Current as the rate of flow of charge; the coulomb and the ampere; counting electrons

Potential difference as energy transferred per unit charge; the volt

Resistance R = V/I, the ohm, and Ohm's law as a special case

Electrical power and energy: P = VI, W = VIt, P = I²R, P = V²/R, kilowatt-hours

Drift velocity and the derivation of I = nqvA

How the number density of charge carriers explains the resistivities of metals, semiconductors and insulators

Every circuit question rests on three quantities: how much charge flows each second, how much energy each coulomb carries, and how hard it is to push the charge through. Current is the first.

This equation is in the formulae list supplied in the exam. For a steady current ΔQ = IΔt; for a changing current, the charge is the area under the current–time graph.

Units: the ampere is an SI base unit. The coulomb is defined from it: 1 C = 1 A s, the charge passing a point when 1 A flows for 1 s.

Charge carriers: free (delocalised) electrons in metals; positive and negative ions in electrolytes and ionised gases; electrons and "holes" in semiconductors.

Charge is quantised: any charge is a whole number of electron charges, Q = Ne, so the number of electrons passing is N = Q/e (learn this; it is not given).

Charge is conserved: it is not used up in a component, so the current into a component equals the current out of it. Measure current with an ammeter in series; an ideal ammeter has zero resistance.

Conventional current is drawn from the positive terminal to the negative terminal round the external circuit.

In a metal the electrons actually move the opposite way; the convention was fixed before the electron was discovered.

WORKED EXAMPLE · QUESTION

A phone charger delivers a steady current of 1.8 A for 45 minutes. e = 1.60 ×10⁻¹⁹ C. Calculate (a) the charge that flows and (b) the number of electrons this represents.

SOLUTION

Time in seconds: t = 45 ×60 = 2700 s.

(a) ΔQ = IΔt = 1.8 ×2700 = 4.9 × 10³ C (4860 C).

(b) N = Q/e = 4860 ÷1.60 ×10⁻¹⁹ = 3.0 × 10²² electrons.

This equation is in the formulae list supplied in the exam. Across a component such as a resistor or lamp, energy is transferred from the charge carriers to the component, for example as thermal energy.

The volt: 1 V = 1 J C⁻¹. A p.d. of 1 V means 1 J is transferred for every coulomb that passes.

P.d. is measured with a voltmeter in parallel with the component; an ideal voltmeter has infinite resistance, so it draws no current.

For a single electron, the energy transferred is W = eV, which gives the electronvolt used on 2.5. The energy given to each coulomb by a source is its e.m.f. (2.9).

WORKED EXAMPLE · QUESTION

The p.d. across the lamp of an LED torch is 3.6 V and the current in it is 0.25 A. The torch is switched on for 20 minutes. e = 1.60 ×10⁻¹⁹ C. Calculate (a) the charge that passes through the lamp, (b) the energy transferred to the lamp and (c) the energy transferred by each electron.

SOLUTION

(a) t = 20 ×60 = 1200 s, so ΔQ = 0.25 ×1200 = 300 C.

(b) W = VQ = 3.6 ×300 = 1.1 × 10³ J (1080 J).

(c) The same equation for one electron's charge: W = 3.6 ×1.60 ×10⁻¹⁹ = 5.8 × 10⁻¹⁹ J (= 3.6 eV).

R = V/I is in the formulae list supplied in the exam. It defines resistance for any component at any instant: use the values of V and I at the same operating point. Resistance measures how much p.d. is needed per ampere: the larger R, the smaller the current for a given p.d.

Ohm's law is a special case: an ohmic conductor has a resistance that stays the same whatever the current, and its I–V graph is a straight line through the origin.

R = V/I is not Ohm's law. It holds for every component; Ohm's law adds the claim that R is constant.

The condition matters: if a wire heats up, its resistance rises and I is no longer proportional to V. Filament lamps, thermistors and diodes are non-ohmic (2.7).

WORKED EXAMPLE · QUESTION

A student uses the circuit above to measure the current in a coil of constantan wire for several p.d.s. The coil is kept in a beaker of water at room temperature.

Show that the coil obeys Ohm's law and determine its resistance.

SOLUTION

V/I for every row: 11.9, 12.0, 11.9, 12.0, 12.0, 12.0 Ω. All values agree to within 1%, so V/I is constant: I ∝V at constant temperature, and Ohm's law is obeyed.

Resistance: the mean of the ratios, or the inverse gradient of the line: R = 12 Ω (12.0 Ω).

The water bath keeps the temperature constant, and small currents also limit heating.

Plotting the graph is the stronger test: a straight line through the origin shows proportionality, not just a straight line.

Here P is power (W), W energy transferred (J) and t time (s); all four are in the formulae list supplied in the exam. They follow from the definitions: W = VQ and Q = It, so W = VIt and P = W/t = VI. Substituting V = IR gives I²R; substituting I = V/R gives V²/R.

Use P = I²R when you know the current (components in series share a current); use P = V²/R when you know the p.d. (appliances in parallel on the same supply).

All the quantities must refer to the same component: the p.d. across it, the current in it and its resistance.

Kilowatt-hours: energy in kWh = power in kW × time in hours; 1 kWh = 3.6 × 10⁶ J (1000 W × 3600 s; learn it).

WORKED EXAMPLE · QUESTION

A fan heater is rated at 2.5 kW when connected to a 230 V supply. It runs for 3.5 hours. Calculate (a) the current, (b) the resistance of the heater when operating and (c) the energy transferred, in kWh and in J.

SOLUTION

(a) I = P/V = 2500 ÷230 = 11 A (10.87 A).

(b) R = V²/P = 230² ÷2500 = 21 Ω (21.2 Ω).

(c) E = 2.5 kW ×3.5 h = 8.75 kWh; ×3.6 ×10⁶: 8.8 kWh = 3.2 × 10⁷ J.

Check: I²R = 10.87² ×21.16 = 2500 W, the rated power.

WORKED EXAMPLE · QUESTION

A workshop takes 25 kW from a supply through a cable whose total resistance is 0.15 Ω. Compare the power wasted in the cable when the power is delivered at 240 V and when it is delivered at 2400 V.

SOLUTION

At 240 V: I = P/V = 25 000 ÷240 = 104.2 A. The cable's current is known, so use I²R: P = 104.2² ×0.15 = 1.6 kW (1628 W, 6.5% of 25 kW).

At 2400 V: I = 10.42 A; P = 10.42² ×0.15 = 16 W (0.065% of 25 kW).

Ten times the p.d. → one tenth of the current → one hundredth of the power wasted (1.6 kW → 16 W).

The p.d. across the cable at 240 V is only IR = 15.6 V. Using 240²/0.15 = 384 kW mixes the supply p.d. with the cable's resistance, an impossible answer.

The free electrons in a metal move randomly at high speed, colliding with the vibrating ions of the lattice; with no p.d. there is no net movement. A p.d. creates an electric field in the wire. Superimposed on the random motion, the electrons gain a small average velocity along the wire: the drift velocity v. The number density n is the number of charge carriers per unit volume (m⁻³).

Deriving I = nqvA

The volume of wire swept through the cross-section in Δt is A ×vΔt. It contains nAvΔt carriers, so the charge passing is ΔQ = nAvΔt ×q. Dividing by Δt:

Here n is in m⁻³, q the charge on each carrier (C), v the drift velocity (m s⁻¹) and A the cross-sectional area (m²). The equation is in the formulae list supplied in the exam, but be ready to derive it as above.

In a series circuit the current is the same everywhere, so where a wire is thinner the electrons drift faster: v ∝1/A.

Drift velocities in metal wires are tiny, typically fractions of a millimetre per second, because n is enormous.

A lamp still lights almost at once when you close a switch: the electric field is set up round the whole circuit almost instantly, so the free electrons everywhere start drifting together. No electron has to travel from the switch to the lamp.

WORKED EXAMPLE · QUESTION

A copper connecting lead has a diameter of 0.80 mm and carries a current of 2.5 A. Copper has 8.5 ×10²⁸ free electrons per m³. e = 1.60 ×10⁻¹⁹ C. Calculate (a) the drift velocity of the electrons and (b) the time for an electron to drift along the 1.5 m length of the lead.

SOLUTION

Area: radius 0.40 mm = 0.40 ×10⁻³ m, so A = π(0.40 ×10⁻³)² = 5.03 ×10⁻⁷ m².

(a) v = I/nqA = 2.5 ÷(8.5 ×10²⁸ ×1.60 ×10⁻¹⁹ ×5.03 ×10⁻⁷) = 3.7 × 10⁻⁴ m s⁻¹.

(b) t = 1.5 ÷3.66 ×10⁻⁴ = 4.1 × 10³ s (about 68 minutes).

Resistivity is a property of the material (the resistance of a 1 m cube of it); its equation R = ρl/A is on 2.8. Here the question is why it varies so much. Rearranging I = nqvA: for a given current in a given sample, the fewer carriers there are, the faster each must drift, and a larger p.d. is needed to drive them. So a material with a small n has a large resistance and a large resistivity.

Metals: roughly one or more free electrons per atom, n ~10²⁸–10²⁹ m⁻³, so very low resistivity.

Semiconductors (silicon, germanium): far fewer free carriers at room temperature, so a much higher resistivity. Their n rises steeply as temperature rises or light falls on them, the basis of thermistors and LDRs (2.10).

Insulators: almost all electrons are bound to atoms; n is negligibly small, so the resistivity is enormous.

WORKED EXAMPLE · QUESTION

A rectangular semiconductor sample is 4.0 mm wide and 0.50 mm thick, and carries a current of 15 mA. The number density of charge carriers in it is 2.0 ×10²² m⁻³, each with charge 1.60 ×10⁻¹⁹ C. For copper, n = 8.5 ×10²⁸ m⁻³. (a) Calculate the drift velocity in the sample. (b) Calculate the drift velocity in a copper strip of the same cross-section carrying the same current. (c) Explain what the comparison shows.

SOLUTION

Area: A = 4.0 ×10⁻³ ×0.50 ×10⁻³ = 2.0 ×10⁻⁶ m².

(a) v = I/nqA = 0.015 ÷(2.0 ×10²² ×1.60 ×10⁻¹⁹ ×2.0 ×10⁻⁶) = 2.3 m s⁻¹ (2.34 m s⁻¹).

(b) The same equation with copper's n: 5.5 × 10⁻⁷ m s⁻¹.

(c) The ratio 2.34 ÷5.51 ×10⁻⁷ = 4.25 ×10⁶ equals 8.5 ×10²⁸ ÷2.0 ×10²².

With millions of times fewer carriers, each must drift millions of times faster; driving them needs a far larger p.d., which is why the semiconductor's resistivity is so much greater.

Unit 1 power and efficiency

P = W/t and the kilowatt-hour are the same ideas as before; P = VI is the electrical input power to motors, heaters and lamps whose efficiency you calculate.

7 Series and parallel circuits; 2.8 Resistivity and potential dividers

Conservation of charge (currents at a junction) and of energy (p.d.s round a loop) build directly on the definitions of current and p.d. here, and I–V graphs for lamps, diodes and thermistors extend the resistance section. R = ρl/A puts numbers on the material property explained by n.

10 Resistance, temperature and light

Thermistors and LDRs are explained by changes in n and in lattice vibrations, using I = nqvA.

1. Show that 1 V = 1 J C⁻¹ is the same as 1 W A⁻¹.

2. Use the current–time graph in the first section to find the mean current over the 8.0 s.

3. A 60 Ω resistor carries 0.20 A. Calculate the p.d. across it and the power it dissipates, two ways.

4. Explain why a lamp lights almost immediately when a switch is closed, even though the drift velocity is less than 1 mm s⁻¹.

5. Two wires of the same metal are joined in series. Wire B has half the diameter of wire A. Compare the currents and the drift velocities.

6. A 9.0 V battery drives 0.15 A through a lamp for 2.0 minutes. Calculate the energy transferred and the number of electrons that pass.

1. 1 W A⁻¹ = 1 J s⁻¹ ÷ 1 C s⁻¹ = 1 J C⁻¹. (This is V = P/I from P = VI.)

2. Charge = area = 2.4 C; mean current = 2.4 ÷8.0 = 0.30 A.

3. V = IR = 12 V; P = VI = 12 ×0.20 = 2.4 W, and I²R = 0.20² ×60 = 2.4 W.

4. Closing the switch sets up an electric field throughout the circuit almost instantly. Free electrons are already present everywhere in the wires and the lamp filament, so they all start drifting at once, including those already in the filament.

5. The currents are equal (series). B's area is one quarter of A's, and n and q are the same, so the drift velocity in B is 4 times that in A.

6. W = VIt = 9.0 ×0.15 ×120 = 162 J. Q = It = 18 C, so N = 18 ÷1.60 ×10⁻¹⁹ = 1.1 × 10²⁰ electrons.