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Currents

Study H2 Physics charge carriers, e.m.f., potential difference and electrical power, then sinusoidal A.C., r.m.s. values and diode rectification.

  • GCE A-Level H2 Physics 2027
  • 6 lessons
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Before you begin

Current measures charge flow. E.m.f. and potential difference measure energy per charge. Use these ideas to understand power, sinusoidal A.C. and rectification.

Be comfortable with: O-Level current electricity, and sine graphs from simple harmonic motion.

Learning goals
  • Relate current to charge flow, number density and drift velocity.
  • Apply potential difference, e.m.f. and electrical power relationships.
  • Represent sinusoidal A.C. and use peak and r.m.s. values.
  • Analyse mean power in resistive A.C. loads and half-wave rectification.
Syllabus statements covered
  • show an understanding that electric current is the rate of flow of charge and solve problems using I = Q/t
  • derive and use the equation I = nAvq for a current-carrying conductor, where n is the number density of charge carriers and v is the drift velocity
  • recall and solve problems using the equation for potential difference in terms of electrical work done per unit charge, V = W/Q
  • recall and solve problems using the equations for electrical power P = VI, P = I²R and P = V²/R
  • distinguish between electromotive force (e.m.f.) and potential difference (p.d.) using energy considerations
  • show an understanding of and use the terms period, frequency, peak value and root-mean-square (r.m.s.) value as applied to an alternating current or voltage
  • represent a sinusoidal alternating current or voltage by an equation of the form x = x₀ sin ωt
  • deduce that the mean power in a resistive load is half the maximum (peak) power for a sinusoidal alternating current
  • distinguish between r.m.s. and peak values, and recall and use Iᵣₘₛ = I₀/√2 and Vᵣₘₛ = V₀/√2 for the sinusoidal case
  • explain the use of a single diode for the half-wave rectification of an alternating current

How the ideas connect

Current and electron drift in a metalA straight section of metal wire. A labelled conventional-current arrow points right. Three electrons are labelled e minus, and a labelled electron-drift arrow points left. Electron drift is opposite to conventional current.Conventional currente−e−e−Electron drift
In a metal, electron drift opposes conventional current. Random thermal motion is not drawn. This is a schematic wire section, not a complete circuit.
  1. Follow the charge

    Current is charge crossing a point per second. In a metal, electron drift opposes conventional current.

    Step 1
  2. Account for energy

    A source supplies energy per charge; a component transfers energy per charge. Power measures energy transferred per second.

    Step 2
  3. Compare A.C. heating

    Connect peak, period and frequency, then use r.m.s. for resistor heating. A diode changes the waveform by rectifying it.

    Step 3
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Lessons

Work through them in order.

Practise and check

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Topic reference

Quick reference

Charge per second

I = Q/t for a steady current.

1 A means 1 C crosses a point each second. For magnitudes, I = nAvq relates current to mean drift speed.

Common pitfall

Electrons move opposite to conventional current. Charge is not consumed by a component.

Read charge flow and drift →

Energy per charge

V = W/Q and P = VI

P.d. measures energy transferred per charge in a component. E.m.f. measures energy supplied per charge by a source. 1 V = 1 J C⁻¹.

Common pitfall

Use the voltage across and current through the same component in P = VI.

Read the energy explanation →

Peak or r.m.s.?

Iᵣₘₛ = I₀/square root of 2 for a sinusoid.

The r.m.s. current gives the same mean heating as a steady D.C. current in the same resistor.

Common pitfall

The peak value goes in the sine equation. The peak divided by √2 rule does not apply to every waveform.

Explore the waveform simulation →
Complete quick-reference table
IdeaRelationNotes
currentI = Q/trate of flow of charge
drift velocityI = nAvqn is the number of carriers per unit volume
potential differenceV = W/Qelectrical energy to other forms, per unit charge
electrical powerP = VI = I²R = V²/RV, I and R for the same component
sinusoidal A.C.x = x₀ sin ω tω = 2π f = 2π/T
r.m.s. valueIᵣₘₛ = I₀/(square root of 2), Vᵣₘₛ = V₀/(square root of 2)sinusoids only
mean power in a resistor⟨P⟩ = (1/2)P₀ = Iᵣₘₛ²Rsinusoidal supply

Core knowledge to remember

Definitions and core knowledge
  • Electric current is the rate of flow of charge. Conventional current is in the direction positive charge would flow.
  • The potential difference across a component is the energy transferred from electrical energy to other forms per unit charge passing through it.
  • The electromotive force of a source is the energy transferred from other forms to electrical energy per unit charge passing through the source.
  • The root-mean-square value of an alternating current is the steady direct current that would produce the same mean power in the same resistor.
  • For a sinusoidal supply, the mean power in a resistive load is half the maximum power.
  • A single diode gives half-wave rectification: current in one direction only, in pulses at the supply frequency, with zero current for half of each cycle.

Common mistakes

More mistakes to check
  1. Drawing electron drift in the same direction as conventional current.
  2. Treating e.m.f. and p.d. as the same thing because both are measured in volts.
  3. Putting the r.m.s. value in front of the sine in x = x₀ sin ω t, or the peak value into the power equations as if it were r.m.s.
  4. Using I₀/square root of 2 for a waveform that is not a full sinusoid, such as a rectified output.
  5. Calling a half-wave rectified output steady D.C.