Currents: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 15 · 9478 Topic 15 · Read time: ~10 minutes
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Key Formulas
Current and drift velocity
Current is the rate of flow of charge, I = Q/t.
In a conductor, I = nAvq: n is the number density of charge carriers, A the cross-sectional area, v the average drift velocity and q the charge on each carrier.
Drift velocities in metals are tiny, around 10⁻⁴ m s⁻¹, because n is enormous. A semiconductor has a much smaller n, so for the same current its carriers drift much faster.
Trick: For a wire that narrows, I is the same throughout (series), so v is larger where A is smaller.
P.d., e.m.f. and power
Potential difference is the electrical work done per unit charge, V = W/Q: energy converted from electrical to other forms in a component.
E.m.f. is the energy converted from other forms (chemical, for example) into electrical energy per unit charge by a source. Both are measured in volts; the difference is the direction of energy conversion.
Power: P = VI = I²R = V²/R.
Alternating current
A sinusoidal a.c. is written x = x₀ sin ωt, with peak value x₀, period T and frequency f = 1/T.
The r.m.s. value is the steady d.c. value that produces the same mean power in a resistor. For a sinusoid, I_rms = I₀/√2 and V_rms = V₀/√2.
For a resistive load, the mean power is half the peak power, P_mean = ½I₀²R.
Mains supplies are quoted as r.m.s. values: 230 V r.m.s. has a peak of about 325 V.
Trap: Using peak values where the question means r.m.s. Unless stated otherwise, quoted a.c. values are r.m.s.
Half-wave rectification
A single diode in series with a load lets current through only when it is forward biased. The output is one half of each cycle, with zero during the other half: half-wave rectification.
The output is one-directional but still pulses, so it is d.c. only in the sense that it never reverses.
Worked Example
A copper wire of cross-sectional area 1.0 × 10⁻⁶ m² carries 3.0 A. Copper has 8.5 × 10²⁸ free electrons per m³. Electron charge = 1.60 × 10⁻¹⁹ C.
(a) Calculate the drift velocity of the electrons. [2]
v = I / (nAq) = 3.0 ÷ (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.60 × 10⁻¹⁹)
= 2.2 × 10⁻⁴ m s⁻¹
(b) A 40 Ω heater is connected to a 230 V r.m.s. supply. Calculate the peak voltage and mean power. [3]
V₀ = √2 × 230 = 325 V
P_mean = V_rms² / R = 230² ÷ 40 = 1.32 × 10³ W
(c) State the peak power dissipated in the heater. [1]
Peak power = 2 × mean power = 2.65 × 10³ W
Why this question is a good test of the topic: it shows how slow drift velocity is, and checks that students use r.m.s. values for mean power and know peak power is twice the mean.
Frequently Asked Questions
What is drift velocity?
The average velocity of charge carriers along a conductor when a current flows. In metals it is very small, about 10⁻⁴ m s⁻¹.
What is the difference between e.m.f. and p.d.?
E.m.f. is energy converted into electrical energy per unit charge by a source. P.d. is electrical energy converted into other forms per unit charge in a component.
What is the r.m.s. value of an alternating current?
The steady direct current that would give the same mean power in a resistor. For a sinusoid, I_rms = I₀/√2.
How does mean power relate to peak power for a sinusoidal a.c.?
For a resistive load, the mean power is half the peak power.
What is half-wave rectification?
Using a single diode so current flows only in one half of each a.c. cycle, giving a one-directional, pulsing output.
r.m.s. and peak values still confusing?
Small group A Level H2 Physics classes at TGC Academy's Bishan, Bukit Timah and Potong Pasir centres, helmed by Andrew Seah, MOE Award-Winning Teacher and Marshall Cavendish textbook author, with a team of tutors.