuit

It's when steady state, not during starting which is the subject here.