Current as the rate of flow of charge, the equation $I = nAvq$, potential difference and EMF as energy per unit charge, and electrical power and energy.

What this dot point is asking

Edexcel wants you to define current as the rate of flow of charge, use the carrier equation $I = nAvq$ both qualitatively and in calculations, distinguish potential difference from EMF as two kinds of "energy per unit charge", and calculate electrical power and energy using $P = VI$ and its Ohm's law variants. This sits at the start of the Electric circuits topic and underpins everything that follows about resistance, internal resistance and potential dividers.

The answer

Current as the rate of flow of charge

By convention, current direction is the direction of flow of positive charge. In a metal the carriers are actually electrons drifting the opposite way, so conventional current and electron flow point in opposite directions. The total charge that has flowed in a time $t$ is the area under a current-time graph, $Q = \int I \, dt$, which for a steady current is simply $Q = It$.

The carrier equation $I = nAvq$

Consider a conductor of cross-sectional area $A$ containing $n$ free charge carriers per unit volume, each of charge $q$, drifting with mean speed $v$. In a time $\Delta t$ every carrier moves a distance $v \Delta t$, so all carriers within a cylinder of volume $A v \Delta t$ pass a chosen cross-section. The number of carriers is $n A v \Delta t$ and the charge they carry is $\Delta Q = n A v q \, \Delta t$. Dividing by $\Delta t$:

Because $n$ in a metal is of order $10^{28}$ m$^{-3}$, the drift velocity is tiny, a fraction of a millimetre per second, even for everyday currents. The lights come on almost instantly not because electrons race along the wire, but because the electric field that pushes them is established at close to the speed of light throughout the circuit.

A useful corollary: for a fixed current, $v \propto 1/A$. Where a wire narrows, the same current flows through a smaller area, so the carriers must drift faster. This is exactly analogous to a river speeding up where its channel narrows.

Potential difference and EMF

Both potential difference and EMF measure energy per unit charge, $V = \frac{W}{Q}$, and are measured in volts (joules per coulomb). The distinction is about energy direction:

Electrical power and energy

The rate of energy transfer is power. Since $V$ is energy per coulomb and $I$ is coulombs per second, their product is energy per second:

The kilowatt-hour is a practical energy unit, the energy used by a $1$ kW device in one hour: $1 \text{ kWh} = 1000 \times 3600 = 3.6 \times 10^{6}$ J.

Examples in context

Domestic wiring. A $3$ kW kettle on a $230$ V UK mains supply draws $I = P/V = 3000/230 = 13$ A, which is why kettle plugs use a $13$ A fuse. The same current in the thin element wire produces a high drift speed and rapid heating, while in the thicker supply cable the larger area keeps the heating per metre low.

Microelectronics. In a copper interconnect on a chip with cross-section of order $10^{-13}$ m$^2$, even a microamp gives a drift velocity comparable to a macroscopic wire because $v \propto I/A$. Designers limit current density $J = I/A = nvq$ to avoid electromigration, where fast-drifting electrons physically displace metal atoms and break the track.

Try this

Q1. Define electric current. [1 mark]

• Cue. The rate of flow of electric charge, $I = \Delta Q / \Delta t$.

Q2. A $12$ V battery delivers $0.50$ A to a lamp for $2.0$ minutes. Find the energy transferred. [2 marks]

• Cue. $W = VIt = 12 \times 0.50 \times 120 = 720$ J.

Q3. Explain why the drift velocity of electrons in a connecting wire is very small even though the lamp lights almost instantly. [3 marks]

• Cue. $n$ is of order $10^{28}$ m$^{-3}$, so from $I = nAvq$ a modest current needs only a tiny $v$; the field that drives the electrons is established almost instantly along the whole circuit.