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Current flows through a resistor (with no direction since it is not a vector/can flow from any potential)?

I thinks it is with no direction since it is not a vector. Is that right?

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closed as unclear what you're asking by Chris White, Nathaniel, Waffle's Crazy Peanut, Emilio Pisanty, akhmeteli Jul 5 '13 at 22:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Note that the current density is a vector. –  Qmechanic Jul 4 '13 at 0:12

3 Answers 3

A current is like the flow of a fluid, so it's a vector field. We tend to think of resistors as one dimensional when we draw circuit diagrams, but of course they aren't.

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4  
Should we be careful with the use of current vs current density? The current density is a vector field defined proportional to the electric field, whereas the current is defined as the integral of the component of the current density through some particular cross-sectional area. –  Will Jul 3 '13 at 18:46
    
Oops, yes, good catch. –  John Rennie Jul 4 '13 at 9:24
    
@JohnRennie Isn't current a scalar, though? It doesn't follow the Vector Law of Addition, neither can you divide it into components like vectors. –  mikhailcazi Jul 4 '13 at 16:12
    
@mikhailcazi How does it not follow vector addition? Show me. See my solution. –  Will Jul 4 '13 at 16:19
    
@Will Well if you take wires arranged in a closed polygon, vector sum says the resultant current should be zero, but that's not true - a circuit is a closed polygon. And current does flow. –  mikhailcazi Jul 4 '13 at 16:24

I feel really bad for the OP getting these downvotes. The question (if worded better) is a really good question. There is obviously a lot of confusion over this - given the range of solutions posted. I think the main confusion comes from two places: (1) using current density in place of current (whether consciously or not), (2) Not realizing that a 1-D vector is a scalar.


Current density:

For simplicity, assume we have a conductor with a constant conductance, $\sigma$. Also for simplicity, let's assume that the electric field is time-independent, but could vary in space. The electric field, $\vec{E}(\vec{x})$, is a vector field, that is, at each point in space, we associate a vector space. The relationship between the electric field and the current density in this simplified model is $$\vec{J}(\vec{x}) = \sigma~\vec{E}(\vec{x})$$ From this equation, we see that $\vec{J}(\vec{x})$, the current density, is also a vector field.


Current:

The current is defined with reference to some area as $$I_A = \int_A \vec{J}(\vec{x})\cdot d\vec{A}$$ In circuits, the area we care about is the cross-sectional area of the wire. With a DC source the electric field is constant throughout the wire (if AC, still constant spatially but has time dependence), that is $\vec{E}(\vec{x}) = E_0~\hat{z}$ (assuming the wire is along the $z$-axis). This means that the current density is $$\vec{J} = \sigma~E_0~\hat{z}$$ and so the current is (area of interest is the cross-section of the wire, with $d\vec{A} = dA\hat{z}$) $$I = \int dA~\sigma E_0~\hat{z}\cdot\hat{z} = A\sigma ~ E_0$$ In this idealized situation, we see that $I$ can be treated as a 1-D vector, so far as the vector space axioms are concerned. We also note that a 1-D vector is a scalar! Note that I am not saying scalar field or vector field.


Let's see what happens when we aren't confined to a wire and have spatial varying electric fields.

Take some area through a region in a conductor (with constant $\sigma$ as before) where the electric field has spatial dependence $\vec{E}(\vec{x})$. The current associated with this electric field and area is $$I_A = \sigma\int_A \vec{E}(\vec{x})\cdot d\vec{A}$$ With a little thought, one can see that this doesn't change the argument in our wire case. The current defined this way still produces a 1-D vector, whose direction indicates which way the total flow of positive charges (by convention) is flowing, through the prescribed surface. And again, a 1-D vector IS a scalar.


Note: in this situation, adding currents is physically identical to adding electric fields (as $\sigma$ is assumed constant).

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Electric current has a direction and that is the direction in which positive charge flows.

In circuit theory, electric current enters the positive terminal of a resistor and exits the negative terminal.

If electric current did not have a direction, how would we determine the polarity of a magnetic dipole due to a current loop?

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