9. Electricity

Current is I=∆Q/∆t: the rate at which charge passes a point, measured in amperes, with 1 A=1 C s⁻¹.

Choose a direction convention and distinguish conventional current from the motion direction of negative electrons.

If 12 C passes a wire section in 3.0 s, the current is 4.0 A.

Current is not consumed by a component in a series circuit; charge flow rate is continuous around the circuit.

Electric charge occurs in discrete amounts q=ne, where n is an integer and e≈1.60×10⁻¹⁹ C.

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.

The quantisation statement does not mean every macroscopic measurement visibly jumps by e; huge carrier counts make charge appear continuous.

For constant current, charge transferred in time t is Q=It; for changing current, use the area under an I–t graph.

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.

Do not use minutes directly with amperes, and do not assume Q=It for a changing current without integrating or finding graph area.

For charge carriers of number density n, cross-sectional area A, drift speed v and charge magnitude q, I=Anvq.

The formula counts carriers crossing the section each second; use the conductor’s actual area and the carrier type appropriate to the material.

A thinner wire with the same carrier density and drift speed carries less current because A is smaller.

Drift speed is not the same as the electromagnetic signal speed, and n is a volume number density, not total carrier count.

Potential difference V is the energy transferred per coulomb when charge moves between two points: V=energy per charge.

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.

Potential difference is not current and is not “used up” as charge circulates; components have voltage drops or rises.

The potential difference across a component is V=W/Q, so W=VQ is the energy transferred when charge Q crosses it.

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.

A 12 V label is not 12 J total; it means 12 J per coulomb under the specified operating conditions.

For a component, power P=VI. Combining with V=IR gives P=I²R and P=V²/R, with the appropriate measured voltage and current.

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.

Do not use P=V²/R for a non-ohmic device with a fixed resistance assumption unless its operating-point resistance is known.

Resistance is R=V/I at an operating point, measured in ohms. It describes how much potential difference is needed for a given current.

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.

Resistance is not “used up” by a component and is not necessarily constant for every material or device.

Ohm’s law states that current is directly proportional to potential difference for a conductor at constant physical conditions: V=IR.

Check temperature and other conditions before applying it. Rearrange only after identifying which quantity is unknown.

A 10 Ω resistor at 5.0 V carries 0.50 A, provided its temperature remains effectively constant.

V=IR defines resistance at a point, but Ohm’s law requires a straight-line proportional relationship over the tested range.

For an ohmic metal kept at constant temperature, current is proportional to voltage, so the I–V graph is a straight line through the origin.

The gradient of an I–V graph is 1/R, whereas the gradient of a V–I graph is R. State which axes are used.

Doubling voltage on a constant-temperature 4 Ω wire doubles current from 0.5 A to 1.0 A.

A curved I–V graph does not mean the device has no resistance; it means resistance changes with operating point.

Increasing current heats a filament, raising its temperature and increasing resistance because lattice vibrations scatter charge carriers more strongly.

Read the curved I–V characteristic rather than assuming V/I is fixed. Cooling on reduction can make the path history relevant.

At higher voltage a lamp’s current rises less than proportionally, showing a larger operating-point resistance.

The lamp is not non-ohmic because current stops; its resistance changes as temperature changes.

A component obeys Ohm’s law only when its V–I relation is proportional under constant conditions, giving constant resistance.

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.

“Resistance” existing for a device does not prove that V/I stays constant across all voltages.

The resistance of a uniform conductor is R=ρL/A, where ρ is resistivity, L length and A cross-sectional area.

Use the conductor’s material through ρ, and account for geometry: longer wires resist more; thicker wires resist less.

Doubling a wire’s length doubles R, while doubling its diameter quarters R because area scales with diameter squared.

Resistivity is a material property, not the same as the resistance of one particular wire.

A light-dependent resistor’s resistance decreases as light intensity increases, because illumination creates more mobile charge carriers in its semiconductor material.

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.

An LDR responds to light intensity, not simply to elapsed time or temperature; the direction of voltage change depends on circuit placement.

For an NTC thermistor, resistance decreases when temperature increases because carrier availability and scattering conditions change in the semiconductor.

The sensor direction must be combined with circuit topology to predict whether an output voltage rises or falls.

In a divider with the thermistor on top, heating it lowers its resistance and lowers the output across it.

“Thermistor resistance decreases” is not enough to predict a voltage without specifying which voltage is measured and where the thermistor sits.