\(\)

Ohm’s law is the linear relation between the potential and the flow of charge.

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## Current

Current is **defined** charge moved per unit time
$$
\shaded{
I = \frac{\Delta Q}{\Delta t}
}
\tag{Current}
$$

The convention is that the current \(I\) is in the direction of positive charges

Keep in mind, that energy moves through electro-magnetic fields, as we will see when we learn about the Poynting vector. Meanwhile, the electric current is an very effective abstraction model.

## Deriving Ohm’s law

### Drift velocity

As we have seen, in static conditions

Charges inside a conductor always move automatically in such a way that they kill the electric field inside. It will reach this equilibrium in a very short time.

If we keep adding electric charge to one end of the conductor, it will will not achieve equilibrium. We do this by applying a potential difference across a conductor. This *potential difference then generates an electric field along and inside the conductor*. The free electrons (with charge \(\rm e\)) will experience a force \(F\)
$$
F = \rm{e} \,E
\nonumber
$$

The free electrons are accelerated by the electric field, and gain momentum \(p\) $$ \frac{\Delta p}{\Delta t} = -q_e\,E \label{eq:gainrate} $$

We call the average net velocity in the direction opposite to the field, drift velocity \(v_d\). The collision of free electrons and ions will absorb all this drift momentum. The average momentum is the mass of the electron times the drift velocity $$ \frac{\Delta p}{\Delta t} = m_e\,v_d \label{eq:lossrate} $$

The rate of gain \(\eqref{eq:gainrate}\) and loss of momentum \(\eqref{eq:lossrate}\) must match $$ \begin{align} -q_e\,E &= m_e\,v_d \nonumber \\ \Rightarrow v_d &= – \frac{q_e E \tau}{m_e} \label{eq:vd} \end{align} $$

#### For copper

E.g. a copper \(\left(\rm{Cu}\right)\) conductor at room temperature

- the free electrons in copper have an average speed, \(v_e\approx 10^6\,\rm(m)/\rm{s}\). A chaotic thermal motion due to the temperature.
- An electron moving through a conductor suffers about \(10^{14}\) collisions with ions per second. So, the time between collations, \(\tau\approx 3\times 10^{-14}\,\rm{s}\)
- The number of free electrons, \(n\approx 10^{29}\,/\rm{m}^3 \)

Apply a potential difference \(V=10\,\rm{V}\) over length \(l=10\,\rm{m}\). Then inside the conductor the electric field is \(E = 1\,\rm{V/m}\), causing a drift velocity $$ \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} v_d &\approx \frac{(\numexp{1.6}{-19})\,(\numexp{3}{-19})}{10^{-30}} \\ &\approx 4\times 10^{-3}\,\rm m/\rm s \approx 0.5\,\rm{cm}/\rm s \end{align*} $$

The thermal motion is a million of meters per second, but due to the electric field, they only advance along the wire at \(\approx 0.5\,\rm{cm}/\rm s\). Note that the electrons on the “far end” of the wire start moving around the same time as the ones at the “start” of the wire.

### Ohm’s law on a napkin

When we apply a potential over a conductor, we create an electric field. The free electrons in the conductor will move until the electric field is zero. Because we keep the potential difference, they can’t succeed. The linear relationship between current and potential is called Ohm’s law. To avoid quantum mechanics, we’ll derive Ohm’s law in a crude way.

Given a wire with cross section \(A\), length \(l\) and potential difference \(V\).

In time \(\Delta t\), a free electron moves distance \(\Delta x\). The shaded wire segment, in the picture below, represents the free electrons passing though cross section \(A\). The shaded volume is \(A\,\Delta x\).

Let \(n\) be the number of free electrons per volume unit, and \(q_e\) be the charge of one electron. Then, the number of charge crossing through cross section \(A\) is $$ \Delta Q = -q_e\,n\,A\,\Delta x \nonumber $$

Substituting this back in the expression for the current $$ \begin{align*} I &= -\frac{q_e\,n\,A\,\Delta x}{\Delta t} & \left(\frac{x}{\Delta t}=v_d\right) \\ &= -q_e\,n\,A\,v_d & \left(\text{Eq. } \ref{eq:vd}\right) \\ &= q_e\,n\,A\,\frac{q_e\,E}{m_\rm{e}} \tau \\ &= \underbrace{\frac{{q_e}^2\,n\,\tau}{m_\rm{e}}}_{=\sigma} A\, E \end{align*} $$

Conductivity \(\sigma\) only depends on the material. Note: we often use resistivity \(\rho=\dfrac{1}{\sigma}\) $$ \begin{align*} I &= \sigma\,A\,E & \left(E=\frac{V}{l} \right) \\ &= \sigma\, A\, \frac{V}{l} \end{align*} $$

Rearranging terms $$ \begin{align*} V &= \underbrace{\frac{l}{\sigma\,A}}_{=R} \,\, I \end{align*} $$

**Ohm’s law** follows
$$
\begin{array}{l}
\shaded{
V = I\,R
} \quad \left[\Omega\right]
\tag{Ohm’s law} \label{eq:ohmslaw} \\
\text{where } R = \frac{l}{\sigma\,A} = \frac{l\,\rho}{A}
\end{array}
$$

The unit for resistance \(R\) is volt per ampere. We call that ohm, \(\Omega\).

Ohm’s law:

- resistance is proportional with the length of the wire (like water flowing through a longer pipe)
- resistance is inversely proportional with the area of the wire (like water flowing through a pipe with a wider cross section).
- Often also holds for insulators.

#### For copper

For copper, the conductivity \(\sigma\) at 300 K is $$ \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} \sigma_\text{Cu} &= \frac{e^2\,n\,\tau}{m_e} \\ &\approx \frac{\left(\numexp{1.6}{-19}\right)^2\,10^{29}\,(\numexp{3}{-14})}{10^{-30}} \\ &\approx \numexp{7.7}{7} \end{align*} $$

A \(1\,\rm{mm}^2\) copper wire of \(10\,\rm{m}\) has a resistance of roughly $$ \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} R &= \frac{l}{\sigma\,A} \\ &\approx \frac{10}{\numexp{7.7}{7}\,10^{-6}} \\ &\approx 0.13\,\Omega \end{align*} \nonumber $$

### Conductors vs. insulators

Assume a chunk of material as pictured below. What is the resistance?

For a good conductor, such as \(\rm{Au}, \rm{Ag}, \rm{Cu}\) $$ \begin{align*} \sigma &\approx 10^8 \\ \Rightarrow \rho &\approx \frac{1}{\sigma} = 10^{-8} \\ \Rightarrow R &\approx \frac{l\,\rho}{A} = \frac{1\,10^8}{10^{-6}} = 10^{-2}\,\Omega \end{align*} $$

For a good insulator, such as glass, porcelain $$ \begin{align*} \sigma &\approx 10^{-14} \\ \Rightarrow \rho &\approx \frac{1}{\sigma} = 10^{14} \\ \Rightarrow R &\approx \frac{l\,\rho}{A} = \frac{1\,10^{-14}}{10^{-6}} = 10^{20}\,\Omega \end{align*} $$You see 22 orders of magnitude difference between good conductors and insulators.

If I make the potential difference \(1\,\rm{V}\), then the current $$ \begin{align*} I_\text{cond.} &= \frac{V}{R} = \frac{1}{10^{-2}} = 100\,\rm{A} \\ I_\text{ins.} &= \frac{V}{R} = \frac{1}{10^{20}} = 10^{-20}\,\rm{A} \end{align*} $$

### The limit of Ohm’s law

The conductivity is a strong function of the temperature. If you increase the temperature, then the speed of the free electrons goes up and the time between collisions \(\tau\) goes down \(\Longrightarrow\) the conductivity will go down, and the resistance \(R\) will go up. In other words: when you heat up a substance the resistance goes up.

The moment the resistance becomes a function of the temperature, we call the breakdown of Ohm’s law.

One could say that \(R\) is \(R(T)\), but that is a poor way of saving the law, because the temperature \(T\) itself is a function of the current $$ \left\{ \begin{align*} R &= \frac{V}{I} = f(T) \\ T &= f(I) \end{align*} \right. $$

Notes:

- It is fortunate that the \(R\) goes up as the \(T\) goes up, otherwise a light bulb would self destroy.
- In general we will assume that Ohm’s law hold, and the heat produced does not play an important role.

## Networks of resistors

### Series

Suppose we have the circuit shown below with resistors \(R_1\) and \(R_2\) in series, and potential difference \(V\). What is the current, and potential difference over each resistor?

The current is the same for both resistors. Apply Ohm’s law for each resistor to find the voltage drop over each resistor $$ \left\{ \begin{align*} V_1 &= I\,R_1 \\ V_2 &= I\,R_2 \\ \end{align*} \right. $$

The total voltage is sum of the voltage drops over each resistor $$ \def\triangleq{\overset{\Delta}{=}} \begin{align*} V &= V_1 + V_2 \\ &= I\,R_1 + I\,R_2 & \text{(Ohm’s law)} \\ \Rightarrow I &= \frac{V}{R_1+R_2} \triangleq I = \frac{V}{R_{eq}} \end{align*} $$

So \(R_1\) and \(R_2\) in series behave as $$ \shaded{ R_{eq} = R_1 + R_2 } \tag{R in series} \label{eq:series} $$

### Parallel

Suppose we have the circuit shown below with resistors \(R_1\) and \(R_2\) in parallel, and potential difference \(V\). What is the current through each resistor, and the total current?

The voltage is the same over both resistors. Apply Ohm’s law for each resistor to find the current $$ \left\{ \begin{align*} I_1 &= \frac{V}{R_1} \\ I_2 &= \frac{V}{R_2} \end{align*} \right. $$

The total current is sum of the current through each resistor $$ \def\triangleq{\overset{\Delta}{=}} \begin{align*} I &= I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2} \\ &= V\left(\frac{1}{R_1}+\frac{1}{R_2}\right) \triangleq V\frac{1}{R} \\ \end{align*} $$

So \(R_1\) and \(R_2\) in parallel behave as $$ \shaded{ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} } \tag{R in parallel} \label{eq:parallel} $$

Or, taking the reciprocal on both sides $$ \shaded{ R_{eq} = \frac{R_1R_2}{R_1+R_2} } \tag{R in parallel} \label{eq:parallel2} $$