Conductors have an effectively infinite supply of charge, so when they are placed in an electric field, a **separation of charge** occurs. A battery with a potential drop across the ends creates such an electric field; when the ends are connected with a wire, charge will flow across it. The term given to the flow of charge is **electric current**, and it is measured in Amperes (A) --- Coulombs per second. Current is analogous to a river of water, but instead of water flowing, charge does.

**Current** is the number of Coulomb's that flow by per second. Thus 1 Amp of current is equivalent to saying that 1 C of electric charge is passing every second (i.e. the rate of change of charge is 1 C/s).

**Voltage** is the electrical energy density (energy divided by charge) and differences in this density (voltage) cause electric current in the circuit. **Batteries** and **power supplies** often provide a voltage difference across the ends of a circuit, but other **voltage sources** exist. Using the water analogy that current is a river, then differences in voltage can be thought of as pipes coming out of a water dam at different heights. The lower the pipe along the dam wall, the larger the water pressure, thus the higher the voltage. If that pipe is connected then current will flow. Current will be larger for the pipe with the greatest pressure (i.e. the lowest pipe on the dam wall).

\begin{align*}I = \frac{\Delta q}{\Delta t}\end{align*} ; current is the rate at which charge passes by; the units of current are Amperes \begin{align*}(1 \ A = 1 \ C/s)\end{align*}

#### Example

Somehow, you are able to see the electrons passing through a wire as current flows. Over the course of 5 seconds, you count \begin{align*}1.5*10^{20}\end{align*} electrons pass a single point in the wire. How much charge passed you and the current in the wire?

We can use the known charge of an electron to answer the first part of this problem.

\begin{align*} q&=1.5*10^{20}\:\text{electrons}*1.6*10^{-19}\:\text{C/electron}\\ q&=24\:\text{C}\\ \end{align*}

Now we use the equation above to determine the current

\begin{align*} I&=\frac{\Delta q}{\Delta t}\\ I&=\frac{24\:\text{C}}{5\:\text{s}}\\ I&=4.8\:\text{A}\\ \end{align*}

### Interactive Simulation

### Review

- The current in a wire is 4.5 A.
- How many coulombs per second are going through the wire?
- How many electrons per second are going through the wire?

- Which light bulb will shine brighter? Which light bulb will shine for a longer amount of time? Draw the schematic diagram for both situations. Note that the objects on the right are batteries, not resistors.

### Review (Answers)

- a. \begin{align*}4.5\mathrm{C}\end{align*} b. \begin{align*}2.8 \times 10^{19}\end{align*} electrons
- left = brighter, right = longer