voltage of an open switch in an imcomplete device

E

ERICA

Guest
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.
 
ERICA wrote:
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.
Click to expand...
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.
 
I read in sci.electronics.design that ERICA <homecat2001@yahoo.com.hk>
wrote (in <f0eca228.0404100000.737b5d53@posting.google.com>) 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.
Click to expand...
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.
Click to expand...
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.


--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
 
Robert Baer <robertbaer@earthlink.net> wrote in message >
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.
Click to expand...
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?
 
ERICA wrote:
there is still one question suddenly comes up. how can voltage flow in
the circuit when the circuit is imcompleted?
Click to expand...
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
 
ERICA wrote:

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?
Click to expand...
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.

--
John Popelish
 
ERICA wrote:
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.
Click to expand...
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.
 
"ERICA" <homecat2001@yahoo.com.hk> wrote in message
news:f0eca228.0404100000.737b5d53@posting.google.com...
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
^
Click to expand...
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.
Click to expand...
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.
Click to expand...
Hope I cleared that up for you. :)

Cheers!
Rich
 
On Sat, 10 Apr 2004 14:30:03 GMT, John Miller wrote:

ERICA wrote:
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?
Click to expand...
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.
--
Best Regards,
Mike
 
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