Re: emf, potential, voltage

[Bob Sciamanda <trebor@VELOCITY.NET>]

If n = # of carriers per unit volume,
q = charge per carrier,
A=cross sectional area of a cylindrical wire,
v= carrier speed,

Then the current in this wire is I = nqAv

It is easy to arrange n,q and A to be the same in both resistor leads.
Then if I measure I_in = I_out, this model says v_in = v_out.

Some basic assumptions are of course involved, but they are not outlandish.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
www.velocity.net/~trebor
----- Original Message -----
From: "brian whatcott" <inet@INTELLISYS.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, March 13, 2001 12:03 AM
Subject: Re: emf, potential, voltage


> At 23:20 3/12/01 -0500, Bob Sciamanda wrote:
> > Consider any two terminal device with a current of I amperes flowing
> > through it down a potential hill of V volts.  That means there is an
> > electrostatic field in the device which is giving energy to the charge
> > carriers as they pass through the device, down the potential hill.  If
this
> > were the whole story, the carriers would emerge from the device with
> > increased kinetic energy, and the current out of the device would exceed
the
> > current into the device.
> /snip/
> > Bob Sciamanda (W3NLV)
> > Physics, Edinboro Univ of PA (em)
> > trebor@velocity.net
> > www.velocity.net/~trebor
>
>
> Consider a stream of vehicles entering an on ramp at 10 miles per hour.
> Let us call the current or 'traffic flow rate' x vehicles per hour.
>
> Before they  merge on the thoroughway, vehicles match speed with the
> through way traffic travelling at 30 miles per hour.
>  What is their flow rate at this time?
>
> (They have gained kinetic energy, but the on ramp exit current has
> *not* increased, the traffic flow rate is *still* x vehicles per hour)
>
>
>
> brian whatcott <inet@intellisys.net>  Altus OK
>                 Eureka!