In Current and Voltage you met current as the flow of charge around a circuit. Now we measure it. Because charge is conserved, counting the charge that passes a point tells you a great deal about the whole circuit.
Why it works
Stand at one point in a wire and count the charge going past. The current is how much passes per second. In a metal the moving charges are electrons, and they drift slowly, typically well under a millimeter per second, but the push that moves them spreads through the circuit almost at once, so a lamp lights the instant the switch closes. Charge cannot pile up anywhere in a steady circuit, so on a single path the same current passes every point, before and after each part.
Key idea
$$I = \frac{\Delta Q}{\Delta t}, \qquad 1\,\text{A} = 1\,\text{C/s}$$ Conventional current points the way positive charge would flow: out of the battery's $+$ terminal, around the circuit, and back into its $-$ terminal. In wires, electrons drift the opposite way.
Worked examples
Worked example: Finding a current
$3.0\,\text{C}$ of charge passes a point in $1.5\,\text{s}$. What is the current?
- $I = \dfrac{\Delta Q}{\Delta t} = \dfrac{3.0\,\text{C}}{1.5\,\text{s}} = 2.0\,\text{A}$.
Answer: $2.0\,\text{A}$
Worked example: Charge and electrons
A current of $0.50\,\text{A}$ flows for $2.0$ minutes. How much charge passes, and how many electrons is that?
- $\Delta Q = I\,\Delta t = (0.50\,\text{A})(120\,\text{s}) = 60\,\text{C}$.
- $N = \dfrac{60\,\text{C}}{1.60 \times 10^{-19}\,\text{C}} = 3.75 \times 10^{20}$ electrons.
Answer: $60\,\text{C}$; $3.75 \times 10^{20}$ electrons
Common mistake
Thinking current gets used up as it passes through a bulb, so less comes out than goes in. The charge that goes in must come out; otherwise it would pile up inside the bulb. What the bulb uses is energy: each coulomb leaves at a lower potential than it entered. Check: put an ammeter on each side of the bulb; both read the same.