amazon.co.ukA-Level Physics electricity revision: current, pd, resistance, drift velocity, EMF and internal resistance, Kirchhoff's laws, potential dividers, resistivity and practical circuits with derivations - AQA 7408 3.5, OCR A H556 4.1-4.2, Edexcel 9PH0 Topic 3.
Current is the rate of flow of charge: I = ΔQ/Δt (AQA 7408 3.5.1.1; OCR A H556 4.1; Edexcel 9PH0 Topic 3), measured in amperes. Potential difference is work done per unit charge, V = W/Q, in volts - 1 V means each coulomb gains 1 J of energy passing through. Resistance R = V/I in ohms; Ohm's law states that for an ohmic conductor at constant temperature, I ∝ V (straight I-V line through the origin) - the constant temperature condition is part of the law, so quote it.
Drift velocity derivation (I = nqvA): in a wire of cross-sectional area A with n charge carriers per cubic metre, each of charge q moving at drift velocity v, in time Δt all carriers within a length vΔt pass a point. The volume is AvΔt, containing nAvΔt carriers with total charge Q = nqAvΔt. Current I = Q/Δt = nqvA. Typical drift velocities are ~mm/s (slow!), because n is enormous (~10^28-10^29 per m^3 in metals) - lights turn on instantly because the electric field propagates near light speed, not the electrons.
I-V characteristics: ohmic conductor - straight line, constant R; filament lamp - curve flattening at higher V, because current heats the filament, lattice ions vibrate more, electrons collide more often and R increases; diode - zero current in reverse and below the forward threshold (~0.7 V for silicon), then steep conduction. Resistivity: R = ρL/A, in Ω m - resistance rises with length, falls with area; ρ is a material property. Semiconductors (thermistors, LDRs) behave oppositely to metals: resistance falls as temperature rises because more charge carriers are released, outweighing increased lattice vibration.
A copper wire carries 5.0 A with n = 8.5 × 10^28 carriers per m^3 and area 1.0 mm^2. Find the drift velocity.
Line 1: I = nqvA, so v = I/(nqA) (M1: rearrangement).
Line 2: v = 5.0/(8.5 × 10^28 × 1.60 × 10^-19 × 1.0 × 10^-6) (A1: unit conversion of area).
Line 3: v = 5.0/1.36 × 10^4 = 3.7 × 10^-4 m/s ≈ 0.4 mm/s (A1) - millimetres per second, the standard contrast point.
Kirchhoff's first law (junction): the total current into a junction equals the total current out - conservation of charge (charge cannot pile up at a point) (AQA 7408 3.5.1.4). Second law (loop): around any closed loop the sum of EMFs equals the sum of IR products - conservation of energy (a coulomb gains energy in the source and loses exactly that amount around the loop).
Series rule derivation: the same current I flows through each resistor (junction rule: no charge leaves the path). Total pd = V_1 + V_2 + ... = IR_1 + IR_2 + ... = I(R_1 + R_2 + ...), so R_total = ΣR - resistances add, and pds share in proportion to resistance.
Parallel rule derivation: each branch has the same pd V (both ends of each branch connect to the same two nodes). Currents add: I_total = V/R_1 + V/R_2 + ... = V(1/R_1 + 1/R_2 + ...), so 1/R_total = Σ1/R - and the total is always less than the smallest individual resistance (a standard check). Two equal resistors in parallel halve the resistance; adding branches always lowers total resistance and raises total current.
A 12 V battery of negligible internal resistance drives 4.0 Ω in series with a parallel pair of 6.0 Ω and 3.0 Ω.
Line 1: Parallel pair: 1/R_p = 1/6 + 1/3 = 0.5, R_p = 2.0 Ω (M1 A1).
Line 2: Total: R = 4.0 + 2.0 = 6.0 Ω (A1).
Line 3: Total current: I = 12/6.0 = 2.0 A (A1).
Line 4: pd across parallel pair: V_p = 2.0 × 2.0 = 4.0 V (M1).
Line 5: Branch currents: 4.0/6.0 = 0.67 A and 4.0/3.0 = 1.33 A - sum 2.0 A, checking the junction rule (A1: Kirchhoff check).
A real cell has internal resistance r: the EMF ε drives current through both the external circuit (R) and the cell itself (r), so the energy per coulomb divides: ε = I(R + r) (AQA 7408 3.5.1.6; OCR PAG7; Edexcel CP3). The terminal pd - what the circuit actually sees - is V = IR = ε - Ir: it falls below ε whenever current flows, by the 'lost volts' Ir. High current (low R) → large lost volts → dim lamps and flattened batteries; with no current (open circuit) the terminal pd equals ε.
Maximum power transfer: P = I^2R with I = ε/(R + r); differentiating shows power peaks when R = r - the matched-load condition (efficiency is then only 50%, half the energy lost inside the source).
Measuring ε and r (required practical): vary the load resistance (or a variable resistor) and record V and I with a voltmeter (high resistance, across the cell) and ammeter (low resistance, in series). Rearrange V = ε - Ir as V = -rI + ε: plot V against I - a straight line with gradient -r and y-intercept ε. Using the gradient and intercept averages random error across all data. Errors: the voltmeter draws some current (high-resistance voltmeter minimises this), contact resistance, heating of the cell changing r during the experiment (take readings quickly, switch off between readings), and zero errors on meters (check before starting).
Varying the load gives: I = 0.10 A, V = 1.44 V; I = 0.50 A, V = 1.20 V.
Line 1: V = ε - Ir (M1: equation quoted).
Line 2: Two points determine the line: gradient = (1.20 - 1.44)/(0.50 - 0.10) = -0.60, so r = 0.60 Ω (M1 A1).
Line 3: ε = V + Ir = 1.44 + 0.10 × 0.60 = 1.50 V (A1).
Line 4: Check with the second point: 1.20 + 0.50 × 0.60 = 1.50 V - consistent (A1: verification).
A potential divider is two resistors in series sharing a supply (AQA 7408 3.5.1.5): the same current flows through both, so V_out (across R_2) = IR_2 = V_in × R_2/(R_1 + R_2) - the output is the input scaled by the resistance fraction. Derivation: I = V_in/(R_1 + R_2) from the loop rule; multiply by R_2.
Sensor circuits replace one resistor with a component whose resistance depends on the environment: an LDR (resistance falls as light intensity rises) or a thermistor (resistance falls as temperature rises). Putting the sensor as R_2 makes V_out rise when the sensed quantity rises (light/temperature up → R_2 down → wait: R_2 down means V_out falls - the share follows the resistance). Getting the direction right is the examinable logic: V_out across a component rises when that component's resistance rises relative to the other. So for a light-activated circuit that switches on when dark, put the LDR as R_1 (dark → LDR resistance rises → V_out across R_2 falls... trace it carefully for the required behaviour).
The potentiometer (a continuously variable divider) can compare EMFs: sliding the contact until the galvanometer reads zero gives a balance point where the pd per unit length equals the unknown EMF - no current drawn, so no lost volts, hence more accurate than a voltmeter.
V_in = 12 V, R_1 = 2.0 kΩ, R_2 = 4.0 kΩ; a load of 4.0 kΩ is connected across R_2.
Line 1: Unloaded: V_out = 12 × 4/6 = 8.0 V (M1 A1).
Line 2: Loaded: R_2 ∥ load = (4.0 × 4.0)/(4.0 + 4.0) = 2.0 kΩ (M1: parallel combination).
Line 3: V_out = 12 × 2.0/(2.0 + 2.0) = 6.0 V (A1).
Line 4: Loading pulls the output down - high-resistance loads (or low-resistance divider legs) minimise the loading effect (A1: conclusion).
Power: P = IV, and with V = IR the alternatives P = I^2R (best when current is common, e.g. series) and P = V^2/R (best when pd is common, e.g. parallel) follow by substitution (AQA 7408 3.5.1.1). Energy: E = Pt = IVt, in joules; domestic units are kWh (1 kWh = 3.6 × 10^6 J). Fuses and power ratings: a fuse melts when I^2R_heat exceeds its rating - choose a fuse just above the appliance's normal current: a 230 V, 900 W kettle draws I = P/V ≈ 3.9 A, so a 5 A fuse. Power dissipated in transmission lines: P_loss = I^2R_lines motivates high-voltage, low-current transmission.
Required practicals in this topic: (1) Resistivity of a wire (AQA RP5, Edexcel CP2, OCR PAG6-style): measure diameter at many points with a micrometer (A ∝ d^2 doubles its percentage uncertainty), length with a rule, resistance with an ammeter-voltmeter or multimeter; plot R against L - gradient = ρ/A, so ρ = gradient × A. Length and diameter are the dominant uncertainties; check the wire stays near room temperature (heating changes ρ). (2) EMF and internal resistance (AQA RP6, Edexcel CP3, OCR PAG7): V against I analysis as in section 3. (3) I-V characteristics: vary the supply, record I and V for a lamp, diode and fixed resistor; plot I-V curves - reverse the diode to show one-way conduction; keep the lamp's temperature from changing too fast by taking readings promptly.
A nichrome wire: L = 1.000 ± 0.001 m, d = 0.31 ± 0.01 mm, R measured 5.4 ± 0.1 Ω.
Line 1: A = πd^2/4 = π × (0.31 × 10^-3)^2/4 = 7.55 × 10^-8 m^2 (M1 A1).
Line 2: ρ = RA = 5.4 × 7.55 × 10^-8 = 4.1 × 10^-7 Ω m (A1) - the accepted value for nichrome, ≈1.1 × 10^-6 Ω m, is a little higher; check against data books.
Line 3: Percentage uncertainties: d: 0.01/0.31 = 3.2%, doubled (power rule) = 6.5%; L: 0.1%; R: 0.1/5.4 = 1.9% (M1).
Line 4: Total ≈ 8.5%, so ρ = (4.1 ± 0.3) × 10^-7 Ω m - diameter dominates, measure at more points (A1: evaluation).