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Re: [Phys-L] standard dc circuits

So the fields that cause charges to move are due more to the charfes that are near them than the charges in the battery. The q/rr rules make sense there.

If the majority of the field is due to the charges closest to the charge to be moved, does this work...
Starting at the node prior to the 2 parallel resistors. A charge q is present in the node. I would think the field it creates to move charges in each of the upcoming parallel braches would be the same as the field if there were only one wire leaving that node.
Perhaps the reason there is less current in each of those branches is because of the charges that accumulate at the nearest edge of the lamp or resistor. The field those accumulating charges produce is opposite the field of the charge in node X. The sum of the fields in that branch would be less and so there would be less current.

Am I getting closer?

Sorry my circuit drawing abilities are limited right now. I am using my phone and don't have pc access.

I am comfortable using kirchoff for loops and junctions. I am trying to learn how to analyze things from a fields perspective. I figure it cant hurt, and it is always good to be able to analyze things in different ways. I was thinking about using gravitational field analogies more to address current and potential difference in dc circuits.


Thanks for your help. I appreciate it.

.:. Sent from a touchscreen .:.
Paul Lulai



-------- Original message --------
From: Bob Sciamanda
Date:11/22/2013 4:16 PM (GMT-06:00)
To: Paul Lulai ,Phys-L@Phys-L.org
Subject: Re: [Phys-L] standard dc circuits

The transient situation which occurs just after switching on the circuit is
impossibly complicated to model in any detail. But once things have settled
down to a steady state DC, we have a useful mathematical model of the
situation. Div E = rho/epsilon, j = sigma*E, etc. I don't understand your
reference to the cross-sectional area of the wires - what has this got to do
with anything? The potential difference across a resistor is produced by
the field of charges which have accumulated at the resistor terminals -
which occur because of a discontinuity of conductivity at the resistor
terminals. Follow the math.



Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

-----Original Message-----
From: Paul Lulai
Sent: Friday, November 22, 2013 4:21 PM
To: Phys-L@Phys-L.org ; Bob Sciamanda
Subject: RE: [Phys-L] standard dc circuits

The last line of Bob's statement is what I am trying to clean up for myself.
- because the fields are half as strong.
Why is it that the fields are half as strong at that location? Why does the
cross-sectional area of the wire affect the field produced by the batteries?


-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of LaMontagne, Bob
Sent: Friday, November 22, 2013 2:22 PM
To: Phys-L@Phys-L.org; Bob Sciamanda
Subject: Re: [Phys-L] standard dc circuits

I don't know what level of student you are considering, but let's assume a
traditional Intro Physics -College or High School.

You describe a simple battery with three lamp filaments. Assuming no
resistance in the connecting wires and a 1.5 Volt battery, the filament of
the single resistor will have a 1 Volt potential drop across it. Each of
the two lamps connected in parallel will have a 0.5 Volt potential across
them. The field in the single resistor can be found by dividing the 1.0
Volts by the length of the filament. The field in the parallel resistors
will be 0.5 Volts divided by the length of the individual filaments (assume
all the same for simplicity). The current in each will be the conductivity
multiplied by the cross sectional area of the individual filaments and then
multiplied by the field. That gives the current in each of the parallel
resistors which is half the current in the single resistor - because the
fields are half as strong.

Bob at PC

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Bob
Sciamanda
Sent: Friday, November 22, 2013 2:26 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] standard dc circuits

It might help to realize that the fields are produced by charges which
accumulate on wire surfaces AND wherever there is a change in
conductivity, eg at resistor terminals.



Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

-----Original Message-----
From: Paul Lulai
Sent: Friday, November 22, 2013 2:11 PM
To: Phys-L@Phys-L.org
Subject: [Phys-L] standard dc circuits

I have a question about the junction rule for simple dc circuits.
I have a battery connected to 1 lamp, then split to 2 lamps in
parallel and back to the battery.
For easy numbers, if 1 amp flows through the first lamp, then 0.5 amps
flow through each lamp in parallel (if all lamps are ideal).
Using the junction rule, I know that current into each junction or
node must equal the current out of each junction or node. It is
simply a conservation of charge in that area.
How does that reconcile with a fields approach? I might have some
incorrect notions below. I am open to correction.
The batteries produce a field that move charges already present in the
wire.
If all wires are of the same material, diameter, length (and so on)
why is the current less in the parallel branches than in the wires
before and after the parallel branch? Does something make the field
in the parallel branches smaller than the field in the series portion
before and after the parallel branch? I don't see it.
Thanks for any input.
Paul.

_______________________________________________
Forum for Physics Educators
Phys-l@phys-l.org
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_______________________________________________
Forum for Physics Educators
Phys-l@phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l