Why is conventional current opposite to electron flow?

The convention was fixed before the electron was discovered, when charge carriers were assumed positive. It is kept because all the standard rules and equations are built around it.

Is current used up as it goes round a circuit?

No. Charge is conserved, so the current is identical at every point in a series loop. Energy is transferred, which is why the potential difference drops across components.

What is the difference between e.m.f. and potential difference?

Both are energy per unit charge in volts. E.m.f. is energy given to the charge by a source; p.d. is energy taken from the charge by a component.

Why do insulators barely conduct?

Because n, the number density of free charge carriers, is close to zero. With almost nothing available to carry charge, I = nAvq gives a negligible current.

Back to course

31–32, 37, 41ElectricityFoundation

Charge, Current and Potential Difference

A circuit moves charge around a loop and transfers energy as it goes. $\textbf{Current}$ measures how fast the charge flows; $\textbf{potential difference}$ measures how much energy each coulomb gives up. Keeping those two ideas distinct is most of what makes circuits make sense.

35 min Topic 3 · Electric Circuits

What you'll be able to do

Define current as rate of flow of charge and use $I = \Delta Q / \Delta t$
Define potential difference and e.m.f. in terms of energy per unit charge
Use $I = nAvq$ and interpret each quantity
Explain why a lamp lights immediately despite a very low drift velocity

Current as a rate of flow

Electric current is the rate of flow of charge: $I = \dfrac{\Delta Q}{\Delta t}$ . One ampere is one coulomb per second, and the ampere is an SI base unit — the coulomb is defined from it, not the other way round.

In a metal the charge carriers are free electrons, each carrying $-1.60 \times 10^{-19}\ \mathrm{C}$ . $\textbf{Conventional current}$ is defined as flowing from positive to negative, which is opposite to the actual electron flow — a historical convention that predates the discovery of the electron and is now too entrenched to change.

Charge is $\textbf{quantised}$ : any charge is a whole-number multiple of the elementary charge $e$ . That is why the number of electrons is always a whole number in these calculations.

I = \dfrac{\Delta Q}{\Delta t}

Current is rate of flow of charge.

Q = It

Charge from a steady current.

e = 1.60 \times 10^{-19}\ \mathrm{C}

The elementary charge.

Example 1A current of

250\ \mathrm{mA}

flows for 2 minutes. Find the charge transferred and the number of electrons.

1Convert:

I = 0.250

t = 120

Q = It = 0.250 \times 120 = 30

3Number of electrons:

\dfrac{30}{1.60\times10^{-19}}

Answer

Q = 30\ \mathrm{C}

; about

1.9 \times 10^{20}

electrons.

Tip — Convert milliamps to amps and minutes to seconds before substituting. Mixed units here produce answers wrong by factors of 1000 or 60 with no warning.

Potential difference and e.m.f.

$\textbf{Potential difference}$ across a component is the energy transferred $\textit{from}$ the charge to that component, per coulomb: $V = \dfrac{W}{Q}$ . One volt is one joule per coulomb.

$\textbf{Electromotive force}$ is the same quantity for a source: the energy transferred $\textit{to}$ each coulomb by the cell. Despite its name it is not a force — it is measured in volts, and the misnomer is another historical leftover.

The distinction is direction of energy transfer. A cell gives energy to the charge (e.m.f.); a resistor takes energy from it (p.d.). Both are joules per coulomb.

V = \dfrac{W}{Q}

Potential difference: energy per coulomb delivered.

\varepsilon = \dfrac{W}{Q}

E.m.f.: energy per coulomb supplied.

W = QV = IVt

Energy transferred.

Example 1A 12 V battery drives a current of 2.0 A through a lamp for 5.0 minutes. Find the energy transferred.

t = 300

s, so

Q = 2.0 \times 300 = 600

W = QV = 600 \times 12

Answer

7200\ \mathrm{J}

, or

7.2\ \mathrm{kJ}

The drift velocity equation

Current depends on how many charge carriers there are, how fast they move, and how much charge each carries. Combining these gives $I = nAvq$ , where $n$ is the number density of carriers (per cubic metre), $A$ the cross-sectional area, $v$ the mean drift velocity, and $q$ the charge per carrier.

The derivation is worth seeing: in time $t$ the carriers travel $vt$ , so the volume passing a point is $Avt$ , containing $nAvt$ carriers and therefore a charge $nAvtq$ . Dividing by $t$ gives the current.

The equation explains the difference between conductors, semiconductors and insulators. A metal has an enormous $n$ , around $10^{28}\ \mathrm{m^{-3}}$ ; a semiconductor has far fewer carriers; an insulator has almost none — which is why the same p.d. drives vastly different currents through them.

I = nAvq

Number density, area, drift velocity, charge.

v = \dfrac{I}{nAq}

Rearranged for drift velocity.

Example 1A copper wire of cross-sectional area

1.0 \times 10^{-6}\ \mathrm{m^{2}}

carries 3.0 A. Copper has

n = 8.5 \times 10^{28}\ \mathrm{m^{-3}}

. Find the drift velocity.

v = \dfrac{I}{nAq}

2Denominator:

8.5\times10^{28} \times 1.0\times10^{-6} \times 1.60\times10^{-19} = 1.36\times10^{4}

v = \dfrac{3.0}{1.36\times10^{4}}

Answer

v = 2.2 \times 10^{-4}\ \mathrm{m\,s^{-1}}

— about 0.2 mm per second.

Tip — A drift velocity of a fraction of a millimetre per second is the expected answer for a metal. Getting metres per second means $n$ or $A$ has been mishandled.

Why the lamp lights instantly

At that drift velocity an electron would take hours to travel the length of a mains cable, yet a lamp lights the moment the switch closes. The resolution is that the electrons do not have to travel from the switch to the lamp.

The conductor is already full of free electrons everywhere, including inside the lamp. Closing the switch establishes an electric field along the wire at close to the speed of light, and every electron in the circuit begins drifting almost simultaneously.

So what propagates quickly is the $\textbf{field}$ , not the charge carriers. The electrons themselves also move rapidly in random thermal motion — around $10^{5}\ \mathrm{m\,s^{-1}}$ — with the drift velocity being only a tiny net bias superimposed on that chaos.

Example 1Explain why a lamp lights immediately when a switch is closed, given a drift velocity of about

10^{-4}\ \mathrm{m\,s^{-1}}

1Free electrons are already present throughout the circuit, including in the lamp.

2Closing the switch sets up an electric field along the conductor at near light speed.

3All electrons start drifting almost at once, so the lamp’s electrons move immediately.

AnswerThe electric field propagates almost instantly and the charge carriers are already everywhere, so no electron has to travel from the switch to the lamp.

Equation recap

I = \tfrac{\Delta Q}{\Delta t}

Definition of current.

V = \tfrac{W}{Q}

Potential difference.

W = QV = IVt

Energy transferred.

I = nAvq

Drift velocity equation.

e = 1.60\times10^{-19}\ \mathrm{C}

Elementary charge.

Common mistakes to avoid

Saying current is "used up" by a component.

Charge is conserved and current is the same all round a series loop. Energy is what is transferred.

Treating e.m.f. as a force.

It is energy per unit charge, measured in volts. The name is historical.

Using area in

\mathrm{mm^{2}}

I = nAvq

Convert to

\mathrm{m^{2}}

— a factor of

10^{6}

that silently wrecks the answer.

Claiming electrons travel from the switch to the lamp at high speed.

The field propagates quickly; the electrons drift slowly and are already present throughout.

Key takeaways

Current is rate of flow of charge, $I = \Delta Q/\Delta t$ , measured in amperes.
Potential difference and e.m.f. are both energy per coulomb, differing in direction of transfer.
Charge is quantised in multiples of $e = 1.60\times10^{-19}\ \mathrm{C}$ .
$I = nAvq$ links current to carrier density, area, drift velocity and charge.
Lamps light instantly because the field propagates fast and carriers are already present throughout.

Test yourself

Answer a few questions, then review anything you missed.

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