§ 9.1 Electric current
Key ideas
- Current is the rate of flow of charge, I = Q/t, in amperes.
- Charge is quantised in units of the elementary charge e = 1.6 × 10⁻¹⁹ C.
- In a metal, current is carried by electrons drifting at speed v: I = nAvq, with n the number density of carriers and A the cross-sectional area.
- Drift speeds are tiny (well under 1 mm s⁻¹); the electric field that sets charges moving travels near light speed, which is why a lamp lights at the flick of a switch.
Equations
I = Q / tcurrent = charge ÷ timeA
I = n A v qnumber density × area × drift speed × charge per carrierA
Fig. 1 · I = nAvq: the current through the shaded face counts how many carriers, each with charge q, drift through area A per second. Conventional current runs opposite to the electron drift.
Watch out: for the same current, a thinner section of wire forces a faster drift speed (v = I/nAq): n and q are fixed by the material, so v must rise where A falls.
§ 9.2 Potential difference and power
Key ideas
- Potential difference is the energy transferred per unit charge, V = W/Q; one volt is one joule per coulomb.
- Power has three equivalent forms: P = VI, P = I²R, P = V²/R; pick the one whose quantities you actually know.
Equations
V = W / Qp.d. = energy transferred ÷ chargeV
P = V I = I²R = V²/Rthree equivalent power forms, linked by V = IRW
Watch out: in P = I²R, doubling the current quadruples the dissipation. This square is the whole argument for high-voltage transmission: halve I and the cable loss falls by four.
§ 9.3 Resistance and resistivity
Key ideas
- Resistance R = V/I. Ohm's law: at constant temperature, I is proportional to V for a metallic conductor.
- Resistivity ρ is the material property in R = ρL/A; its unit is Ω m.
- I-V characteristics: metallic conductor, straight line through the origin; filament lamp, curve flattening as it heats; diode, conduction in one direction beyond a threshold; thermistor resistance falls as temperature rises; LDR resistance falls as light level rises.
Equations
R = V / Iresistance = p.d. ÷ currentΩ
R = ρ L / Aresistivity × length ÷ cross-sectional areaΩ
Fig. 2 · Geometry sets the resistance: double the length and R doubles; double the diameter and the area quadruples, so R falls to a quarter.
Watch out: resistivity's unit is Ω m, not Ω/m, and halving a wire's diameter quarters the area A = πd²/4, multiplying R by four, not two.