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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.
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_______________________________________________
Forum for Physics Educators
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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