voltage of an open switch in an imcomplete device

[ERICA]

if there is a circuit with a resistor , a 6v battery and a switch,
when the switch is opened, there is no current passes through in the
circuit, by V=IR, THE voltage across the switch should be 0, but why
the voltage across it is 6?

I have think of it and i think maybe air act as a resistor of infinity
resisrance so it shares all the voltage of the battery. but as
voltage= potential difference = work done to bring a charge from 1
point to another, if the circuit is imcompeted, there will be no work
done, the voltage will be 0.
there will be contradition.

 

[Robert Baer]

if there is a circuit with a resistor , a 6v battery and a switch,
when the switch is opened, there is no current passes through in the
circuit, by V=IR, THE voltage across the switch should be 0, but why
the voltage across it is 6?

I have think of it and i think maybe air act as a resistor of infinity
resisrance so it shares all the voltage of the battery. but as
voltage= potential difference = work done to bring a charge from 1
point to another, if the circuit is imcompeted, there will be no work
done, the voltage will be 0.
there will be contradition.

For a very brief period of time, when the wires were connected to the
battery, work was done, current flowed, and the capacitance of them was
charged up to equi-potentoal of the terminals.
No contradiction.
Theoretically, one could ise some sort of radio receiver and detect
the radiation of the EMF generated by that very brief current flow.

 

[John Woodgate]

I read in sci.electronics.design that ERICA <[email protected]>
wrote (in <[email protected]>) about
'voltage of an open switch in an imcomplete device', on Sat, 10 Apr
2004:

if there is a circuit with a resistor , a 6v battery and a switch,
when the switch is opened, there is no current passes through in the
circuit, by V=IR, THE voltage across the switch should be 0, but why
the voltage across it is 6?

I have think of it and i think maybe air act as a resistor of infinity
resisrance so it shares all the voltage of the battery.

Yes, that is correct.

but as
voltage= potential difference = work done to bring a charge from 1
point to another, if the circuit is imcompeted, there will be no work
done, the voltage will be 0.
there will be contradition.

Moving a charge from one pole of the switch to the other *completes* the
circuit for the time that the charge is moving. Moving charge is
current; when the switch is closed, a lot of charge is moving through
it.

 

[ERICA]

For a very brief period of time, when the wires were connected to the
battery, work was done, current flowed, and the capacitance of them was
charged up to equi-potentoal of the terminals.
No contradiction.
Theoretically, one could ise some sort of radio receiver and detect
the radiation of the EMF generated by that very brief current flow.

but the circuit is not completed, if this is the case, by V=IR, when
there is voltage through the circuit and if an ammeter is connected to
the circuit, there will be reading but i don't think there will be
reading.
moreover,the desire of a resistor is not the same as a capacitor, can
it be charged up as that of a capacitor?

there is still one question suddenly comes up. how can voltage flow in
the circuit when the circuit is imcompleted?

 

[John Miller]

there is still one question suddenly comes up. how can voltage flow in
the circuit when the circuit is imcompleted?

That may be at the root of the misunderstanding. It is current that flows,
not voltage.

In thinking about your (open) circuit, start by eliminating the switch. (As
long as the switch is open, it doesn't exist.) You can still measure the
voltage aross the battery and resistor, yes? Why is that so?

--
John Miller
Email address: domain, n4vu.com; username, jsm

Civilization is the limitless multiplication of unnecessary necessities.
-Mark Twain

 

[John Popelish]

but the circuit is not completed, if this is the case, by V=IR, when
there is voltage through the circuit and if an ammeter is connected to
the circuit, there will be reading but i don't think there will be
reading.
moreover,the desire of a resistor is not the same as a capacitor, can
it be charged up as that of a capacitor?

there is still one question suddenly comes up. how can voltage flow in
the circuit when the circuit is imcompleted?

Voltage is the potential that makes current happen (charge flow)
through resistance. If you somehow released an electron very near the
surface of the more negative contact of the switch (light shining on
metal is one way to do this) and there wasn't any air molecules in the
way, it would move through empty space toward the more positive
contact with a smooth acceleration, and arrive with the energy that
the electric field between the contacts (voltage difference) gave it.
The fact that there are no (or very few) electrons making this trip
when there is air between the contacts does not eliminate the
'potential' for such acceleration.

Your thinking of the air spaced contacts as an infinite resistance
that is included in the circuit is correct. When the resistance has a
magnitude of infinity and the voltage is finite, the current through
that resistance, by Ohm's law comes out to be zero.

 

[Fred Bloggs]

if there is a circuit with a resistor , a 6v battery and a switch,
when the switch is opened, there is no current passes through in the
circuit, by V=IR, THE voltage across the switch should be 0, but why
the voltage across it is 6?

I have think of it and i think maybe air act as a resistor of infinity
resisrance so it shares all the voltage of the battery. but as
voltage= potential difference = work done to bring a charge from 1
point to another, if the circuit is imcompeted, there will be no work
done, the voltage will be 0.
there will be contradition.

In circuit theory the ideal switch never dissipates energy because, if
it is open, the current is zero, and, if it is closed, the voltage loss
is zero. In either of the two cases, V x I =0. In the case of your
particular conundrum with the open switch, the circuit has been
interrupted and it is clear that I=0 throughout. The battery therefore
supplies no energy to the circuit, and therefore the circuit dissipates
no energy, and none of the circuit elements, including the switch,
dissipates energy. You are justified in saying the switch has 6V across
it because if you place a voltmeter with non-infinite but very large
input impedance across the switch, then most of the battery voltage will
appear across the meter. And as the meter impedance is allowed to
approach the ideal infinity, the voltage reading approaches identity
with the battery voltage.

 

[Rich Grise]

if there is a circuit with a resistor , a 6v battery and a switch,
when the switch is opened, there is no current passes through in the
circuit, by V=IR, THE voltage across the switch should be 0, but why
the voltage across it is 6?

I have think of it and i think maybe air act as a resistor of infinity
resisrance so it shares all the voltage of the battery. but as
voltage= potential difference = work done to bring a charge from 1

^
No. Potential is just that - potential. There is no work being
done when nothing is moving, as in when the switch is open.

point to another, if the circuit is imcompeted, there will be no work
done, the voltage will be 0.

No, the voltage will be the battery voltage. Take the negative terminal
of the battery, the resistor, and one switch contact. If there is no
current flowing, these will all be at the same voltage, or potential.
This will be <the battery voltage> more negative than at the positive
terminal of the battery, which is also at the same potential as the
other terminal of the switch. Ergo, the difference in potential
from one switch contact to the other is exactly the same as the
difference in potential from one battery terminal to the other -
resistance doesn't enter into it, because the current is zero, and
therefore the voltage "drop" (E = IR) is zero. But that's just for
the conductors. Clearly, an open switch is not a conductor (or the
equivalent, has infinite resistance) so the switch "drops" essentially
all of the voltage.

there will be contradition.

Hope I cleared that up for you.

Cheers!
Rich

 

[Active8]

That may be at the root of the misunderstanding. It is current that flows,
not voltage.

In thinking about your (open) circuit, start by eliminating the switch. (As
long as the switch is open, it doesn't exist.) You can still measure the
voltage aross the battery and resistor, yes? Why is that so?

because the volt meter has a very high resistance relative to the
resistor in the circuit and therefor more voltage is dropped across
it - if the meter is connected across the switch terminals. If
connected across the batt, the answer should be obvious as well as
if it's just connected across the R.

 

[ERICA]

Moving a charge from one pole of the switch to the other *completes* the
circuit for the time that the charge is moving. Moving charge is
current; when the switch is closed, a lot of charge is moving through
it.

then as you said, there will always be current flow in the
circuit...then in daily life, when the switch of an electric apparatus
is closed, it apparatus will still function?

 

[ERICA]

^
No. Potential is just that - potential. There is no work being
done when nothing is moving, as in when the switch is open.


This will be <the battery voltage> more negative than at the positive
terminal of the battery, which is also at the same potential as the
other terminal of the switch.

Ergo, the difference in potential
from one switch contact to the other is exactly the same as the
difference in potential from one battery terminal to the other -
resistance doesn't enter into it, because the current is zero, and
therefore the voltage "drop" (E = IR) is zero. But that's just for
the conductors. Clearly, an open switch is not a conductor (or the
equivalent, has infinite resistance) so the switch "drops" essentially
all of the voltage.


Hope I cleared that up for you.

Cheers!
Rich

thanks a lot i think i can understand the above part except this part:

No, the voltage will be the battery voltage. Take the negative terminal
of the battery, the resistor, and one switch contact. If there is no
current flowing, these will all be at the same voltage, or potential.

why?

 

[John Woodgate]

I read in sci.electronics.design that ERICA <[email protected]>
wrote (in <[email protected]>) about
'voltage of an open switch in an imcomplete device', on Sun, 11 Apr
2004:

then as you said, there will always be current flow in the
circuit...then in daily life, when the switch of an electric apparatus
is closed, it apparatus will still function?

Unless it has developed a fault, yes.

 

[Robert Baer]

but the circuit is not completed, if this is the case, by V=IR, when
there is voltage through the circuit and if an ammeter is connected to
the circuit, there will be reading but i don't think there will be
reading.
moreover,the desire of a resistor is not the same as a capacitor, can
it be charged up as that of a capacitor?

there is still one question suddenly comes up. how can voltage flow in
the circuit when the circuit is imcompleted?

*Voltage* does not flow; *current* does.
I spoke of a *very* brief period of time when (any conductive)
component is connected to a battery terminal, period.
All of them have a resistance, capacitance (to everything else around
it - including the batery) and inductance.
Have you never heard of an antenna????

 

[Rich Grise]

thanks a lot i think i can understand the above part except this part:

why?

I'm afraid at this point I'm going to have to refer you to the books - I
gave it my best shot and don't know how to explain it any differently.

Good Luck!
Rich

 

[Mark Fergerson]

I'm afraid at this point I'm going to have to refer you to the books - I
gave it my best shot and don't know how to explain it any differently.

Lemme try.

Erica, think about the battery with nothing connected to
it. There's a potential between the terminals but no current
flow because there's no path for current to flow through;
you got that already.

Now attach a wire to the negative terminal; immediately
before you do, the wire is at whatever potential it had
(probably it grabbed some electrons from your hand, or gave
it some, depending on what potential your hand was at
compared to the battery terminal before you picked the wire
up), and when you attach it, a miniscule current flows until
the wire is at the negative terminal's potential. It's
basically become an extension of the terminal. Got that?

Now attach a resistor to the wire; same thing happens as
with the wire, except that a tiny current will flow through
the resistor until the voltage across it goes to zero. It'll
take a tiny bit longer than the wire did, but you get the
same end result; both ends of the resistor are at the same
potential as the wire and the negative battery terminal. If
they weren't, current would flow. Hence, the resistor is
effectively the same kind of extension of the battery
terminal as the wire. Got that?

Now attach the switch; same as above _except_ the switch,
being open, can't carry any current, so the unconnected
contact still has whatever potential it had before (the term
for that condition is "floating"). The contact attached to
the resistor is now at the negative battery terminal's
potential because there isn't any potential across the
resistor or the wire. Got that?

Whether you attach the other contact of the switch to the
positive terminal of the battery or not (as long as the
switch remains open), the negative terminal of the battery,
the wire, the resistor, and the first switch contact are all
at the same potential. Again, if they weren't, current would
flow until they were.

When you connect the other contact of the switch to the
positive battery terminal, that contact becomes an extension
of the positive terminal, bringing the full potential of the
battery across the space between the contacts. When you
close the switch, you move the extension of the positive
terminal to the end of the resistor so that all the battery
potential appears across it, and now there's a path for
current to flow through.

Did that help?

Mark L. Fergerson

 

[ERICA]

Lemme try.

Erica, think about the battery with nothing connected to
it. There's a potential between the terminals but no current
flow because there's no path for current to flow through;
you got that already.

Now attach a wire to the negative terminal; immediately
before you do, the wire is at whatever potential it had
(probably it grabbed some electrons from your hand, or gave
it some, depending on what potential your hand was at
compared to the battery terminal before you picked the wire
up), and when you attach it, a miniscule current flows until
the wire is at the negative terminal's potential. It's
basically become an extension of the terminal. Got that?

Now attach a resistor to the wire; same thing happens as
with the wire, except that a tiny current will flow through
the resistor until the voltage across it goes to zero. It'll
take a tiny bit longer than the wire did, but you get the
same end result; both ends of the resistor are at the same
potential as the wire and the negative battery terminal. If
they weren't, current would flow. Hence, the resistor is
effectively the same kind of extension of the battery
terminal as the wire. Got that?

Now attach the switch; same as above _except_ the switch,
being open, can't carry any current, so the unconnected
contact still has whatever potential it had before (the term
for that condition is "floating"). The contact attached to
the resistor is now at the negative battery terminal's
potential because there isn't any potential across the
resistor or the wire. Got that?

Whether you attach the other contact of the switch to the
positive terminal of the battery or not (as long as the
switch remains open), the negative terminal of the battery,
the wire, the resistor, and the first switch contact are all
at the same potential. Again, if they weren't, current would
flow until they were.

When you connect the other contact of the switch to the
positive battery terminal, that contact becomes an extension
of the positive terminal, bringing the full potential of the
battery across the space between the contacts. When you
close the switch, you move the extension of the positive
terminal to the end of the resistor so that all the battery
potential appears across it, and now there's a path for
current to flow through.

Did that help?

Mark L. Fergerson

that helps a lots, thanks you very much but there i still have a
question about the extention of the positive terminal. As current is
the flow of electrons, the current should flow from the negative
terminal to the positive one in fact but due to some historical
reason, we said it flow in the opposite direction as the electron. If
we extent the positive terminal , there should be no electrons flow
and thus the wire cannot bring the full potential of the battery
across the space between the contacts..
a strange question but hope it won't disturb you.

 

[Mark Fergerson]

that helps a lots, thanks you very much but there i still have a
question about the extention of the positive terminal. As current is
the flow of electrons, the current should flow from the negative
terminal to the positive one in fact but due to some historical
reason, we said it flow in the opposite direction as the electron.

Which way "current" flows is indeed a matter of
convention (it goes back to Benjamin Franklin making the
wrong guess, basically). I still think in terms of
electrons, but that's just a matter of personal prejudice
from early training.

If
we extent the positive terminal , there should be no electrons flow
and thus the wire cannot bring the full potential of the battery
across the space between the contacts..
a strange question but hope it won't disturb you.

Not at all.

Now we have to dig into the battery itself and figure out
how it gets that potential difference between its terminals.
That takes us into chemistry, and if you do a little
Googling for "battery chemistry" you'll find that it comes
down to electrons; follow the negative terminal down into
the guts of the battery and you'll find a chemical with more
electrons than it needs to maintain neutrality (it's
negatively-charged), similarly for the positive terminal
you'll find a chemical with fewer electrons than it needs to
maintain neutrality (positively charged).

Those chemicals are separated by a material that will
pass electrons in only one direction which is why the
battery doesn't self-discharge internally (that's for the
"ideal" battery you see discussed in beginners' textbooks;
real batteries do indeed slowly self-discharge because
real-world stuff never meets the ideal). So the chemicals
have potential fields surrounding them that can affect other
electrons, pulling them in or pushing them away.

(You know that conductors are made of metals which
conveniently have very loosely bound "conduction" electrons,
right?)

If you bring those two chemicals together, you should see
that the electrons in the negative material will happily
fill the vacancies in the positive material, restoring the
neutrality both materials want, allowing the potential
fields of each to disappear.

Apparently it seems to you that attaching a conductor to
the negative material allows some of the excess electrons to
"spill" into the conductor, producing the negative potential
along the conductors length, but it seems impossible that
attaching a conductor to the positive material does the same
thing without moving any charge carriers, since there aren't
any positively-charged equivalents to electrons in normal
use. If you think about it, if that were true you couldn't
get the positive potential out to the positive terminal of
the battery at all.

But that doesn't matter because what really happens is
that the electrons in the negatively-charged chemical don't
actually go anywhere (except for the transient [look that
word up] when the conductor is attached and whatever charge
it had before equalizes with the terminal charge); their
cumulative electric field reaches out into the conductor
which is why I used the phrase "extending the terminal". The
opposite happens at the positive end; the positive potential
of the chemical reaches into the conductor just the same as
at the negative end.

If you like, you can think of the electrons in the
conductor(s) attached to the negatively-charged material as
"leaning away from" the material, and the electrons in the
conductor(s) attached to the positive material as "leaning
towards" the positive material. When the conductors are
allowed to touch (through a resistor or whatever), they all
fall in the same direction like dominoes and then electrons
start going places; we have current until all the vacancies
in the positive material are filled.

(Mind you the above is a mixture of analogy and some very
simplified chemistry and physics, not a physically accurate
explanation. You'll have to learn a lot more than you can
from newsgroups to get anything resembling a true picture
including such things as the behavior of "holes" in
semiconductors.)

Now, don't you wish you'd paid more attention in your
Chemistry and Physics classes? I sure wish I had, instead I
had to relearn all this stuff later when I started doing
electronics as a hobby. ;>)

BTW, one last thing; your question better belonged in
sci.electronics.basics or maybe .misc as it isn't a design
issue (for most people, anyway). If you have more specific
questions about chemistry or physics, try sci.chem or
sci.physics and you'll get more specific answers as those
groups are populated by specialists who can catch and
correct errors and inaccuracies in replies like mine.

Mark L. Fergerson

 

[John Woodgate]

I read in sci.electronics.design that Mark Fergerson <[email protected]>

BTW, one last thing; your question better belonged in sci.electronics.basics or
maybe .misc as it isn't a design issue (for most people, anyway). If you have
more specific questions about chemistry or physics, try sci.chem or sci.physics
and you'll get more specific answers as those groups are populated by
specialists who can catch and correct errors and inaccuracies in replies like
mine.

I don't agree. Fundamental questions like this really require the
attention of high-grade adepts. The 'basics' newsgroup is for practical
solutions, not abstract philosophical questions.

 

[Mark Fergerson]

I read in sci.electronics.design that Mark Fergerson <[email protected]>



I don't agree. Fundamental questions like this really require the
attention of high-grade adepts. The 'basics' newsgroup is for practical
solutions, not abstract philosophical questions.

Hmm. Well, nobody else suggested she take it elsewhere,
so I'll plead nolo contendre.

My disclaimer notwithstanding, did I screw anything else
up so badly as to mislead Erica? You can't possibly mean to
class me with "high-grade adepts" like Rich Grise. ;>)

Mark L. Fergerson

 

[Rich Grise]

John Woodgate wrote:

Hmm. Well, nobody else suggested she take it elsewhere,
so I'll plead nolo contendre.

My disclaimer notwithstanding, did I screw anything else
up so badly as to mislead Erica? You can't possibly mean to
class me with "high-grade adepts" like Rich Grise. ;>)

Huh? Wha? ....

Must be talking about my keyboard fingering ;-)

Thanks,
Rich