Ohm’s law is one of the fundamental laws of electricity, and connects the Resistance to the Voltage and Current
Definition
Formula
Ohm's Law
V = I R
##### Terms
* $V$ = [Electrical Potential](Electrical%20Potential.md) difference (i.e. voltage) (in $V$)
* $I$ = [Current](Current.md) flowing through device (in $\text{A}$)
* $R$ = [Resistance](Resistance.md) of the circuit. (in $\Omega$)
### Derivation using [Drude Model of Conduction](Drude%20Model%20of%20Conduction.md)
[Drude Model of Conduction](Drude%20Model%20of%20Conduction.md) states that electrons move in random directions, but the net movement can be achieved through the use of an [Electric Field](Electric%20Field.md).
Using [Newton's second law of motion](Newton's%20Laws%20of%20Motion.md#Second%20Law%20Law%20of%20Acceleration), and the definition of an electric field:
F = q\times \vec{E} = m\times \vec{a} \to \vec{a} = q \dfrac{\vec{E}}{m}
\vec{a}{e} = \dfrac{\vec{E}q{e}}{m_{e}}
Electrons in this field experience this acceleration, until they collide with another electron or particle. If we assume the average time between collisions is $\tau$, then
v_{drift} = \vec{a}{e}\times \tau = \dfrac{q{e}\vec{E}\tau}{m_{e}}
J = q_{e}\rho_{e}v_{drift} \to v_{drift} = \dfrac{J}{q_{e}\rho_{e}}
\dfrac{J}{q_{e}\rho_{e}} =\dfrac{q_{e}\vec{E}\tau}{m_{e}}
J = \dfrac{q_{e}^2\vec{E}\tau}{m_{e}}
Which shows that $J \propto \vec{E}$ (which was kind of obvious anyway, but still.)
[Conductivity & Resistivity](Conductivity%20&%20Resistivity.md) is defined to simplify this equation:
J = \sigma \vec{E} = \dfrac{\vec{E}}{\rho}
$J \propto \sigma \text{(conductivity)} \propto \dfrac{1}{\rho} (\dfrac{1}{\text{resistivity}})$
Finally, we can now expand this model up to a larger scale.
Assume we have a conducting line of cylinder of length $l$ and area $A$
The electric field can be represented as the gradient of the [Electrical Potential](electrical%20potential.md) difference over the length $l$:
\vec{E} = \dfrac{\Delta V}{l}
J = \dfrac{I}{A}
And we define resistance to be $R = \dfrac{l}{A\sigma}$
Finally:
J = \dfrac{\vec{E}}{\rho}
\dfrac{I}{A} = \dfrac{\Delta V}{l}\sigma$$$$I = \dfrac{\Delta VA}{l}\sigma = \dfrac{\Delta V}{R}
So we get $\Delta V = IR$, proving Ohm's law
### Ohmic Vs Non-Ohmic
Any device that obeys Ohm's law, i.e. has a constant gradient for a voltage current graph is said to be **ohmic**. Conversely, any device that has a non constant gradient, i.e. doesn't obey Ohm's law, is a **non-ohmic device**.
![500](images-1.png#BC)