Current

Claimed by spencer

Explanation of Current Through a Wire

The Main Idea

The electric current is the rate of flow of electric charge through an area. It is important to note that the charge in the circuits you will be dealing with are not in equilibrium, but instead, in the steady state. The distinction between equilibrium and the steady state is as follows. In equilibrium, a metal contains no mobile charges in motion (average drift velocity = 0). In the steady state, the particles may be moving however they please, but their average drift velocity stays constant.

Another important distinction to make is the difference between the electron current and the conventional current. The electron current is denoted by a lowercase i, describes the number of electrons per second to enter or leave a section of a conductor, and has units of electrons per second. The direction of the electron current is described as the direction the electrons inside a wire are moving. This is true in most cases since electrons, not protons, are the mobile charge carriers in circuits.

Negative Charge Carriers

On the other hand, the conventional current is denoted by an uppercase I, describe the amount of charge that passes a cross-section of a wire per second, and therefore has units of coulombs per second, or amperes (abbreviated as A). Since the conventional current is described by a positive charge, the direction of the conventional current will be opposite to the direction of the electron current. If the mobile charges were positive, they would move in the direction of the conventional current.

Positive Charge Carriers

Before scientists knew that electrons were the common charge carriers, they discovered current. And, with only two choices, they chose to treat the moving charges as positive. This may seem annoying at first, but there are some benefits to using the conventional current over the directly opposed "electron current". For one, it flows from the positive end of a battery toward the negative end, and from high to low energy.

A Mathematical Model

Assume uniform density [math]\displaystyle{ n }[/math] of electrons (uniform current) in a wire of area [math]\displaystyle{ A }[/math]

[math]\displaystyle{ i=density⋅volume/seconds =nA v }[/math]

If a charge distribution of density [math]\displaystyle{ ρ }[/math] moves with the velocity [math]\displaystyle{ v }[/math], the charge per unit time through [math]\displaystyle{ ΔA }[/math] is [math]\displaystyle{ ρv⋅nΔA }[/math] [math]\displaystyle{ Δq=ρv⋅nΔAΔt }[/math]. [math]\displaystyle{ ρ = Nq }[/math]

The charge per unit time is then [math]\displaystyle{ ρv⋅nΔS }[/math], from which we get the current density to be [math]\displaystyle{ Nqv }[/math]

The current I through the surface is [math]\displaystyle{ I=∫Nqv⋅dA }[/math]

[math]\displaystyle{ I = |q|nAv }[/math]

[math]\displaystyle{ i = nAv }[/math]

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

One type of problem you are sure to encounter involves a battery and resistor. You can read more about that section on the RC page of this wiki, but here's how to solve one now:

Using your formula sheet, notice that [math]\displaystyle{ I = {|ΔV|\over R} }[/math]. where R is the resistance in the circuit and |ΔV| is the voltage.

If you're given that the 50V battery is connected to a 100 [math]\displaystyle{ Ω }[/math] resistor, you simply substitute the values into your equation. [math]\displaystyle{ I = {50V\over 100 Ω} }[/math] [math]\displaystyle{ I = 0.5A }[/math]

Middling

Current will not always be what you solve for, though. In this problem, let's use the equations we've discussed to find the drift velocity [math]\displaystyle{ v }[/math]. Given the current is [math]\displaystyle{ 1A }[/math], the electron density is [math]\displaystyle{ 8 ⋅ 10^{27}m^{-3} }[/math], and the diameter of our wire is [math]\displaystyle{ 1mm }[/math].

First, let's find the area [math]\displaystyle{ {0.001\over 2}^{2}⋅π=7.85⋅10^{-7} }[/math]

Then, plug in the rest: [math]\displaystyle{ 1=|-1.6⋅10^{-19}|⋅5⋅10^{28}⋅7.85⋅10^{-7}⋅v, v = 9.9477.85⋅10^{-4}{m\over s} }[/math]

References

https://en.wikipedia.org/wiki/Electric_current

http://www.feynmanlectures.caltech.edu/II_13.html#Ch13-S2

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html