9. Electricity

Conventional current is defined in the direction positive charge would move. Positive carriers move with conventional current; negative carriers move in the opposite direction. In a metal wire, electrons drift from the negative terminal towards the positive terminal, opposite to conventional current.

The current magnitude tells how rapidly charge crosses the section: 1 ampere means 1 coulomb per second. A larger number of carriers crossing each second, or a larger charge per carrier, gives a larger current.

Use the sign to identify positive or negative carriers and interpret n as a count, not a continuously adjustable fraction.

A charge of −3.2×10⁻¹⁹ C corresponds to two excess electrons.

Use seconds and coulombs, and state whether Q is magnitude or signed charge according to the chosen direction.

A 0.50 A current flowing for 4.0 minutes transfers Q=120 C.

A wire carries 1.2 A with A = 4.7 × 10⁻⁷ m², n = 8.5 × 10²⁸ m⁻³ and q = 1.60 × 10⁻¹⁹ C. v = I/(Anq) = 1.2/[(4.7 × 10⁻⁷)(8.5 × 10²⁸)(1.60 × 10⁻¹⁹)] = 1.9 × 10⁻⁴ m s⁻¹.

For series sections made of the same material, I, n and q are the same. Therefore v ∝ 1/A: halving wire diameter makes area one quarter as large and drift speed four times larger.

Specify the component and direction of energy transfer. A source supplies energy; a resistor or motor transfers it to other stores.

A 6 V battery supplies 6 J to each coulomb passing through the external circuit ideally.

Use joules and coulombs, and identify whether W is supplied or dissipated. The sign depends on the chosen direction.

Moving 5 C through a 12 V motor transfers 60 J to mechanical and thermal stores.

Choose the form that uses known quantities and distinguish input power from useful output. These equations assume the component’s voltage-current relation at that operating point.

A 6 Ω resistor carrying 2 A dissipates P=I²R=24 W, also equal to VI when V=12 V.

For a non-ohmic device the ratio can change with voltage, current or temperature, so call it operating-point resistance when appropriate.

A component carrying 0.50 A at 4.0 V has resistance 8.0 Ω at that point.

A component has 6.0 V across it and carries 0.25 A. Its resistance at that operating point is R = 6.0/0.25 = 24 Ω. If that resistance is constant, 3.0 V would produce 0.125 A.

From an I-V graph, choose the required point and calculate R = V/I using its coordinates. The gradient of an I-versus-V graph is 1/R only when the graph is a straight line through the origin.

Sketch from the origin outward: decide whether the component is symmetric under voltage reversal; decide whether resistance stays constant, increases, or falls; then convert that into I-V steepness. On these axes, greater resistance means smaller I/V and therefore a less-steep line from the origin to the operating point.

As voltage magnitude increases, current still increases, but less than proportionally. Therefore V/I grows and an I-V characteristic becomes less steep away from the origin. The curve is approximately symmetric for positive and negative voltage because either current direction heats the filament.

If current decreases, heating falls, the filament cools and its resistance decreases. The changing resistance—not a change in electron charge—is why a filament lamp does not have a straight constant-temperature metal characteristic during normal operation.

Test proportionality experimentally and identify variables such as temperature. Use a non-linear model or a local resistance when the graph curves.

A fixed resistor can be approximately ohmic over its rated range, while a filament lamp fails the test as it warms.

A wire has ρ = 1.7 × 10⁻⁸ Ω m, L = 2.0 m and diameter 0.50 mm. Radius r = 0.25 × 10⁻³ m, so A = πr² = 1.96 × 10⁻⁷ m². Hence R = (1.7 × 10⁻⁸ × 2.0)/(1.96 × 10⁻⁷) = 0.17 Ω (2 s.f.).

For unchanged material and temperature, R ∝ L/A. Doubling length doubles R. Doubling diameter makes area four times larger, so R becomes one quarter. Rearrangement gives ρ = RA/L when resistivity is required.

Use it as a variable sensor: the circuit output depends on how the LDR is arranged with other resistors and the supply.

In a potential divider, brighter light lowers the LDR resistance and changes the share of supply voltage across the other component.

A qualitative graph of resistance R against temperature T slopes downward throughout the stated range and is usually curved rather than linear. It starts from the stated resistance at the initial temperature and remains above R = 0.

With a fixed supply directly across the thermistor, heating lowers R and increases current. In a series circuit or potential divider, first update the thermistor resistance, then use the actual topology to determine current, power or the measured output voltage.