Current across a wire with zero potential difference
$begingroup$
If there was a circuit connected with a $50 ,Omega$ resistor and a $5 , rm V$ battery and we measured the voltage across two points of the wire that have no resistor or battery, does it mean the voltage is zero? Then, according to $V = I R$, is the current also zero?
Assume that the wire has negligible resistance.
voltage current resistors
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
$endgroup$
add a comment |
$begingroup$
If there was a circuit connected with a $50 ,Omega$ resistor and a $5 , rm V$ battery and we measured the voltage across two points of the wire that have no resistor or battery, does it mean the voltage is zero? Then, according to $V = I R$, is the current also zero?
Assume that the wire has negligible resistance.
voltage current resistors
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
$endgroup$
3
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
3
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
6
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
2
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46
add a comment |
$begingroup$
If there was a circuit connected with a $50 ,Omega$ resistor and a $5 , rm V$ battery and we measured the voltage across two points of the wire that have no resistor or battery, does it mean the voltage is zero? Then, according to $V = I R$, is the current also zero?
Assume that the wire has negligible resistance.
voltage current resistors
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
$endgroup$
If there was a circuit connected with a $50 ,Omega$ resistor and a $5 , rm V$ battery and we measured the voltage across two points of the wire that have no resistor or battery, does it mean the voltage is zero? Then, according to $V = I R$, is the current also zero?
Assume that the wire has negligible resistance.
voltage current resistors
voltage current resistors
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
edited Feb 27 at 6:22
Rodrigo de Azevedo
16610
16610
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
asked Feb 26 at 6:33
Ali JinnahAli Jinnah
172
172
3
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
3
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
6
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
2
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46
add a comment |
3
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
3
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
6
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
2
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46
3
3
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
3
3
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
6
6
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
2
2
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
[in a plain wire] does it mean the voltage is zero?
Yes, the voltage across both ends of an ideal wire is always zero.
[given U = R * I] is the current also zero?
No, it means that the current can have an arbitrary value. Because in ...
0 V = 0 Ohm * x Ampere
... x can have any value.
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
$endgroup$
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
|
show 1 more comment
$begingroup$
If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:
$$ I = U/R = 0.1 A $$
Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.
Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:
$$ V = R cdot I = 1 nOmega cdot 0.1 A = 100 pV $$
Of course, measuring such a voltage of 100 pico-Volt would be a challenge.
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
$endgroup$
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
|
show 5 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
[in a plain wire] does it mean the voltage is zero?
Yes, the voltage across both ends of an ideal wire is always zero.
[given U = R * I] is the current also zero?
No, it means that the current can have an arbitrary value. Because in ...
0 V = 0 Ohm * x Ampere
... x can have any value.
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
$endgroup$
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
|
show 1 more comment
$begingroup$
[in a plain wire] does it mean the voltage is zero?
Yes, the voltage across both ends of an ideal wire is always zero.
[given U = R * I] is the current also zero?
No, it means that the current can have an arbitrary value. Because in ...
0 V = 0 Ohm * x Ampere
... x can have any value.
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
$endgroup$
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
|
show 1 more comment
$begingroup$
[in a plain wire] does it mean the voltage is zero?
Yes, the voltage across both ends of an ideal wire is always zero.
[given U = R * I] is the current also zero?
No, it means that the current can have an arbitrary value. Because in ...
0 V = 0 Ohm * x Ampere
... x can have any value.
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
$endgroup$
[in a plain wire] does it mean the voltage is zero?
Yes, the voltage across both ends of an ideal wire is always zero.
[given U = R * I] is the current also zero?
No, it means that the current can have an arbitrary value. Because in ...
0 V = 0 Ohm * x Ampere
... x can have any value.
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
answered Feb 26 at 6:51
Nikolai RuheNikolai Ruhe
33617
33617
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
|
show 1 more comment
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 26 at 23:43
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
$begingroup$
@SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value.
$endgroup$
– user60561
Feb 27 at 3:51
1
1
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@user60561 Well, theoretically, yes. But here's 500 Amp vs wrench
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 3:58
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors.
$endgroup$
– user60561
Feb 27 at 4:45
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
$begingroup$
@user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks.
$endgroup$
– Sergiy Kolodyazhnyy
Feb 27 at 4:55
|
show 1 more comment
$begingroup$
If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:
$$ I = U/R = 0.1 A $$
Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.
Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:
$$ V = R cdot I = 1 nOmega cdot 0.1 A = 100 pV $$
Of course, measuring such a voltage of 100 pico-Volt would be a challenge.
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
$endgroup$
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
|
show 5 more comments
$begingroup$
If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:
$$ I = U/R = 0.1 A $$
Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.
Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:
$$ V = R cdot I = 1 nOmega cdot 0.1 A = 100 pV $$
Of course, measuring such a voltage of 100 pico-Volt would be a challenge.
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
$endgroup$
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
|
show 5 more comments
$begingroup$
If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:
$$ I = U/R = 0.1 A $$
Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.
Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:
$$ V = R cdot I = 1 nOmega cdot 0.1 A = 100 pV $$
Of course, measuring such a voltage of 100 pico-Volt would be a challenge.
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
$endgroup$
If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:
$$ I = U/R = 0.1 A $$
Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.
Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:
$$ V = R cdot I = 1 nOmega cdot 0.1 A = 100 pV $$
Of course, measuring such a voltage of 100 pico-Volt would be a challenge.
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
edited Feb 26 at 10:09
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
answered Feb 26 at 10:02
Frank from FrankfurtFrank from Frankfurt
2103
2103
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
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– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
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– Frank from Frankfurt
Feb 26 at 13:08
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I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
|
show 5 more comments
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
$endgroup$
– MCG
Feb 26 at 12:28
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
@MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits.
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:03
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
$endgroup$
– MCG
Feb 26 at 13:06
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
@MCG: Where do you read zero?
$endgroup$
– Frank from Frankfurt
Feb 26 at 13:08
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
$begingroup$
I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
$endgroup$
– MCG
Feb 26 at 13:12
|
show 5 more comments
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3
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
Feb 26 at 6:37
3
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
Feb 26 at 7:19
6
$begingroup$
NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some.
$endgroup$
– eckes
Feb 26 at 9:31
2
$begingroup$
What is the smallest voltage you can measure, and how does it compare to 'negligable'?
$endgroup$
– Sean Houlihane
Feb 26 at 14:23
$begingroup$
@eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance"
$endgroup$
– user60561
Feb 27 at 4:46