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CIE 9702 Physics · AS · Topic 9

Electricity

Clear, syllabus-mapped CIE 9702 Physics revision notes on electricity: explanations, worked examples and exam technique, then a free targeted practice drill.

CIE 9702 PhysicsASFree revision notes
Contents: 9 sections

Syllabus points

Current

Current is the rate of flow of charge:

I = Q / t

measured in amperes. The ampere is a base unit; the coulomb is derived from it, as 1 C = 1 A s. Questions ask which of the two is the base unit, and it is the ampere.

Conventional current flows from positive to negative, which is opposite to the direction the electrons actually move. That convention is fixed and questions rely on it.

Charge is quantised: any charge is a whole number multiple of e = 1.60 × 10⁻¹⁹ C.

Drift velocity

I = Anvq

where A is the cross-sectional area, n the number of charge carriers per unit volume, v the mean drift velocity and q the charge on each carrier.

The equation says a great deal once you read it as a proportionality. For a given current:

Drift velocities in a metal are surprisingly small, of order 10⁻⁴ m s⁻¹. The lamp lights immediately anyway, because the electric field is established through the circuit at nearly the speed of light and all the electrons start moving at once.

Potential difference and e.m.f.

Potential difference is the energy transferred from electrical to other forms per unit charge:

V = W / Q

The volt is one joule per coulomb.

Electromotive force is the energy transferred from other forms to electrical per unit charge, by a source such as a cell. It has the same unit and the same defining equation, and the difference is the direction of the energy conversion. E.m.f. is not a force, despite the name.

Both definitions are per unit charge, not per unit time, and "the energy per unit charge" is the phrasing the mark scheme wants.

Power

Three forms, all equivalent through V = IR:

P = VI = I²R = V²/R

Choosing between them is a matter of what the question gives you. Two of them are worth thinking about carefully:

Those two conclusions point in opposite directions, and a question that asks which component is brightest is asking you to notice which arrangement you are in.

Energy transferred: W = VIt.

Resistance and Ohm's law

R = V / I

measured in ohms. Every component has a resistance at any given moment; the question is whether it is constant.

Ohm's law: the current through a conductor is proportional to the potential difference across it, provided physical conditions such as temperature remain constant. A component obeying this is ohmic, and its I-V graph is a straight line through the origin.

I-V characteristics

Metallic conductor at constant temperature: a straight line through the origin. Ohmic. The gradient is 1/R.

Filament lamp: a curve that flattens as V increases. As the current rises the filament gets hotter, the lattice ions vibrate with greater amplitude, the electrons collide with them more often, and the resistance increases. Non-ohmic. The curve bends towards the voltage axis.

Diode: almost no current in reverse bias, and in forward bias almost no current until about 0.6 V, then a steeply rising current. Non-ohmic, and it has a very high resistance one way and a low resistance the other.

Thermistor (negative temperature coefficient): resistance falls as temperature rises, because more charge carriers are released. Note this is the opposite of a metal, and the reason is different: in a metal n is fixed and the collisions increase, while in a semiconductor n rises sharply.

Light-dependent resistor: resistance falls as light intensity rises, for the same reason.

The gradient of an I-V graph is not the resistance unless the line passes through the origin. Resistance at a point is V/I, the ratio of the coordinates, not the local gradient. That distinction is examined directly on the lamp characteristic.

Resistivity

R = ρL / A

where ρ is the resistivity in Ω m, a property of the material, independent of the shape of the sample.

So resistance rises with length and falls with area. Doubling the length doubles the resistance; doubling the diameter quadruples the area and quarters the resistance.

Worked example. A wire has resistance R. A second wire of the same material has the same mass but twice the length. What is its resistance?

Same material and same mass means the same volume, V = AL. Doubling L must therefore halve A.

R' = ρ(2L) / (A/2) = 4R.

The trap is answering 2R by changing the length and forgetting that the same mass drawn out longer must also be thinner. Questions that specify equal mass or equal volume are always testing this.

Common mistakes

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