Current across a wire with zero potential differencePotential difference across one resistor with and without...
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Current across a wire with zero potential difference
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$begingroup$
If there was a circuit connected with a 50ohm resistor and a 5V battery and we measure the voltage across the two point of the wire which has no resistor or e cell connected to it,does it mean the voltage is zero? Then acc to V=IR is the current also zero? (Assume that e wire has negligible resistance)
voltage current resistors
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If there was a circuit connected with a 50ohm resistor and a 5V battery and we measure the voltage across the two point of the wire which has no resistor or e cell connected to it,does it mean the voltage is zero? Then acc to V=IR is the current also zero? (Assume that e wire has negligible resistance)
voltage current resistors
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
3
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
If there was a circuit connected with a 50ohm resistor and a 5V battery and we measure the voltage across the two point of the wire which has no resistor or e cell connected to it,does it mean the voltage is zero? Then acc to V=IR is the current also zero? (Assume that e wire has negligible resistance)
voltage current resistors
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If there was a circuit connected with a 50ohm resistor and a 5V battery and we measure the voltage across the two point of the wire which has no resistor or e cell connected to it,does it mean the voltage is zero? Then acc to V=IR is the current also zero? (Assume that e wire has negligible resistance)
voltage current resistors
voltage current resistors
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Ali JinnahAli Jinnah
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Ali JinnahAli Jinnah
161
asked 6 hours ago
Ali JinnahAli Jinnah
161
161
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ali Jinnah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
3
$begingroup$
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
3 hours ago
add a comment |
1
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
3
$begingroup$
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
3 hours ago
1
1
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
3
3
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
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 6 hours ago
Nikolai RuheNikolai Ruhe
24616
$endgroup$
add a 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 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
$endgroup$
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
add a comment |
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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 6 hours ago
Nikolai RuheNikolai Ruhe
24616
$endgroup$
add a 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 6 hours ago
Nikolai RuheNikolai Ruhe
24616
$endgroup$
add a 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 6 hours ago
Nikolai RuheNikolai Ruhe
24616
$endgroup$
[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 6 hours ago
Nikolai RuheNikolai Ruhe
24616
answered 6 hours ago
Nikolai RuheNikolai Ruhe
24616
answered 6 hours ago
Nikolai RuheNikolai Ruhe
24616
answered 6 hours ago
Nikolai RuheNikolai Ruhe
24616
24616
add a comment |
add a 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 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
$endgroup$
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
add a 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 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
$endgroup$
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
add a 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 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
$endgroup$
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 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
edited 2 hours ago
answered 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
answered 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
answered 3 hours ago
Frank from FrankfurtFrank from Frankfurt
1523
1523
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
add a comment |
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
$begingroup$
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
39 mins ago
$begingroup$
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
39 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
@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
4 mins ago
$begingroup$
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
1 min ago
$begingroup$
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
1 min ago
add a comment |
Ali Jinnah is a new contributor. Be nice, and check out our Code of Conduct.
Ali Jinnah is a new contributor. Be nice, and check out our Code of Conduct.
Ali Jinnah is a new contributor. Be nice, and check out our Code of Conduct.
Ali Jinnah is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Current is through the wire. Voltage is measured across two points.
$endgroup$
– JRE
6 hours ago
$begingroup$
In the limit, $I=frac{V}{R}=frac{0}{0} $, which is indeterminate.
$endgroup$
– Chu
5 hours ago
3
$begingroup$
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
3 hours ago