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Home > Archive > Electrical Engineering > April 2007 > voltage drop
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| conrad 2007-04-20, 8:25 pm |
| If you have a series circuit with two resistors on that
circuit, and your power supply is a 6 volt battery, then
given that both resistors are 1 ohm, you would end up
with 3 amps. This means that across each resistor,
you would have a 3 volt drop. My question is this, the
voltage drop represents some voltage lost, right? If that
is the case, then by the time current has moved past
the second resistor, you would have 0 volts! In which case,
at this given point past the resistor, suppose you had another
inch of wire for the free electrons to move across, they wouldn't
be able to! Because we no longer have any voltage.
Is my thinking wrong? What is at conflict, is this. This is the
model taught in my electronics class, yet realistically, this
can't happen as those free electrons do find their way to
the positive end of the battery so that the chemical reaction
in the battery can take place again, to give up free electrons
all over again. So what is the problem here?
--
conrad
| |
| no_one 2007-04-20, 8:25 pm |
| you are wrong for two reasons:
1) there is either a real, finite, resistance to the wire in which case you
have additional "resistors" in your voltage divider affecting your
calculation.
2) there is 0 resistance in the wire and therefore it takes 0 volts to move
current (superconductor).
"conrad" <conrad@lawyer.com> wrote in message
news:1177110393.846695.242660@o5g2000hsb.googlegroups.com...
> If you have a series circuit with two resistors on that
> circuit, and your power supply is a 6 volt battery, then
> given that both resistors are 1 ohm, you would end up
> with 3 amps. This means that across each resistor,
> you would have a 3 volt drop. My question is this, the
> voltage drop represents some voltage lost, right? If that
> is the case, then by the time current has moved past
> the second resistor, you would have 0 volts! In which case,
> at this given point past the resistor, suppose you had another
> inch of wire for the free electrons to move across, they wouldn't
> be able to! Because we no longer have any voltage.
>
> Is my thinking wrong? What is at conflict, is this. This is the
> model taught in my electronics class, yet realistically, this
> can't happen as those free electrons do find their way to
> the positive end of the battery so that the chemical reaction
> in the battery can take place again, to give up free electrons
> all over again. So what is the problem here?
>
> --
> conrad
>
| |
| conrad 2007-04-20, 8:25 pm |
| On Apr 20, 4:45 pm, "no_one" <no_...@verizon.net> wrote:
> you are wrong for two reasons:
>
> 1) there is either a real, finite, resistance to the wire in which case you
> have additional "resistors" in your voltage divider affecting your
> calculation.
>
> 2) there is 0 resistance in the wire and therefore it takes 0 volts to move
> current (superconductor).
Ah. So if after the two resistors, you experienced two voltage
drops, both 3 volts, adding up to the total voltage of
a 6 volt battery, then after the second resistor, when you
have zero volts, and because the resistance is really
low in the wire, current is still able to flow without
the voltage?
--
conrad
| |
| no_one 2007-04-20, 9:25 pm |
| think of water pipes with a couple of restrictions; the restrictions
represent the resistors. in the return leg of the pipes (downstream side)
your pressure relative to the outside of the pipe is low or zero and yet the
water still flows.
"conrad" <conrad@lawyer.com> wrote in message
news:1177116944.294392.216420@p77g2000hsh.googlegroups.com...
> On Apr 20, 4:45 pm, "no_one" <no_...@verizon.net> wrote:
>
> Ah. So if after the two resistors, you experienced two voltage
> drops, both 3 volts, adding up to the total voltage of
> a 6 volt battery, then after the second resistor, when you
> have zero volts, and because the resistance is really
> low in the wire, current is still able to flow without
> the voltage?
>
> --
> conrad
>
| |
| Ben Miller 2007-04-20, 9:25 pm |
| "conrad" <conrad@lawyer.com> wrote in message
news:1177116944.294392.216420@p77g2000hsh.googlegroups.com...
> Ah. So if after the two resistors, you experienced two voltage
> drops, both 3 volts, adding up to the total voltage of
> a 6 volt battery, then after the second resistor, when you
> have zero volts, and because the resistance is really
> low in the wire, current is still able to flow without
> the voltage?
That is not what he said. In case 1, the wire adds a small resistance (let's
say 0.1 Ohms for our example). Therefore, the current is now 6/2.1=2.857A,
and the voltage across the resistors is 5.714V. The remaining voltage is
across the wire.
In case 2, the wire is "ideal" with zero ohms resistance, and there is no
voltage across it. It therefore doesn't exist, and the circuit is the same
as the original. The current is 3A and the voltage across the resistors is
6V.
No matter what the wire resistance, by Ohm's law there will always be some
voltage across it if current is flowing.
Ben Miller
--
Benjamin D. Miller, PE
B. MILLER ENGINEERING
www.bmillerengineering.com
| |
| Chugga Chug 2007-04-24, 9:25 am |
| > If you have a series circuit with two resistors on that
> circuit, and your power supply is a 6 volt battery, then
> given that both resistors are 1 ohm, you would end up
> with 3 amps. This means that across each resistor,
> you would have a 3 volt drop. My question is this, the
In a real-world circuit there is NEVER "no resistance" (0 Ohms). There is
always a finite resistance, and an accompanying voltage drop. This may be
micro-volts, pico-volts, or whatever, but as long as any current flows there
will always be a voltage.
Zero Ohms is not practical. We can get rather close to it, but never zero.
There is therfore always a voltage difference, as long as current flows.
Electrons are also a lot more sensitive to voltage than a 1K-Ohm AVO. They
still get the general idea as to where to go :-)
| |
| Engineer 2007-04-29, 5:25 pm |
| On Apr 20, 9:56 pm, "Ben Miller" <benmil...@worldnet.att.net> wrote:
> "conrad" <con...@lawyer.com> wrote in message
>
> news:1177116944.294392.216420@p77g2000hsh.googlegroups.com...
>
>
> That is not what he said. In case 1, the wire adds a small resistance (let's
> say 0.1 Ohms for our example). Therefore, the current is now 6/2.1=2.857A,
> and the voltage across the resistors is 5.714V. The remaining voltage is
> across the wire.
>
> In case 2, the wire is "ideal" with zero ohms resistance, and there is no
> voltage across it. It therefore doesn't exist, and the circuit is the same
> as the original. The current is 3A and the voltage across the resistors is
> 6V.
>
> No matter what the wire resistance, by Ohm's law there will always be some
> voltage across it if current is flowing.
>
> Ben Miller
>
> --
> Benjamin D. Miller, PE
> B. MILLER ENGINEERINGwww.bmillerengineering.com
True, but still convoluted. I suggest everyone goes back and reads
Thevenin on equivalent generators and source impedance.
Cheers,
Roger
| |
| Don Kelly 2007-04-29, 8:25 pm |
| ----------------------------
"Engineer" <analogdino@rogers.com> wrote in message
news:1177875791.173038.163250@n76g2000hsh.googlegroups.com...
> On Apr 20, 9:56 pm, "Ben Miller" <benmil...@worldnet.att.net> wrote:
>
> True, but still convoluted. I suggest everyone goes back and reads
> Thevenin on equivalent generators and source impedance.
> Cheers,
> Roger
--------------------------------------------
Go back further than Thevenin. What Ben stated has nothing to do with
Thevenin as it is more basic than Thevenin
Thevenin is based on
a) Kirchoff's laws
b)Linearity (which, strictly speaking, is what makes Ohm's Law true. )
a) is the main thing in consideration of series or parallel circuits -or
any combination. Thevenin is secondary- a very useful circuit analysis
approach but hardly fundamental and easily developed from a), b) above.
I would suggest that you, not Ben, are the one introducing unnecessary
convolution.
Cheers
--
Don Kelly dhky@shawcross.ca
remove the X to answer
>
| |
| BFoelsch 2007-04-29, 9:25 pm |
| On Mon, 30 Apr 2007 01:08:31 GMT, "Don Kelly" <dhky@shaw.ca> wrote:
>----------------------------
>
>"Engineer" <analogdino@rogers.com> wrote in message
>news:1177875791.173038.163250@n76g2000hsh.googlegroups.com...
>--------------------------------------------
>Go back further than Thevenin. What Ben stated has nothing to do with
>Thevenin as it is more basic than Thevenin
>
>Thevenin is based on
>
>a) Kirchoff's laws
>b)Linearity (which, strictly speaking, is what makes Ohm's Law true. )
>
>a) is the main thing in consideration of series or parallel circuits -or
>any combination. Thevenin is secondary- a very useful circuit analysis
>approach but hardly fundamental and easily developed from a), b) above.
>
>I would suggest that you, not Ben, are the one introducing unnecessary
>convolution.
>
>Cheers
The problem with this whole silly thread is that the OP (and everbody
else) keeps changing the simplifying assumptions.
The initial premise, that two 3 ohm resistors in series draw 2 amps at
6 volts assumes NO resistance in the connecting wires. Then he changes
the premise and starts to argue about the effect of the wires. If they
are real wires then the current simply can't be 2 amps and the voltage
across the resistors can't be 3 volts.. If the wires are real and the
current is 2 amps then the source can't be 6 volts.
And on and on it goes. This stuff was all figured out years ago,
guys!!
| |
| Don Kelly 2007-04-30, 3:25 am |
|
----------------------------
"BFoelsch" <BFoelsch@comcast.ditch.this.net> wrote in message
news:l0ia33hucdhm0i6dsvrh84qnarguc4vslr@4ax.com...
> On Mon, 30 Apr 2007 01:08:31 GMT, "Don Kelly" <dhky@shaw.ca> wrote:
>
>
> The problem with this whole silly thread is that the OP (and everbody
> else) keeps changing the simplifying assumptions.
>
> The initial premise, that two 3 ohm resistors in series draw 2 amps at
> 6 volts assumes NO resistance in the connecting wires. Then he changes
> the premise and starts to argue about the effect of the wires. If they
> are real wires then the current simply can't be 2 amps and the voltage
> across the resistors can't be 3 volts.. If the wires are real and the
> current is 2 amps then the source can't be 6 volts.
>
> And on and on it goes. This stuff was all figured out years ago,
> guys!!
-----------------------
You are right- it is a silly thread and I have joined in the silliness. Mea
Culpa However, there are a lot if silly statements going around -possibly
because of misinterpretation of what others have said.
If I have misinterpreted- then I am sorry but I do find that calling on
Thevenin is not improving the situation nor is calling on Ohms Law to
explain that in a series circuit, the current is the same at evey point in
the circuit.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
| |
| BFoelsch 2007-04-30, 3:25 am |
| On Sun, 29 Apr 2007 21:49:10 -0400, BFoelsch
<BFoelsch@comcast.ditch.this.net> wrote:
>On Mon, 30 Apr 2007 01:08:31 GMT, "Don Kelly" <dhky@shaw.ca> wrote:
>
>
>The problem with this whole silly thread is that the OP (and everbody
>else) keeps changing the simplifying assumptions.
>
>The initial premise, that two 3 ohm resistors in series draw 2 amps at
>6 volts assumes NO resistance in the connecting wires. Then he changes
>the premise and starts to argue about the effect of the wires. If they
>are real wires then the current simply can't be 2 amps and the voltage
>across the resistors can't be 3 volts.. If the wires are real and the
>current is 2 amps then the source can't be 6 volts.
>
>And on and on it goes. This stuff was all figured out years ago,
>guys!!
Of course, my memory is twisted. Initial premise was that 2 one ohm
resistors in series across 6 volts produce a current of 3 amps.
The rest of my criticism stands, however.
| |
| SuperM@ssiveBlackHoleAtTheCenterOfTheMilkyWayGalax 2007-04-30, 1:25 pm |
| On Sun, 29 Apr 2007 21:49:10 -0400, BFoelsch
<BFoelsch@comcast.ditch.this.net> Gave us:
>On Mon, 30 Apr 2007 01:08:31 GMT, "Don Kelly" <dhky@shaw.ca> wrote:
>
>
>The problem with this whole silly thread is that the OP (and everbody
>else) keeps changing the simplifying assumptions.
>
>The initial premise, that two 3 ohm resistors in series draw 2 amps at
>6 volts assumes NO resistance in the connecting wires. Then he changes
>the premise and starts to argue about the effect of the wires. If they
>are real wires then the current simply can't be 2 amps and the voltage
>across the resistors can't be 3 volts.. If the wires are real and the
>current is 2 amps then the source can't be 6 volts.
>
>And on and on it goes. This stuff was all figured out years ago,
>guys!!
And Ohm's law applies at any point, and across any given element in
the circuit at ALL times!
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