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meaning of voltage drop
|
|
| conrad 2007-04-22, 5:25 pm |
| I've read conflicting definitions for voltage drop.
Some sources describe it as 'voltage lost'
while others 'voltage difference'. I tend to favor
the latter. My reasoning is as follows: If voltage
was lost, then adding say additional resistors
to a series circuit would mean that there would
be a lack of voltage to continue current.
Why I think voltage drop means 'voltage difference'
is for this reason: resistors impede the flow
of current, so resistance goes up, while current
stays the same as opposed to current staying the
same but resistance being very small. The former
implies a smaller voltage while the latter does not.
Given this smaller voltage, we have a reduction in
pressure operating over an electric field. So the time
that it takes to push a charge becomes longer.
This would then be represented by a voltage drop,
or the difference between two points in a circuit
over some load. It represents a disparity in voltage
over time. However, voltage is not lost. Because the
voltage after the resistor, would be represented by
the applied voltage.
Does that sound correct? If not, why not?
--
conrad
| |
| Don Kelly 2007-04-22, 8:25 pm |
| ----------------------------
"conrad" <conrad@lawyer.com> wrote in message
news:1177272496.789216.147520@q75g2000hsh.googlegroups.com...
> I've read conflicting definitions for voltage drop.
> Some sources describe it as 'voltage lost'
> while others 'voltage difference'. I tend to favor
> the latter. My reasoning is as follows: If voltage
> was lost, then adding say additional resistors
> to a series circuit would mean that there would
> be a lack of voltage to continue current.
>
> Why I think voltage drop means 'voltage difference'
> is for this reason: resistors impede the flow
> of current, so resistance goes up, while current
> stays the same as opposed to current staying the
> same but resistance being very small. The former
> implies a smaller voltage while the latter does not.
> Given this smaller voltage, we have a reduction in
> pressure operating over an electric field. So the time
> that it takes to push a charge becomes longer.
> This would then be represented by a voltage drop,
> or the difference between two points in a circuit
> over some load. It represents a disparity in voltage
> over time. However, voltage is not lost. Because the
> voltage after the resistor, would be represented by
> the applied voltage.
>
> Does that sound correct? If not, why not?
>
> --
> conrad
>
------------
Several things don't sound correct.
Voltage drop corresponds to voltage difference across the resistor of
concern. It is the product of current and resistance. Higher resistance with
a given current results in an increase in voltage drop not a decrease as you
imply. Higher current through a given resistance does the same. In both
cases, the "pressure drop" increases. For most situations, resistance and
current are two independent things.
All voltages are voltage differences- that is the voltage at one point with
respect to another point. In a resistor the voltage drop is + in the
direction of conventional current - that is the voltage at the ingoing
terminal is + with respect to the outgoing terminal so the voltage is said
to "drop". Note that I am not mentioning charge here-for most circuit
analysis it is not necessary to refer to it but stick to current which is
defined (conventionally) as the rate of change of +charge with time without
giving a hoot as to the actual charge carriers (+ or -)
Dont get concerned about the time to push a given charge. The electrical
pressure supplied by the source so the pressure typically doesn't change.
wiring resistance (assuming constant wire temperature) is constant. The
current can and will change and depends on the source "pressure" and the
total resistance of load and wiring. Increasing wire resistance for a given
load will reduce the current-In a way the time to push a given quantity of
charge will be longer but don't deal with it that way- consider current
which is the rate of change of charge with time (stick to conventional
current, not electron current as it is easier in the long run -particularly
when you get to AC where the electrons are simply wobbling about and going
nowhere). Don't get hung up with what charge carriers such as electrons are
doing.
If you have a constant voltage source then this source voltage is
distributed between the load and the wiring. Rather than dealing with charge
per se at the present time. look at Kirchoff's Laws. These are the basic
circuit equations which will not lead you down the garden path. Try to
understand what they mean.
Don Kelly dhky@shawcross.ca
remove the X to answer
| |
|
| conrad wrote:
> I've read conflicting definitions for voltage drop.
> Some sources describe it as 'voltage lost'
> while others 'voltage difference'.
+10 A----[1 ohm]---+
R1 |
B
R2 |
Gnd C----[1 ohm]---+
I = 5 amps
The voltage measured from A to B is 5 volts.
The voltage measured from B to C is 5 volts.
Best to call it voltage drop.
> I tend to favor
> the latter. My reasoning is as follows: If voltage
> was lost, then adding say additional resistors
> to a series circuit would mean that there would
> be a lack of voltage to continue current.
That is not correct. The current will continue - it
will just be a smaller value.
R1 R2
+10 A----[1 ohm]---[1 ohm]---+
|
B
|
Gnd C----[1 ohm]-------------+
I = 10/3 (~3.33) amps
The voltage measured from A to B is 20/3 (~6.67) volts.
The voltage measured from B to C is 10/3 (~3.33) volts.
>
> Why I think voltage drop means 'voltage difference'
> is for this reason: resistors impede the flow
> of current, so resistance goes up, while current
> stays the same as opposed to current staying the
> same but resistance being very small.
That is completely bollixed up.
Given a fixed voltage, current will _decrease_ when
resistance increases. The second clause in your sentence
"as opposed to current staying the same but resistance
being very small"
is completely meaningless.
You've gotten completely confused. Focus on the paragraph
below:
A current through a resistor will cause a _voltage drop_
across that resistor. The formula is E = IR
Voltage (E) equals current (I) times resistance (R)
Learn and use the standard term: _voltage drop_.
> The former
> implies a smaller voltage while the latter does not.
> Given this smaller voltage, we have a reduction in
> pressure operating over an electric field. So the time
> that it takes to push a charge becomes longer.
> This would then be represented by a voltage drop,
> or the difference between two points in a circuit
> over some load. It represents a disparity in voltage
> over time. However, voltage is not lost. Because the
> voltage after the resistor, would be represented by
> the applied voltage.
>
> Does that sound correct? If not, why not?
No, for the reasons described above. Also, voltage *is* lost.
+ 6 -a-[R1]-b-[R2]-c-[R3]-d-[R4]-e-[R5]-f-[R6]--- Gnd
Assume all resistors above are equal value.
If you connect the black lead of your meter to gnd and
measure to each of the following points, you will get the
voltages indicated:
a 6V
b 5V
c 4V
d 3V
e 2V
f 1V
You drop 1 volt across each resistor. The electrical
energy is lost from the circuit - it is converted to heat
energy in the resistor. That heat energy does not convert
back to electrical energy in the circuit. Now, that should
be the last time you ponder voltage being "lost" across
a resistor. From now on, refer to it as voltage drop. It
will be more understandable to others, and will help you
to avoid confusion.
Ed
>
> --
> conrad
>
| |
| conrad 2007-04-23, 3:25 am |
| On Apr 22, 7:14 pm, ehsjr <e...@bellatlantic.net> wrote:
> conrad wrote:
>
> +10 A----[1 ohm]---+
> R1 |
> B
> R2 |
> Gnd C----[1 ohm]---+
>
> I = 5 amps
>
> The voltage measured from A to B is 5 volts.
> The voltage measured from B to C is 5 volts.
> Best to call it voltage drop.
>
>
> That is not correct. The current will continue - it
> will just be a smaller value.
>
> R1 R2
> +10 A----[1 ohm]---[1 ohm]---+
> |
> B
> |
> Gnd C----[1 ohm]-------------+
>
> I = 10/3 (~3.33) amps
>
> The voltage measured from A to B is 20/3 (~6.67) volts.
> The voltage measured from B to C is 10/3 (~3.33) volts.
>
>
>
>
> That is completely bollixed up.
> Given a fixed voltage, current will _decrease_ when
> resistance increases. The second clause in your sentence
> "as opposed to current staying the same but resistance
> being very small"
> is completely meaningless.
>
See, I've been told that current remains constant
in a series circuit. If I am not mistaken, you are
implying that current is variable in a series
circuit. Is that correct?
--
conrad
| |
| no_one 2007-04-23, 3:25 am |
|
"conrad" <conrad@lawyer.com> wrote in message
news:1177301405.583598.4510@p77g2000hsh.googlegroups.com...
> On Apr 22, 7:14 pm, ehsjr <e...@bellatlantic.net> wrote:
>
> See, I've been told that current remains constant
> in a series circuit. If I am not mistaken, you are
> implying that current is variable in a series
> circuit. Is that correct?
>
> --
> conrad
>
have you ever studied electrical circuits? I suggest that you do that as
your assertions are getting a bit foolish. FWIW you need to learn about
ohms law and what it implies. The current in a circuit is related by the
resistance and the voltage. Thus if the resistance changes for a fixed
voltage then so does the current. Forget what you have been told; it is
either wrong or you have completely failed to understand it. Get a basic
book of electrical theory and learn, or go to a reputable community college
(or high school class if you are that young) and get an education.
| |
| Floyd L. Davidson 2007-04-23, 3:25 am |
| conrad <conrad@lawyer.com> wrote:
>See, I've been told that current remains constant
>in a series circuit. If I am not mistaken, you are
>implying that current is variable in a series
>circuit. Is that correct?
E = I * R
Where
E is Voltage
I is Current
R is Resistane
It's called Ohm's Law, for the fellow who discovered it.
Google.com can find dozens of tutorials about it.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
Ukpeagvik (Barrow, Alaska) floyd@apaflo.com
| |
| conrad 2007-04-23, 3:25 am |
| On Apr 22, 9:31 pm, "no_one" <no_...@verizon.net> wrote:
> "conrad" <con...@lawyer.com> wrote in message
>
> news:1177301405.583598.4510@p77g2000hsh.googlegroups.com...
>
>
>
>
>
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> have you ever studied electrical circuits? I suggest that you do that as
> your assertions are getting a bit foolish. FWIW you need to learn about
> ohms law and what it implies. The current in a circuit is related by the
> resistance and the voltage. Thus if the resistance changes for a fixed
> voltage then so does the current. Forget what you have been told; it is
> either wrong or you have completely failed to understand it. Get a basic
> book of electrical theory and learn, or go to a reputable community college
> (or high school class if you are that young) and get an education.- Hide quoted text -
>
I know ohm's law and what it implies. If resistance goes up, and
current is constant, then voltage increases. If it goes down and
current is constant, voltage decreases.
But my book and my electronics teacher state that current in a series
circuit is constant. And they also say that voltage in a parallel
circuit is constant. You are
saying that current in a series circuit is not constant. I think that
all this is mostly coated with a bunch of fluffy explanations.
Consider this, I have a series circuit with
two resistors. So let's say my total resistance is 3. If my applied
voltage is say 6, then
my current is 2 amps. Now, let's add a resistor. And let's say my
total resistance is now 4. 6/4 = 3/2 amps. Why should this proportion
keep holding true if I add resistors?
That is the essential question that I am trying to figure out. This
proportion can be proven through empirical means but what about
theoretically. How can it be shown that it should be true? And not
just some fluffy approximation that is crafted to 'save the
phenomenon' so to speak.
--
conrad
| |
| Beachcomber 2007-04-23, 3:25 am |
| On 22 Apr 2007 22:14:31 -0700, conrad <conrad@lawyer.com> wrote:
[color=darkred]
>On Apr 22, 9:31 pm, "no_one" <no_...@verizon.net> wrote:
The current is only constant if your source supply is defined as a
'constant current source' which can exist, but is usually a special
case analysis.. Most simple resistor circuit analysis assumes a
constant voltage source with the current varying from 0 to the
maximum, and dependant upon the total resistance of the circuit.
Formally, if you assume that a voltage source is positive and the
resitive drops are negative.
(sum of votage sources) + (sum of resitive drops) = 0
in a series circuit.
This is true, even if the current approaches zero. If the current is
zero, then the total resistance must be infinite. The source voltage
simply divides across the resistors in proportion to their individual
resistances.
Conversely, if the resistance goes to zero, theoretically the current
will move to infinity, but in practice this never happens because, in
the real world, fuses blow, circuit breakers trip, there is always
some internal resistance. A constant voltage supply cannot be
maintained against a resistance of zero (a direct short).
| |
| SuperM 2007-04-23, 9:25 am |
| On Mon, 23 Apr 2007 04:31:06 GMT, "no_one" <no_one@verizon.net> Gave
us:
>
>"conrad" <conrad@lawyer.com> wrote in message
>news:1177301405.583598.4510@p77g2000hsh.googlegroups.com...
>have you ever studied electrical circuits? I suggest that you do that as
>your assertions are getting a bit foolish. FWIW you need to learn about
>ohms law and what it implies. The current in a circuit is related by the
>resistance and the voltage. Thus if the resistance changes for a fixed
>voltage then so does the current. Forget what you have been told; it is
>either wrong or you have completely failed to understand it. Get a basic
>book of electrical theory and learn, or go to a reputable community college
>(or high school class if you are that young) and get an education.
>
>
http://www.tpub.com/content/neets/index.htm
| |
| SuperM 2007-04-23, 9:25 am |
| On 22 Apr 2007 22:14:31 -0700, conrad <conrad@lawyer.com> Gave us:
>I know ohm's law and what it implies.
Apparently not. Current IS constant in a circuit of a given fixed
voltage input with a given fixed resistance. Change either, and the
current changes. Until you understand that relationship intimately
(Ohm's Law) you will not understand the concept of voltage drop.
A wire IS a resistor. It just has a very low resistance for shorter
lengths. At longer lengths, it is a higher value, and carries with it
a higher "voltage drop". All resistances drop voltage.
If you cannot grasp those basic terms, you need to start over and it
is as simple as that.
| |
| SuperM 2007-04-23, 9:25 am |
| On Sun, 22 Apr 2007 20:31:51 -0800, floyd@apaflo.com (Floyd L.
Davidson) Gave us:
>conrad <conrad@lawyer.com> wrote:
>
>
> E = I * R
>
> Where
>
> E is Voltage
> I is Current
> R is Resistane
>
>
>It's called Ohm's Law, for the fellow who discovered it.
>
>Google.com can find dozens of tutorials about it.
I think he needs to go back to even more basic understandings than
that. He should start at the beginning.
http://www.tpub.com/content/neets/index.htm
| |
| Don Kelly 2007-04-23, 8:25 pm |
|
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
"conrad" <conrad@lawyer.com> wrote in message
news:1177305271.789488.23850@q75g2000hsh.googlegroups.com...
> On Apr 22, 9:31 pm, "no_one" <no_...@verizon.net> wrote:
>
> I know ohm's law and what it implies. If resistance goes up, and
> current is constant, then voltage increases. If it goes down and
> current is constant, voltage decreases.
> But my book and my electronics teacher state that current in a series
> circuit is constant. And they also say that voltage in a parallel
> circuit is constant. You are
> saying that current in a series circuit is not constant. I think that
> all this is mostly coated with a bunch of fluffy explanations.
> Consider this, I have a series circuit with
> two resistors. So let's say my total resistance is 3. If my applied
> voltage is say 6, then
> my current is 2 amps. Now, let's add a resistor. And let's say my
> total resistance is now 4. 6/4 = 3/2 amps. Why should this proportion
> keep holding true if I add resistors?
> That is the essential question that I am trying to figure out. This
> proportion can be proven through empirical means but what about
> theoretically. How can it be shown that it should be true? And not
> just some fluffy approximation that is crafted to 'save the
> phenomenon' so to speak.
>
> --
> conrad
>---------
I think that you have been misled or misunderstood things.
The basic laws that you have to deal with are Kirchoff's Laws (You should
definitely have dealt with these before considering series and parallel
circuits ).
One is the current law which says that the total current into a junction is
0
that is "what goes in comes out"
The other is the voltage law which says that the sum of the voltages around
a closed loop is 0
That is, "if you go around the block and come back to where you started, you
haven't gone anywhere (in the circuit or physics sense) as you are at
exactly the same place as you started."
In the series circuit, at each element, the current "in" is the same as the
current "out". The current into one resistor is the same current as that
coming out of the previous resistor. The current is the same everywhere in
the circuit. It is "common" rather than constant (and if measured at any
point in the series circuit, the current will be the same).
In the parallel case the voltage across each branch is "common"
In your series case KVL says V=V1 +V2 +V3 +....=R1*I1 +R2*I2 +R3*I3 +...
but I1 =I2 =I3 by KCL so
V=(R1+R2+R3+...)I
and I=V/(R1+R2+R3+ ...)
If R1+R2+R3=3 ohms and V=6V then I=2A as you correctly indicate
If another resistor R4 is added in series then the total resistance is 4
ohms and the current is 6/4 A
If you add another R5=6 ohms the total is 10 ohms and the current is 6/10 A
Increase the resistance for a constant source voltage then the current
decreases.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
| |
| conrad 2007-04-23, 9:25 pm |
| On Apr 23, 6:13 pm, "Don Kelly" <d...@shaw.ca> wrote:
> --
>
> Don Kelly d...@shawcross.ca
> remove the X to answer
> ----------------------------"conrad" <con...@lawyer.com> wrote in message
>
> news:1177305271.789488.23850@q75g2000hsh.googlegroups.com...
>
>
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> I think that you have been misled or misunderstood things.
> The basic laws that you have to deal with are Kirchoff's Laws (You should
> definitely have dealt with these before considering series and parallel
> circuits ).
> One is the current law which says that the total current into a junction is
> 0
> that is "what goes in comes out"
> The other is the voltage law which says that the sum of the voltages around
> a closed loop is 0
> That is, "if you go around the block and come back to where you started, you
> haven't gone anywhere (in the circuit or physics sense) as you are at
> exactly the same place as you started."
>
> In the series circuit, at each element, the current "in" is the same as the
> current "out". The current into one resistor is the same current as that
> coming out of the previous resistor. The current is the same everywhere in
> the circuit. It is "common" rather than constant (and if measured at any
> point in the series circuit, the current will be the same).
> In the parallel case the voltage across each branch is "common"
>
> In your series case KVL says V=V1 +V2 +V3 +....=R1*I1 +R2*I2 +R3*I3 +...
> but I1 =I2 =I3 by KCL so
> V=(R1+R2+R3+...)I
> and I=V/(R1+R2+R3+ ...)
> If R1+R2+R3=3 ohms and V=6V then I=2A as you correctly indicate
> If another resistor R4 is added in series then the total resistance is 4
> ohms and the current is 6/4 A
> If you add another R5=6 ohms the total is 10 ohms and the current is 6/10 A
> Increase the resistance for a constant source voltage then the current
> decreases.
> --
>
> Don Kelly d...@shawcross.ca
> remove the X to answer
> ----------------------------- Hide quoted text -
>
> - Show quoted text -
Yeah, this site http://www.tpub.com/neets/book1/chapter3/1-11.htm
says basically the same thing I have been taught about current.
Living the impression that it is 'constant'.
--
conrad
| |
| no_one 2007-04-23, 9:25 pm |
|
"conrad" <conrad@lawyer.com> wrote in message
news:1177378851.268582.82390@n59g2000hsh.googlegroups.com...
> On Apr 23, 6:13 pm, "Don Kelly" <d...@shaw.ca> wrote:
>
> Yeah, this site http://www.tpub.com/neets/book1/chapter3/1-11.htm
> says basically the same thing I have been taught about current.
> Living the impression that it is 'constant'.
>
> --
> conrad
Re read the text; it says the same current flows through all the components,
not that the current is constant. This is consistent with what you have
been told here that the current is "common". Ohms and Kirchoff's laws
apply.
You appear to be working too hard with these introductory concepts; is this
your first class in the subject?
| |
| Chugga Chug 2007-04-24, 9:25 am |
| > I've read conflicting definitions for voltage drop.
> Some sources describe it as 'voltage lost'
> while others 'voltage difference'. I tend to favor
> the latter. My reasoning is as follows: If voltage
> was lost, then adding say additional resistors
> to a series circuit would mean that there would
> be a lack of voltage to continue current.
I think that you will find that a "voltage drop" is the generally accepted
term to indicate that a voltage is developed across a resistive element when
a current is passed through that resistance. The "voltage drop" occurs
across the terminals of the resistance, and this "drop" is taken away from
other elements in the circuit - eg.
100 Ohms is placed across a 100 Volt source. One ampere will flow. An
additional 1 Ohm resistance is connected in series with the 100 Ohm load.
The 1 Ohm resistance will develop a (almost) 1 Volt drop across its
terminals, leaving less than 100 Ohms to be developed across the load.
Current limiting resistors are often refered to as "dropper resistors",
because of what they do.
| |
| do.not@reply.nonet 2007-04-24, 9:25 am |
| > says basically the same thing I have been taught about current.
> Living the impression that it is 'constant'.
>
> --
> conrad
I think the "constant" bit means, is that it is the "same" at
any point in the circuit. So if you have the simple circuit
of a battery, wires and resistor (lightbulb for good ideas)
and you measure at all points around the ciruit, at the same
moment, then you will find all the measurments of current
are the same. Constant in location!
Robert
| |
| Don Kelly 2007-04-26, 3:25 am |
|
"conrad" <conrad@lawyer.com> wrote in message
news:1177378851.268582.82390@n59g2000hsh.googlegroups.com...
> On Apr 23, 6:13 pm, "Don Kelly" <d...@shaw.ca> wrote:
>
> Yeah, this site http://www.tpub.com/neets/book1/chapter3/1-11.htm
> says basically the same thing I have been taught about current.
> Living the impression that it is 'constant'.
>
> --
> conrad
----------------
The reference is OK. It uses "same current" which has a different meaning
than "constant current".
If you add a resistor, the current changes- but wherever you measure it, it
is the same value as at any other point in the series circuit.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
>
| |
| SuperM 2007-04-26, 3:25 am |
| On Thu, 26 Apr 2007 04:46:30 GMT, "Don Kelly" <dhky@shaw.ca> Gave us:
>If you add a resistor, the current changes- but wherever you measure it, it
>is the same value as at any other point in the series circuit.
Yeah... it's kinda a rule around these parts...
It's called Ohm's law, and that's what we have been trying to tell
him all along.
| |
| sforu.satyakam@gmail.com 2007-04-26, 9:25 am |
| the volatge drop and the voltage difference are two different way of
understanding the same thing..
suppose a 5V voltage source and two resistances connected to it in
series to it. if the first resistance shows a voltage 2v across it the
other resistor will show 3V. this way voltage across the resistors
reduces the source voltage drop from 5V to 3V or 2V thus it is termed
as voltage drop.(if open circuited, the source wd hv shown 5V intead
of 3V or 2V)
another way of understanding it is way of voltage difference concept.
if we see the same examle, the source has 5V and the second resistor
shows 3V the voltage drop or potential difference across the first
resistor will be 5-3= 2V
| |
| Don Kelly 2007-04-27, 3:25 am |
| "SuperM" <SuperM@ssiveBlackHoleAtTheCenterOfTheMilkyWayGalaxy.org> wrote in
message news:pmj033dgi91oa852okvvfe7pa867fpqhee@4ax.com...
> On Thu, 26 Apr 2007 04:46:30 GMT, "Don Kelly" <dhky@shaw.ca> Gave us:
>
>
>
> Yeah... it's kinda a rule around these parts...
>
> It's called Ohm's law, and that's what we have been trying to tell
> him all along.
------------------
No- it's NOT Ohms Law. It is a result of Kirchoff's "Current Law" and is
true even where the resistors don't obey Ohm's Law (look up Ohms Law- which
is often misquoted).
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
| |
| SuperM 2007-04-27, 8:25 pm |
| On Fri, 27 Apr 2007 02:45:58 GMT, "Don Kelly" <dhky@shaw.ca> Gave us:
>"SuperM" <SuperM@ssiveBlackHoleAtTheCenterOfTheMilkyWayGalaxy.org> wrote in
>message news:pmj033dgi91oa852okvvfe7pa867fpqhee@4ax.com...
>------------------
>
>No- it's NOT Ohms Law. It is a result of Kirchoff's "Current Law" and is
>true even where the resistors don't obey Ohm's Law (look up Ohms Law- which
>is often misquoted).
I just posted the same PDF learning aid I posted in abse some three
years ago, and it is certainly not mis-stated.
It is a relationship between the resistance of a circuit, and the
voltage applied to it, which results in a specific current through
said resistance.
So yes, multiple resistances follow Kirchoff's law, and the voltage
on each resistive element (including the wires between them) is
referred to as the "drop" on each.
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| Don Kelly 2007-04-28, 3:25 am |
| "SuperM" <SuperM@ssiveBlackHoleAtTheCenterOfTheMilkyWayGalaxy.org> wrote in
message news:m255331qlslsf43j49v3qqm438b6l4as1l@4ax.com...
> On Fri, 27 Apr 2007 02:45:58 GMT, "Don Kelly" <dhky@shaw.ca> Gave us:
>
>
> I just posted the same PDF learning aid I posted in abse some three
> years ago, and it is certainly not mis-stated.
>
> It is a relationship between the resistance of a circuit, and the
> voltage applied to it, which results in a specific current through
> said resistance.
>
> So yes, multiple resistances follow Kirchoff's law, and the voltage
> on each resistive element (including the wires between them) is
> referred to as the "drop" on each.
Quote:
">>If you add a resistor, the current changes- but wherever you measure it,[color=darkred]
That is NOT Ohm's Law which is simply the linear relationship between
voltage and current in an "ohmic" element.
Ohm's law has nothing to do with the fact that the current is the same
everywhere in the loop. KCL does (even where Ohm's Law is meaningless).
That is all that I was trying to say.
I have no problem with use of Ohm's Law to determine the voltage drop across
each element in the circuit -provided that they are linear and actually obey
Ohm's Law (q.v).
If all the elements are ohmic, then linearity exists and you can go on,
(using the far more basic Kirchoff's Laws) to develop the concept of series
and parallel equivalents, Thevenin and all that nice stuff. Without
linearity - all these things go out the window leaving Kirchoff's Laws and
pain in the butt solutions.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
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