Charge, current, voltage and resistance: a quick recap
Chapter 1 starts with five ideas you have met before. Current is the charge that passes a point each second: I = Q ÷ t, and 1 A is 1 coulomb every second. Potential difference is the work done on each coulomb between two points: V = W ÷ Q, and 1 V is 1 joule per coulomb. A source also has an emf, the work it does on each coulomb all the way round the circuit; the next week studies it. Resistance says how strongly a conductor opposes the current: R = V ÷ I in ohms (Ω), and Ohm’s law says that R stays the same while the temperature stays the same. For a wire, R = ρℓ ÷ A: it grows with the length ℓ and with the resistivity ρ of the material, and it falls when the cross-section A is larger. The conductivity σ = 1 ÷ ρ says how easily the material lets the current pass. Finally, by agreement the conventional current leaves the positive pole of a source and runs through the outside circuit to the negative pole, while the electrons really move the other way. In this course every number carries its unit, and results are rounded at the end to three significant figures unless the task says otherwise.
| Quantity | Symbol | Formula | Unit |
|---|---|---|---|
| current | I | Q ÷ t | ampere (A) |
| potential difference | V | W ÷ Q | volt (V) |
| resistance | R | V ÷ I | ohm (Ω) |
| resistivity | ρ | R·A ÷ ℓ | Ω·m |
| conductivity | σ | 1 ÷ ρ | (Ω·m)⁻¹ |
| Quantity | Given | In SI units |
|---|---|---|
| length ℓ | 5 m | 5 m |
| cross-section A | 0.1 mm² | 1 × 10⁻⁷ m² |
| resistivity ρ | 5 × 10⁻⁷ Ω·m | 5 × 10⁻⁷ Ω·m |
| resistance R | ? | 25 Ω |
Draw, with a grown-up, a circuit of one 1.5 V cell, a small torch bulb and two short wires. Mark the positive pole of the cell, then draw arrows for the conventional current from the positive pole round the outside to the negative pole, and arrows for the electrons the opposite way. If you also build it, use only that one cell and one small bulb, keep the wires short, and stop if anything feels warm. The drawing alone is enough.