Electric Current in Conductors
Moving charge, resistance and the circuits built from them.
Electric Current & Current Density
Charge on the move, and how it spreads over a cross-section.
Until now charges sat still. Let them move and you get an electric current: the thing that lights bulbs, turns motors and runs every circuit.
Electric current
Whenever charge passes from one side of a surface to the other, there is a current through that surface. If charge Q crosses in time Δt, the average current is Q/Δt; the current at an instant is the rate of flow:
The SI unit is the ampere (A): one coulomb per second. It is one of the seven SI base units. The direction of a current is the direction in which positive charge moves; if the carriers are negative, the current points opposite to their motion.
Current density
To describe how current is spread over a cross-section we use a vector, the current density.
If the area is tilted so that its normal makes an angle θ with the flow, only the perpendicular part counts: Δi = jΔS cos θ = j⃗·ΔS⃗. For a finite surface:
In a beam of cross-section 2.0 mm², 3.0 × 10¹⁶ electrons pass any cross-section each second. Find (a) the current, (b) the current density.
An electron moves round a circle of radius 5.0 cm at a steady 2.0 × 10⁶ m/s. What is the current at a point on the circle?
Where currents come from
- A hot metal emits electrons that fly through space: a current through any area they cross.
- In a solution, positive ions drift along a field and negative ions against it; both add to the current along the field.
- While a capacitor charges, charge flows along the connecting wires.
- This chapter: free electrons in a metal, pushed by a field inside the conductor.
- 1i = dQ/dt (32.1); 1 A = 1 C/s; direction = motion of positive charge.
- 2Current is not a vector; current density j⃗ is.
- 3Uniform: j = i/S (32.2); in general i = ∫j⃗·dS⃗ (32.3).
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Drift Speed of Electrons
Fast random motion, slow steady drift.
Electrons inside a metal rush about at hundreds of kilometres a second, yet carry a current by creeping along at a fraction of a millimetre per second.
Free electrons and the lattice
A metal contains many loosely bound free (conduction) electrons moving among heavy positive ions, the lattice, which vibrate about fixed positions (more strongly when hotter). Every so often an electron collides with the lattice and sets off in a new random direction with a new random speed, so it follows a zig-zag path. With so many electrons moving at random, as many cross any area one way as the other: no current.
Drift in a field
Apply a field E. Between collisions each electron feels a force eE against the field and picks up extra velocity in that direction, growing steadily until the next collision wipes it out. The random part averages to nothing, but the extra part doesn’t: the electrons slowly drift opposite to the field. If the time between collisions is τ, the extra distance gained per interval is
For a given material at a given temperature τ is fixed, so we can picture the electrons as at rest with no field and moving at a steady drift velocity proportional to E when there is one:
Current and drift speed
Take a wire of cross-section A with n free electrons per unit volume. In time Δt the electrons in a length vdΔt, i.e. nAvdΔt of them, cross a given section, carrying charge nAvdΔt · e. Dividing by Δt:
A copper wire of cross-section 1.5 mm² carries 3.0 A. Copper has 8.5 × 10²⁸ free electrons per m³. Find the drift speed.
A wire of cross-section 0.50 mm² carries 1.0 A and has 8.0 × 10²⁸ free electrons per m³. Find the current density and drift speed.
- 1Free electrons move randomly and collide with the vibrating lattice; no field, no net flow.
- 2In a field they drift against it: v_d = (eτ/2m)E = kE (32.4).
- 3i = nAev_d and j = nev_d (32.5); drift speeds are tiny (≈ 0.1 mm/s).
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Ohm's Law & Resistance
Current density proportional to field; V = Ri.
Combine the drift picture with i = nAevd and one of the most used rules in physics falls out: current density is proportional to the field.
Ohm’s law
Putting vd = (eτ/2m)E into j = nevd:
The content of Ohm’s law is that σ does not depend on the field: double E and j doubles. This holds for metals over a wide range of fields.
Resistance
Take a conductor of length l and uniform cross-section A with potential difference V across its ends. Inside, E = V/l and j = i/A, so i/A = σV/l, i.e. V = (ρl/A) i:
- R is the resistance, in ohms (Ω); 1/R is the conductance.
- Resistivity is in Ω m; conductivity in (Ω m)⁻¹, also written mho m⁻¹.
- A component made to have a chosen resistance is a resistor, drawn as a zig-zag.
Find the resistance of a copper wire 20 m long with a cross-section of 1.0 mm² (ρ = 1.72 × 10⁻⁸ Ω m).
The colour code
Carbon resistors are cylinders with wire leads, labelled by four coloured bands. Read from the end the bands are bunched toward: the first two are digits, the third is the power of ten, the fourth the tolerance (gold ±5 %, silver ±10 %; no fourth band means ±20 %).
A resistor is banded red, violet, orange, gold. What is its value?
- 1Ohm's law: j = σE (32.6), ρ = 1/σ (32.7); σ independent of E.
- 2V = Ri (32.8) with R = ρl/A (32.9); unit ohm; conductance 1/R.
- 3Not valid for semiconductors, electrolytes, insulators…
- 4Colour code: digit, digit, multiplier, tolerance.
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Resistivity & Temperature
Metals resist more when hot; semiconductors less; superconductors not at all.
Warm a metal and it resists more. Warm a semiconductor and it resists less. Cool certain materials far enough and resistance vanishes altogether.
Metals get worse when hot
Heating makes the lattice vibrate harder, so electrons collide more often. τ falls, the drift speed falls, and the resistivity rises. Over modest temperature ranges the rise is close to linear:
The length and area of a wire barely change with temperature, so the resistance follows the same rule: R(T) = R(T₀)[1 + α(T − T₀)].
| Material | ρ at room temperature (Ω m) | α (K⁻¹) |
|---|---|---|
| Silver | 1.47 × 10⁻⁸ | 0.0038 |
| Copper | 1.72 × 10⁻⁸ | 0.0039 |
| Aluminium | 2.63 × 10⁻⁸ | 0.0039 |
| Tungsten | 5.51 × 10⁻⁸ | 0.0045 |
| Iron | 9.71 × 10⁻⁸ | 0.0050 |
| Nichrome | 100 × 10⁻⁸ | 0.0004 |
| Germanium | 0.46 | −0.048 |
| Silicon | ≈ 10³ | −0.075 |
| Fused quartz | 7.5 × 10¹⁷ | – |
Resistivities span more than 25 powers of ten: metals around 10⁻⁸ Ω m are good conductors, materials like fused quartz are insulators, and silicon and germanium sit in between: semiconductors.
A copper coil has a resistance of 20.0 Ω at 20 °C. What is its resistance at 80 °C?
A tungsten filament has 30 Ω at 20 °C and 360 Ω when glowing. Estimate its temperature using the linear rule.
Thermistors
In semiconductors α is negative and often large: warming them frees more charge carriers, and the resistivity drops. (The simple free-electron picture doesn’t apply here.) A small bead of semiconducting metal oxide (of nickel, iron, cobalt, copper…) sealed in epoxy is a thermistor. Put it in a circuit with a cell: as the temperature rises its resistance falls and the current climbs. Thermistors can detect changes of about 10⁻³ °C.
Superconductors
Some materials lose their resistance completely below a critical temperature. Kamerlingh Onnes found this in mercury in 1911 (Tc = 4.2 K). Above Tc the material behaves like a normal metal. A current started in a superconducting loop keeps flowing for years with no emf.
- 1Metals: ρ(T) = ρ(T₀)[1 + α(T − T₀)], α > 0; R follows the same law.
- 2Conductors ~10⁻⁸ Ω m, insulators up to ~10¹⁷ Ω m, semiconductors in between.
- 3Semiconductors: α < 0, large → thermistors.
- 4Superconductors: ρ = 0 below T_c (mercury 4.2 K).
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Emf, Heating & Internal Resistance
Batteries pump charge; resistors turn the work into heat.
A battery is a charge pump. Its emf measures how much work it does on each coulomb, and a resistor turns that work into heat.
How a battery keeps its voltage
Inside a battery some non-electrostatic mechanism (chemistry, in a cell) pushes positive charge toward terminal A and negative toward B with a force F⃗b. As charge piles up, an electrostatic field builds between the terminals and pushes back with F⃗e = qE⃗. Accumulation stops when Fe = Fb.
With nothing connected, Fbd = qEd = qV, so the emf equals the terminal voltage of an open battery. The two are different quantities: emf comes from the non-electrostatic battery force, potential difference from the electrostatic field of the piled-up charges.
Connect a wire across the terminals and charge flows round; the battery mechanism keeps restoring the charge on A and B, maintaining a steady current. Normally positive charge leaves from the + terminal (discharging); driving current into the + terminal instead is charging.
Heat in a resistor
A resistor R joined to a battery by wires of negligible resistance has potential difference V = iR across it. In time t a charge q = it passes through and loses potential energy qV, which turns into thermal energy:
A resistor gives out 600 J in 15 s with 2.0 A through it. (a) Find R. (b) How much heat does it produce in 15 s at 3.0 A?
Internal resistance
As charge q = it passes through a battery, the battery force does work qℰ, but the charge gains only qV of electric potential energy. The difference warms the battery. An ideal battery wastes nothing, so V = ℰ always. A real battery behaves like an ideal emf in series with an internal resistance r:
A 9.0 V battery with internal resistance 1.5 Ω delivers 0.40 A. Find (a) its terminal voltage, (b) the heat produced inside it in one minute.
- 1Emf ℰ = work per unit charge by the battery force (32.10); equals open-circuit terminal voltage.
- 2Heat in a resistor: U = i²Rt, P = i²R = V²/R = Vi (32.11–32.12).
- 3Real battery: ideal ℰ plus internal r; V = ℰ − ir while discharging.
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Kirchhoff's Laws
Conservation of charge and energy, applied to circuits.
Two simple rules, one about charge and one about energy, are enough to solve any circuit of cells and resistors.
The junction law
The total current flowing into any point of a circuit equals the total current flowing out. Equivalently, counting currents toward the point as positive and away as negative, their algebraic sum is zero.
Why: no point in a circuit keeps piling up charge or keeps producing it, so whatever arrives must leave. It is conservation of charge.
The loop law
Going all the way round any closed loop, the potential differences add up to zero. Walk round the loop in one direction and list each change: count a drop in potential as positive and a rise as negative (or the other way round, consistently).
- Through a resistor in the direction of its current: drop iR. Against the current: rise iR.
- Through a cell from − to +: rise ℰ. From + to −: drop ℰ.
Why: the electrostatic force is conservative, so its work round a closed path, and hence the net change of potential, is zero. You end where you started.
Cell ℰ₁ = 10 V in series with R₁ = 2 Ω and cell ℰ₂ = 4 V in series with R₂ = 4 Ω both feed the top of a shared resistor R₃ = 4 Ω, whose bottom joins both cells’ negative terminals. Find all three currents. (This is the widget’s starting circuit.)
- 1Junction law: current in = current out (charge conservation).
- 2Loop law: potential changes round a closed loop sum to zero (conservative field).
- 3Resistor along the current: drop iR; cell − to +: rise ℰ.
- 4Negative results mean the real current is opposite to the guess.
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Series, Parallel & Grouped Cells
R adds in series, 1/R in parallel; cells combine too.
Any two-terminal network of resistors acts like one resistor. Series and parallel rules handle most of them; the same ideas tell us how to group cells.
Equivalent resistance
Apply potential difference V across a network’s two end points; if it draws current i, its equivalent resistance is Req = V/i.
Series
Resistors are in series when the same current passes through each. Round the loop: iR₁ + iR₂ + iR₃ − V = 0, so
Parallel
Resistors are in parallel when the same potential difference is across each. The current splits: i = i₁ + i₂ + i₃ = V/R₁ + V/R₂ + V/R₃, so
Same current. Voltages add. R adds.
Same voltage. Currents add. 1/R adds.
A 12 Ω and a 6 Ω resistor are in parallel between A and C; a 2 Ω resistor joins C to B. Find the resistance between A and B.
How current divides
Across two parallel resistors V is common, so i₁R₁ = i₂R₂: the current divides in inverse ratio of the resistances.
3.0 A enters a 4 Ω and a 12 Ω resistor joined in parallel. How much goes through each?
Grouping cells
Series: two cells (ℰ₁, r₁) and (ℰ₂, r₂) driving R. Round the loop, i = (ℰ₁ + ℰ₂)/(R + r₁ + r₂): they act as one cell of emf ℰ₁ + ℰ₂ and internal resistance r₁ + r₂. Reverse one cell and the emf becomes |ℰ₁ − ℰ₂|.
Parallel: applying the loop law to each cell’s loop through R and eliminating gives one equivalent cell:
Cells of (2.0 V, 0.50 Ω) and (1.5 V, 0.25 Ω) drive a 4.0 Ω resistor. Find the current with the cells (a) in series, (b) in parallel.
- 1R_eq = V/i for the network.
- 2Series: R adds (32.13). Parallel: 1/R adds (32.14).
- 3Parallel current divides inversely as the resistances.
- 4Cells in series: emfs and r's add. In parallel: ℰ₀ = (ℰ₁r₂ + ℰ₂r₁)/(r₁ + r₂), r₀ = r₁r₂/(r₁ + r₂).
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Wheatstone Bridge, Ammeters & Voltmeters
Measuring resistance by a null, and meters that disturb little.
Measuring electricity means putting an instrument into the circuit. The Wheatstone bridge measures resistance by finding a null; ammeters and voltmeters are designed to disturb the circuit as little as possible.
The Wheatstone bridge
Four resistors R₁, R₂, R₃, R₄ form a loop with corners A, C, B, D. A cell is joined across A and B, and a galvanometer (a sensitive meter whose needle swings left or right with the current direction) across C and D.
For one particular set of values no current flows through the galvanometer: the bridge is balanced. Then VC = VD, R₁ and R₂ carry the same current, and so do R₃ and R₄. Comparing the drops, (VA − VC)/(VC − VB) = R₁/R₂ and (VA − VD)/(VD − VB) = R₃/R₄. With VC = VD these ratios are equal:
To measure an unknown R₄, make R₃ variable and adjust it until the galvanometer reads zero; then R₄ = R₂R₃/R₁.
In a bridge, the arms from A are 15 Ω (to C) and R (to D); the arms to B are 30 Ω (from C) and 50 Ω (from D). A 100 Ω resistor joins C and D. For what R is there no current in the 100 Ω?
Ammeters and voltmeters
Both contain a coil hung between magnet poles; current through the coil turns it, and a needle fixed to it moves along a scale by an amount proportional to the current.
Ammeter
It goes in series with the part whose current you want, current entering at the + terminal. A small resistance r, the shunt, is joined in parallel with the coil (resistance Rc), so the coil takes only a fixed fraction of the current, i₁ = ri/(Rc + r). Its total resistance is
A galvanometer is built the same way but with its zero in the middle of the scale, so it shows current in either direction.
An ammeter made of a 90 Ω coil with a 10 Ω shunt is in series with a 51 Ω resistor across a 6.0 V ideal cell. What does it read, and how much current goes through its coil?
Voltmeter
It goes in parallel with the part whose voltage you want, + terminal to the higher potential. A large resistance R is joined in series with the coil, so the coil current is V/(Rc + R), proportional to V. Its resistance Rc + R must be large, so that it draws little current and barely changes the voltage it measures.
A coil of 50 Ω gives full-scale deflection at 1.0 mA. What series resistance turns it into a 0–10 V voltmeter?
- 1Wheatstone bridge balanced: R₁/R₂ = R₃/R₄ (32.15), no galvanometer current.
- 2Ammeter: coil + small parallel shunt, connected in series; low resistance.
- 3Voltmeter: coil + large series resistance, connected in parallel; high resistance.
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The Potentiometer
A voltmeter that draws no current.
An ideal voltmeter would draw no current at all. A potentiometer achieves exactly that by balancing the unknown voltage against a known one until nothing flows.
The stretched-wire potentiometer
A long uniform wire AB (5–10 m, laid in 1 m lengths joined by thick copper strips) is fixed to a board beside a metre scale. A driver circuit, a strong cell with a key and a rheostat, sends a steady current through it, so the potential falls evenly from A to B. A galvanometer is connected to a jockey that can touch the wire at any point P.
To measure the potential difference between points a and b (a higher), join a to A and b, through the galvanometer, to the jockey. Slide the jockey until the galvanometer shows no deflection. At that point P has the same potential as b, so Vab = VAP. Because the wire is uniform, the drop is proportional to length: with V₀ across the whole length L,
Calibration
To know V₀, balance a standard cell of accurately known emf ℰ: if it balances at l₀, then ℰ = V₀l₀/L, so V₀ = Lℰ/l₀ and any later reading is V = (l/l₀)ℰ. (No current flows through the standard cell at balance, so its terminal voltage is its emf.)
On a 10 m wire, a standard cell of 1.018 V balances at 509 cm. An unknown cell balances at 742 cm. Find V₀ and the unknown emf.
Uses
Comparing two emfs
Balance the first cell at l₁, then the second at l₂ (a two-way key swaps them). Since ℰ₁ = V₀l₁/L and ℰ₂ = V₀l₂/L,
Two cells balance at 6.00 m and 4.50 m on the same potentiometer. Find the ratio of their emfs.
Internal resistance of a cell
Connect a known resistance R across the cell through a key K. With K open, no current flows from the cell, so it balances at l with ℰ = V₀l/L. Close K: current ℰ/(R + r) flows and the terminal voltage drops to ℰR/(R + r), balancing at a shorter l′. Dividing the two:
A cell balances at 7.50 m on open circuit and at 6.25 m when shunted by 10 Ω. Find its internal resistance.
- 1Uniform wire, steady current: V ∝ length; at balance no current is drawn.
- 2V = V₀l/L; calibrate with a standard cell.
- 3ℰ₁/ℰ₂ = l₁/l₂; internal resistance r = R(l − l′)/l′.
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Charging & Discharging Capacitors
Exponential filling and emptying, paced by CR.
A capacitor joined to a battery through a resistor doesn’t fill instantly. The charge creeps up along an exponential curve whose pace is set by one number, CR.
Charging
A capacitor C, a cell ℰ and a resistance R (the wires plus the cell’s internal resistance) are joined in series at t = 0. At time t the capacitor holds q and the current is i = dq/dt. Round the loop:
The product CR has the dimensions of time; it is the time constant τ. The final charge ℰC is reached only as t → ∞, but after one time constant q = ℰC(1 − 1/e) = 0.63 ℰC.
A 47 µF capacitor charges through 10 kΩ from a 9 V cell. Find the time constant and the time to reach 90 % of full charge.
Discharging
A capacitor holding Q is joined across a resistance R at t = 0. Now the charge falls, so the current is i = −dq/dt, and the loop law gives q/C − Ri = 0, i.e. R dq/dt = −q/C:
After one time constant 37 % (1/e) of the charge is left, so 63 % of the discharge is done. Again it is complete only at t → ∞, but four or five time constants leave less than 2 %.
A 200 µF capacitor holding 1.0 mC is discharged through 5.0 kΩ. How much charge remains after 3.0 s?
- 1Charging: q = ℰC(1 − e^(−t/CR)) (32.16); current ℰ/R e^(−t/CR).
- 2Discharging: q = Q e^(−t/CR) (32.17).
- 3Time constant τ = CR: 63 % done after one τ; ~99 % after about 5τ.
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Atmospheric Electricity
The planet as a leaky capacitor charged by storms.
The Earth and its atmosphere form a giant leaky capacitor, charged by thunderstorms and discharged by fair weather everywhere else.
The fair-weather field
The ground carries a negative charge, a little under 1 nC per square metre, spread fairly evenly. Above it there is a downward electric field of about 100 V/m: over a flat desert, the potential rises by roughly 100 V for every metre you climb. Higher up the field weakens, becoming negligible around 50 km, where the potential is about 400 kV above the ground.
A current in the air
Cosmic rays ionise air molecules, and dust picks up charge by friction, so the air holds both positive and negative ions. The field pushes positive ions down and negative ions up: a vertical current of about 3.5 × 10⁻¹² A through every square metre. The density of ions grows with height and thinner air lets them travel further between collisions, so the air conducts better higher up; by about 50 km it is highly conducting. Picture a conducting shell at +400 kV, 50 km up, leaking current down to the Earth.
Taking 3.5 × 10⁻¹² A/m² over the Earth’s surface (area 5.1 × 10¹⁴ m²), find the total fair-weather current.
The ground’s charge density is about 0.9 nC/m². Roughly how long would 1800 A take to neutralise the whole Earth’s charge?
Thunderstorms recharge the battery
Rising and sinking air carries water vapour into the cold upper atmosphere (about −10 °C at 3–4 km, −20 °C at 6–7 km), where it forms droplets and ice. A mature thunderstorm is roughly 7–8 km across, with its base 3–4 km up and its top at 6–7 km. Its top holds excess positive charge and its base a dense negative charge, putting the cloud base 20–100 MV below the ground’s potential: far larger than, and opposite to, the 400 kV fair-weather difference.
The field below the cloud breaks down the air, and negative charge leaps to the ground: lightning. One stroke delivers about 20 C, and a cloud can recharge for the next in about 5 s. Meanwhile the storm’s positive top leaks up to the conducting layer. Thousands of storms a day thus pump negative charge into the Earth and positive charge upward, acting as the emf; the fair-weather air is the load that discharges it.
- 1Ground: ≈ −1 nC/m²; fair-weather field ≈ 100 V/m downward; ≈ 400 kV up to 50 km.
- 2Ions carry ≈ 3.5 × 10⁻¹² A/m², ≈ 1800 A in total, which would neutralise the Earth in minutes.
- 3Thunderstorms (cloud bases at −20 to −100 MV) and lightning (≈ 20 C a stroke) keep the system charged.