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I need help to understand what is going on in this simple circuit. Consider all voltage source, resistor and wire are ideal. And also assume that the wire has the same crosssection area along its length (A = const (m^2)).
The voltage between two points A and B is defined as path integral from A to B of electric field.
[tex]V_{AB} = \int ^B_A \overrightarrow {E}.\overrightarrow {dl}[/tex]
The current is equal to each other everywhere in the wire and the wire has the same crosssection area => the current density is also the same everywhere.
Acoording to Ohm's law:
[tex]\overrightarrow {J}=\sigma \overrightarrow {E}[/tex]
Because J ≠ 0, σ ≠ 0 => E ≠ 0 (1) everywhere in the wire (also very point from A to B)
But I also know that: [tex]V_{AB} = 0 [/tex]
[tex]V_{AB} = E\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} = 0 [/tex]
If I choose the path is the straight line from A to B => integral from A to B of dl ≠ 0
[tex]\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} ≠ 0 [/tex]=> E = 0 (2)
(1) and (2) can't exist simultaneously. Can you point out what I am wrong?
The voltage between two points A and B is defined as path integral from A to B of electric field.
[tex]V_{AB} = \int ^B_A \overrightarrow {E}.\overrightarrow {dl}[/tex]
The current is equal to each other everywhere in the wire and the wire has the same crosssection area => the current density is also the same everywhere.
Acoording to Ohm's law:
[tex]\overrightarrow {J}=\sigma \overrightarrow {E}[/tex]
Because J ≠ 0, σ ≠ 0 => E ≠ 0 (1) everywhere in the wire (also very point from A to B)
But I also know that: [tex]V_{AB} = 0 [/tex]
[tex]V_{AB} = E\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} = 0 [/tex]
If I choose the path is the straight line from A to B => integral from A to B of dl ≠ 0
[tex]\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} ≠ 0 [/tex]=> E = 0 (2)
(1) and (2) can't exist simultaneously. Can you point out what I am wrong?
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