Current Electricity

An electric current is the rate of flow of electric charge across a cross-section of a conductor. If a charge Δq flows in a time interval Δt, the current is I = Δq/Δt (instantaneous current I = dq/dt). Current is a scalar quantity, and by convention its direction is taken as the direction of flow of positive charge, which is opposite to the actual direction of electron drift in a metal.

In a metal, free (conduction) electrons move randomly with very high thermal speeds (~10⁵ m/s) due to collisions with the vibrating positive ions of the lattice, but this random motion causes no net current since velocities in all directions cancel out. When an electric field E is applied, each electron experiences a force −eE and accelerates between collisions. Because collisions randomise the velocity gained, the electrons acquire a small net average velocity opposite to E, called the drift velocity v_d, superimposed on their thermal motion.

If τ is the average time between successive collisions (relaxation time), the average drift velocity is v_d = eEτ/m, where m is the electron mass. If there are n free electrons per unit volume in a conductor of cross-sectional area A, the number of electrons crossing the area in time Δt is nAv_dΔt, so the current is I = nAev_d. The current density J = I/A = nev_d is a vector along the direction of current flow. Combining these relations with v_d = eEτ/m gives J = (ne²τ/m)E, i.e. current density is directly proportional to the applied electric field — this microscopic result is the origin of Ohm's law.

Ohm's law, in its usual circuit form, states that the potential difference V across a conductor is directly proportional to the current I flowing through it, provided the physical conditions (mainly temperature) remain constant: V = IR, where R is the resistance of the conductor. Conductors that obey this linear V–I relationship are called ohmic conductors (e.g., metals); devices like diodes, thermistors and electrolytic cells with back e.m.f. are non-ohmic, showing a non-linear V–I graph.

Worked Example: A copper wire has a free-electron density n = 8.5 × 10²⁸ m⁻³ and a cross-sectional area A = 1.0 × 10⁻⁶ m². Find the drift velocity of electrons when the wire carries a current of 1.5 A.

Using I = nAev_d:

v_d = I/(nAe) = 1.5/(8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹)

Denominator = 8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹ = 1.36 × 10⁴

v_d = 1.5/1.36 × 10⁴ ≈ 1.1 × 10⁻⁴ m/s (about 0.11 mm/s)

This tiny drift speed, compared with electrons' thermal speeds of ~10⁵ m/s, shows why current appears to flow almost instantaneously — it is the electric field (and the resulting signal) that propagates fast through the wire, not the individual electrons.

The resistance R of a conductor of length L and uniform cross-sectional area A is given by R = ρL/A, where ρ (rho) is the resistivity (or specific resistance) of the material — a property that depends only on the material and its temperature, not on the conductor's dimensions. From the microscopic current equation of the previous section, resistivity can be expressed as ρ = m/(ne²τ), showing that materials with more free electrons or a longer relaxation time between collisions (fewer/weaker scattering events) have lower resistivity and hence conduct better.

Resistivity of metals increases with temperature because increased thermal vibration of the lattice ions reduces the relaxation time τ (more frequent electron collisions), even though the electron density n stays essentially constant. Over a limited temperature range this variation is nearly linear and is written as ρ_T = ρ₀[1 + α(T − T₀)], where ρ₀ is the resistivity at a reference temperature T₀ and α is the temperature coefficient of resistivity. Since R ∝ ρ for a given conductor, the same relation holds for resistance: R_T = R₀[1 + α(T − T₀)]. Metals have a small positive α (resistance increases with temperature); semiconductors and insulators have a negative α (resistance decreases with temperature, because thermally released charge carriers increase n faster than τ decreases); alloys such as manganin and nichrome have very small α, which is why they are used to make standard resistors and heating elements.

When a current I flows through a resistor under a potential difference V, electrical work is done on the charge carriers, which is dissipated as heat due to collisions — this is Joule heating. The electrical power delivered is P = VI, and using Ohm's law this can be written equivalently as P = I²R = V²/R. The total electrical energy consumed in time t is W = Pt; the commercial "unit" of electrical energy is the kilowatt-hour (1 kWh = 3.6 × 10⁶ J).

Worked Example 1 (Power/resistance): An electric bulb is rated "100 W, 220 V". Find its resistance and the current it draws at rated voltage.

R = V²/P = (220)²/100 = 48400/100 = 484 Ω

I = P/V = 100/220 ≈ 0.455 A

Worked Example 2 (Temperature dependence): A copper wire has resistance 5.0 Ω at 20 °C. If its temperature coefficient of resistivity is α = 4.0 × 10⁻³ °C⁻¹, find its resistance at 80 °C.

R₈₀ = R₂₀[1 + α(80 − 20)] = 5.0 × [1 + (4.0 × 10⁻³ × 60)] = 5.0 × [1 + 0.24] = 5.0 × 1.24 = 6.2 Ω

When resistors are connected in series, the same current I flows through each one, and the total potential difference across the combination is the sum of the individual drops: V = IR₁ + IR₂ + ... + IR_n = I(R₁ + R₂ + ... + R_n). The equivalent resistance is therefore R_eq = R₁ + R₂ + ... + R_n, which is always greater than the largest individual resistance.

When resistors are connected in parallel, each resistor is connected across the same pair of points, so each has the same potential difference V across it, while the total current I entering the combination splits into individual branch currents that add up: I = I₁ + I₂ + ... + I_n = V/R₁ + V/R₂ + ... + V/R_n. Dividing through by V gives the equivalent resistance from 1/R_eq = 1/R₁ + 1/R₂ + ... + 1/R_n, which is always smaller than the smallest individual resistance.

Worked Example 1 (Series): Three resistors of 1 Ω, 2 Ω and 3 Ω are connected in series to a 12 V battery of negligible internal resistance. Find the total resistance, the current, and the potential drop across each resistor.

R_eq = 1 + 2 + 3 = 6 Ω; I = V/R_eq = 12/6 = 2 A (same in all three, since series)

V₁ = IR₁ = 2 × 1 = 2 V; V₂ = IR₂ = 2 × 2 = 4 V; V₃ = IR₃ = 2 × 3 = 6 V (sum = 12 V, as expected)

Worked Example 2 (Parallel): Resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 12 V battery of negligible internal resistance. Find the equivalent resistance, the current from the battery, and the current through each resistor.

1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R_eq = 1 Ω

Total current I = V/R_eq = 12/1 = 12 A

I₁ = 12/2 = 6 A, I₂ = 12/3 = 4 A, I₃ = 12/6 = 2 A (sum = 12 A, as expected)

A cell maintains a potential difference between its terminals due to internal electrochemical action. The maximum potential difference it can provide, measured when no current is drawn (open circuit), is called its electromotive force (emf), ε. However, a real cell always has some internal resistance r due to the electrolyte and electrode material. When the cell drives a current I through an external circuit, some potential is dropped across r itself, so the terminal voltage is V = ε − Ir (while discharging). During charging (current forced into the cell by an external source), the terminal voltage instead exceeds the emf: V = ε + Ir. The maximum current a cell can deliver occurs when the external resistance is zero (short circuit): I_max = ε/r.

For n identical cells connected in series (all aiding), the equivalent emf is ε_eq = ε₁ + ε₂ + ... + ε_n and the equivalent internal resistance is r_eq = r₁ + r₂ + ... + r_n. For two cells connected in parallel, the equivalent emf and internal resistance satisfy ε_eq/r_eq = ε₁/r₁ + ε₂/r₂ and 1/r_eq = 1/r₁ + 1/r₂; parallel combination is useful when a larger current (not larger voltage) is needed, since r_eq decreases.

For circuits with multiple sources or loops that cannot be reduced by simple series-parallel rules, Kirchhoff's laws are used. The junction (current) rule states that at any junction, the sum of currents entering equals the sum of currents leaving (ΣI = 0 with sign convention) — this follows from conservation of charge, since charge cannot accumulate at a junction in steady state. The loop (voltage) rule states that around any closed loop in a circuit, the algebraic sum of potential changes (emfs and IR drops) is zero (ΣΔV = 0) — this follows from conservation of energy, since the electric field is conservative.

Worked Example: A battery of emf 12 V and internal resistance 0.5 Ω is connected to an external resistor of 4.5 Ω. Find the current drawn and the terminal voltage of the battery.

I = ε/(R + r) = 12/(4.5 + 0.5) = 12/5 = 2.4 A

Terminal voltage V = ε − Ir = 12 − (2.4 × 0.5) = 12 − 1.2 = 10.8 V (check: V = IR = 2.4 × 4.5 = 10.8 V ✓)

A Wheatstone bridge consists of four resistors P, Q, R and S arranged in a diamond (rhombus) shape: P and Q form one pair of adjacent arms and R and S form the other pair, with a battery connected across one pair of opposite junctions and a galvanometer connected across the other pair of opposite junctions. The bridge is said to be balanced when no current flows through the galvanometer, which happens when the potential at both galvanometer terminals is equal.

Applying Kirchhoff's laws to this balanced condition (with zero galvanometer current, the same current flows through P and Q, and separately the same current flows through R and S) gives the simple balance condition: P/Q = R/S. If three of the four resistances are known and the bridge is balanced, the fourth (unknown) resistance can be calculated precisely. The bridge (null-deflection) method is far more accurate than a simple ammeter–voltmeter method because it does not depend on the accuracy of the meters used, only on detecting zero deflection in the galvanometer.

Worked Example: In a Wheatstone bridge, the arms P, Q and R have resistances 10 Ω, 20 Ω and 15 Ω respectively. Find the resistance S required to balance the bridge.

Using the balance condition P/Q = R/S:

10/20 = 15/S ⟹ S = 15 × 20/10 = 300/10 = 30 Ω

The bridge is balanced when S = 30 Ω.