Current Electricity · TOPIC LESSON
Follow charge from microscopic drift to resistance, networks and electrical heating.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
An electric current is a rate of charge transfer. If charge ΔQ crosses a specified cross-section during time Δt, the average current is ΔQ/Δt. Instantaneous current is dQ/dt, and the charge transferred over an interval is the area under the current–time graph. A current is not the amount of charge stored in a wire, and a circuit does not consume charge as it operates.
Conventional current points in the direction positive charge would move. In a metal the mobile carriers are electrons, so their average drift direction is opposite the conventional current. In an electrolyte or semiconductor, more than one carrier type can contribute. Oppositely charged carriers moving in opposite directions can make currents in the same conventional direction.
A steady current has the same charge flow rate through successive sections of an unbranched conductor because charge is not continually accumulating between them. If cross-sectional area changes, the current can stay the same while current density and drift speed change. At a junction, the algebraic current sum is zero under the usual negligible-charge-accumulation circuit approximation.
Current density J describes current per oriented area locally. For uniform flow perpendicular to a cross-section, I = JA. More generally I = ∫J·dA. Current through a chosen surface is signed; current density is a vector. This distinction matters when comparing wires of different widths or adding contributions from multiple carrier species.
I = dQ/dt; Qtransferred = ∫I dt
For a signed current, the integral is net signed charge transfer. Total absolute transferred charge may require integrating |I|.
I = neAvd
Magnitude for one carrier species with number density n, charge magnitude e, uniform area A and drift-speed magnitude vd.
ORIGINAL WORKED EXAMPLE
A wire carries I(t) = (2 + 3t) A for 0 ≤ t ≤ 4 s. How much charge crosses a section, and what is the average current?
Answer 32 C; average current 8 A.
02 / UNDERSTAND & APPLY
Conduction electrons already move rapidly and randomly even without an applied field. Their random velocity contributions average to zero in an equilibrium metal. An electric field adds a small directed drift to that distribution. The drift speed can be tiny compared with microscopic random speeds; a lamp does not wait for one particular electron to travel all the way from its battery.
The circuit responds through changes in electromagnetic fields propagating through the circuit and its surroundings. Carrier drift and signal propagation are different phenomena. A useful analogy is a pipe already full of water: pushing at one end can establish motion throughout without waiting for the first molecule to traverse the entire pipe, although the actual electrical mechanism is electromagnetic.
In the elementary relaxation-time model, an electron accelerates between scattering events under force −eE. With average relaxation time τ, its mean drift velocity is −eτE/m. Multiplying by charge density gives J = ne²τE/m. This yields conductivity σ = ne²τ/m and resistivity ρ = m/(ne²τ). The simple model captures a proportional relation, not every microscopic property of real materials.
Mobility μ is drift-speed magnitude divided by field magnitude: μ = eτ/m in this model. Then conductivity is neμ for one carrier species. In semiconductors with electrons and holes, conductivity can include both neμe and peμh terms. Carrier density, mobility and scattering mechanisms can all change with temperature, so a single metallic trend is not universal.
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ORIGINAL WORKED EXAMPLE
A wire carries 3.2 A through area 1.0 mm². Take carrier density n = 8.0 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C. Find drift speed.
Answer 2.5 × 10⁻⁴ m s⁻¹, or 0.25 mm s⁻¹.
03 / UNDERSTAND & APPLY
Resistance R is the ratio of potential difference to current for a specified operating state. An ohmic conductor has a linear V–I relation with constant R when temperature and other physical conditions are fixed. A filament lamp heats substantially as current rises and is therefore not well described by one constant resistance across all voltages. A diode is also non-ohmic for a different physical reason.
Resistivity ρ describes the material’s local response in an isotropic ohmic model: E = ρJ. For a uniform wire of length L and cross-sectional area A, E = V/L and J = I/A, giving R = ρL/A. A longer wire offers more resistive length, while a larger area provides more parallel conduction paths. Resistivity has unit Ω m; resistance has unit Ω.
Conductance is 1/R and conductivity is 1/ρ. Keep these pairs distinct. Doubling the area of a wire halves its resistance without changing its intrinsic resistivity at the same temperature. Two materials can have the same resistance through different geometries while having very different resistivities.
For a nonuniform conductor sliced along the current path, small segment resistance is dR = ρ(x) dx/A(x), and these series segments add by integration. If conduction paths lie side by side across the same voltage difference, add their conductances instead. The direction of current flow determines whether the same material pieces are in series or parallel.
V = IR; R = ρL/A; J = σE
Constant-R Ohm’s law requires fixed physical conditions; the uniform-wire formula assumes uniform material and cross-section.
ORIGINAL WORKED EXAMPLE
A uniform wire is stretched to twice its original length, with its volume and resistivity unchanged. How does its resistance change?
Answer Resistance quadruples.
04 / UNDERSTAND & APPLY
For many metals over a limited temperature interval, R(T) ≈ R₀[1 + α(T − T₀)], where α is the temperature coefficient referred to T₀. This is a local linear approximation, not a law valid from absolute zero to melting. Increased lattice scattering often raises a metal’s resistivity as temperature rises.
Semiconductors often show decreasing resistance with increasing temperature because thermally generated carrier density increases strongly. The balance between carrier density and mobility matters. Alloys may have relatively small temperature coefficients, which is useful for stable resistors. A superconductor below its transition is outside the ordinary finite-resistivity model.
When a problem gives a temperature coefficient, identify the reference temperature and whether geometry changes are to be neglected. If thermal expansion is appreciable, both resistivity and L/A can vary. In most elementary numerical problems the supplied resistance coefficient already represents the intended effective change and should not be supplemented with an extra geometric correction unless requested.
Electrical heating can create feedback: at a fixed voltage, an increase in resistance reduces electrical power V²/R; at a fixed current it raises power I²R. Which variable is held fixed therefore changes the thermal response. This is the same invariant-first reasoning used in many other parts of physics.
ORIGINAL WORKED EXAMPLE
A resistor is 20.0 Ω at 20°C and has α = 0.0040 °C⁻¹ referred to that temperature. Estimate its resistance at 70°C.
Answer 24.0 Ω.
05 / UNDERSTAND & APPLY
Resistors are in series when the same current must pass through them without branching between. Their voltage drops add, so equivalent resistance is the sum. Resistors are in parallel when both ends connect to the same two nodes. Their voltage differences match while currents add, so conductances add. Visual placement on a page does not determine the connection.
For two parallel resistors, Req = R₁R₂/(R₁ + R₂), which must be smaller than either positive resistance. For series, Req must exceed each positive component. These inequalities quickly expose arithmetic or topology mistakes. A wire shorting a resistor makes the voltage across that resistor zero in the ideal circuit, while a disconnected branch carries no current.
Voltage division in a series chain follows Vi = Vtotal Ri/ΣR. Current division in parallel follows Ii = V/Ri, so the smaller resistor carries more current. These simple divider formulas assume the output is not loaded by another branch. Adding a measuring device or load can change the equivalent network and the division.
Network reduction is a tool, not a substitute for identifying nodes. A bridge network may not permit immediate series/parallel simplification. In that case use circuit equations rather than forcing a combination based on which lines look adjacent. The present section builds the conduction model; full multi-loop and source analysis belongs to circuit analysis.
ORIGINAL WORKED EXAMPLE
A 6 Ω resistor is in parallel with a 3 Ω resistor; that pair is in series with 4 Ω across an ideal 12 V source. Find total current and power in the 3 Ω branch.
Answer Total current 2 A; 3 Ω branch power 16/3 W.
06 / UNDERSTAND & APPLY
Potential difference is energy transfer per unit charge. A charge amount dQ passing through voltage V transfers energy V dQ. Dividing by time gives electrical power P = VI. For an ohmic resistor with the passive sign convention, substituting V = IR yields P = I²R = V²/R, always nonnegative for R > 0. The energy appears mainly as thermal energy in a resistive element.
The forms I²R and V²/R are equivalent for the same operating state but predict different comparisons depending on what is fixed. At fixed current, a larger resistance dissipates more power. At fixed voltage, a smaller resistance dissipates more power. A claim that “more resistance always means more heat” omits the circuit constraint.
For constant power, transferred energy is Pt. For changing voltage or current, integrate VI over time. A kilowatt-hour is energy, not power: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. Appliance labels usually describe rated operation at a specified voltage, not a resistance guaranteed to be constant at all temperatures.
When a component can deliver power, signs matter. Current entering the terminal marked at higher potential gives positive absorbed power under the passive convention. A discharging source can have negative absorbed power, indicating delivery to the circuit. In a complete idealised circuit, source output equals the sum of load and internal dissipations when stored field energy is not changing.
ORIGINAL WORKED EXAMPLE
Treat a 120 V, 60 W heater as a constant resistor. Find its resistance and its power at 60 V.
Answer 240 Ω; 15 W at 60 V.
ORIGINAL WORKED EXAMPLE
A 1.5 kW heater runs for 40 minutes. Find energy in kWh and joules.
Answer 1.0 kWh = 3.6 MJ.
07 / UNDERSTAND & APPLY
For a fixed delivered power, increasing transmission voltage reduces current because I = P/V in a simple DC or unity-power-factor model. Resistive line loss I²Rline then falls with the square of current. This is why high-voltage transmission is useful; it does not imply that arbitrarily high voltage is free of insulation, safety or conversion constraints.
A fuse protects a circuit by opening when current causes excessive heating under its designed conditions. It is placed in series with the protected path. Its rating is not an assertion that all current below the number is harmless in every wire or appliance; conductor size, installation and time response matter. In exam models, use the specified current limit and total circuit demand.
Current density explains why a thin connection can heat more strongly than a thick one carrying the same current. For a uniform ohmic medium the local power dissipated per volume is J·E = ρJ². A narrow section has greater J = I/A, so its local heating can become severe even though the same total current flows everywhere in the series path.
If a problem asks which bulb is brighter, clarify the idealisation. In elementary constant-resistance comparisons, brightness is often associated with dissipated power. Real filament brightness also depends on temperature and spectral emission. Two rated bulbs in series cannot be compared by their printed wattages alone; infer their rated resistances under the specified common voltage, then use the new circuit current.
08 / UNDERSTAND & APPLY
Check each result’s dimensions and identify the held-fixed quantity before comparing powers or drift speeds.
ORIGINAL PRACTICE · 1
90 C.
Q = It = 0.50 × 180.
ORIGINAL PRACTICE · 2
Both double.
J = I/A and vd = I/(neA). Charge conservation preserves current, not current density.
ORIGINAL PRACTICE · 3
0.05 Ω.
R = ρL/A = 10⁻⁷/(2 × 10⁻⁶).
ORIGINAL PRACTICE · 4
4:1.
Series equivalent is 2R; parallel is R/2.
ORIGINAL PRACTICE · 5
It becomes one third.
Use P = V²/R, since voltage is fixed.
ORIGINAL PRACTICE · 6
It triples.
Use P = I²R, since current is fixed.
ORIGINAL PRACTICE · 7
5 Ω.
Power is 20 W; R = P/I² = 20/4.
ORIGINAL PRACTICE · 8
It stays unchanged.
σ = neμ, so the factors cancel.
ORIGINAL PRACTICE · 9
Rsmall:Rlarge = 4:1.
Area scales as radius squared; resistance is inversely proportional to area.
ORIGINAL PRACTICE · 10
0.10 A; 1 W; 30 J.
I = V/R, P = VI, and E = Pt.
ORIGINAL PRACTICE · 11
Current density includes the signed carrier charge.
J = nqvdrift; both q and the electron drift direction are negative relative to the field, yielding J along E.
ORIGINAL PRACTICE · 12
Current halves; line loss becomes one quarter.
I = P/V and Ploss = I²Rline under the stated idealised conditions.
09 / UNDERSTAND & APPLY
The chain is field to drift, drift to current density, geometry to resistance, and voltage/current to power. Each step carries an assumption. The Drude-like model assumes a relaxation-time description; the uniform-wire formula assumes a uniform geometry; Ohm’s law assumes an appropriate operating regime; the simple power formulas refer to resistive dissipation.
In a calculation, first decide whether current, voltage, temperature, volume or material is held fixed. Then choose the corresponding relation. This prevents the most common comparison errors: forgetting the area change in a stretched wire, treating static charge as current, reversing electron and conventional directions, or using I²R as if current were unchanged after a network modification.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
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