current in coil

[stefan]

Hi
When I power up a DC coil the current increases rather slowly, so they
say. But is this true for the very, very first moment? You do need a
current to get a counter EMF or???? At atomic level, is the current
equal everywhere in the coil?

 

[John Larkin]

On 11 Jul 2003 04:17:20 -0700, stefan.haesen@pandora.be (stefan)
wrote:

Hi
When I power up a DC coil the current increases rather slowly, so they
say. But is this true for the very, very first moment? You do need a
current to get a counter EMF or???? At atomic level, is the current
equal everywhere in the coil?

I = T*V/L

When T=0, I=0. So they say.

John

 

[John Popelish]

stefan wrote:

Hi
When I power up a DC coil the current increases rather slowly, so they
say. But is this true for the very, very first moment? You do need a
current to get a counter EMF or???? At atomic level, is the current
equal everywhere in the coil?

You need a rate of change of current to get an EMF across a coil.
This works, even when the current passes through zero as it changes,
or includes zero as a starting point. From the lumped component
viewpoint, the current is the same everywhere in the coil. From a
fields viewpoint, the applied voltage passes through the wire as a
wave, at the local speed of light. This wave has to charge all the
stray capacitances as it proceeds, so things get a little complicated
on the very small and very fast scale.

--
John Popelish

 

[Pieter Hoeben]

On 11 Jul 2003 04:17:20 -0700, stefan.haesen@pandora.be (stefan)
wrote:

Hi
When I power up a DC coil the current increases rather slowly, so they
say. But is this true for the very, very first moment?
Yes for ideal coils. No for coils with paracitic capacitance.

You do need a
current to get a counter EMF or????
U = L* dI/dT

When the current changes (dI) in the time (dT), a voltage will be
enerated depending on the inductance L. Because you place a voltage
on the coil, a current will be generated, increasing in time (assuming
an ideal coil with 0 ohm resistance): dI/dT = U/L depending on the
voltage and inductance. As you place the voltage on the coil, the
voltage of the counter emf IS the voltage you place on the coil? In
other words, the current changes so the coil counter emk's up to you
supplied voltage.

When there is a resistor in the coil, the counter EMK will drop with
the current, counter EMK = (Supplied voltage) - I * R.
Once the highest current is reached (after several L*R times), the
counter EMK will be almost zero, as the current doesn't change
anymore.

At atomic level, is the current equal everywhere in the coil?
Yes

Regards,

Pieter Hoeben
email: pieter@hoeben.com

 

[William J. Beaty]

stefan.haesen@pandora.be (stefan) wrote in message news:<47ea2dab.0307110317.54b9f31c@posting.google.com>...

When I power up a DC coil the current increases rather slowly, so they
say. But is this true for the very, very first moment?

Are you talking about an ideal inductor? Real inductors are
extremely complicated. They're waveguides and radio antennas
and tuned circuits and inductors all in one. As soon as you
stop seeing them as ideal inductors, all the complexity comes
boiling out.

In the first few hundred picoseconds, waves of EM energy
are distributing a voltage-pattern through all the parts
of the inductor. Yet usually we can't turn the power
supply on in a matter of picoseconds. If it takes
hundreds of thousands of picoseconds for the voltage to
rise, then the energy-waves are too small to notice.

You do need a
current to get a counter EMF or????

You need an EMF to cause a current to begin rising.
Voltage is associated with e-fields and with the
push/pull upon electrons, so voltage causes current.

At atomic level, is the current equal everywhere in
the coil?

In ideal inductors, yes. In real inductors, no, not
in the first fractions of nanoseconds while waves are
bouncing around between different parts of the coil.
(Imagine sound waves bouncing around inside a long
plastic tube. Now imagine that you slowly raise the
speed of air flow in that tube without creating any
sound waves.)