Ohm's law and power in practice

Ohm's law links voltage (U, volts), current (I, amperes) and resistance (R, ohms): U = R · I. If you know two of them, you can always find the third.

Example: A 230 V heating element with 50 Ω resistance draws I = 230/50 = 4.6 A and dissipates P = 230 · 4.6 ≈ 1058 W.

For inductive or capacitive loads (motors, fluorescent fittings) current and voltage are phase-shifted. The real (active) power becomes P = U · I · cos φ, where cos φ is the power factor.

Resistors rarely sit alone. Two basic configurations cover most: in series, the same current flows through all resistors, and voltage divides between them; in parallel, voltage is the same across all while current divides.

Example (series): Three resistors of 10, 20 and 30 Ω in series give R = 60 Ω. At 230 V the circuit draws I = 230/60 ≈ 3.83 A, and the voltage divides as 38.3 V, 76.7 V and 115 V — the largest voltage drop across the largest resistance.

Example (parallel): Two resistors of 10 Ω and 40 Ω in parallel give R = (10 · 40)/(10 + 40) = 400/50 = 8 Ω — note that the result is always smaller than the smallest individual resistance.

Kirchhoff's two laws are the tools for circuits that are more than a simple series or parallel. Current law (junction law): the sum of currents entering a junction equals the sum leaving — no charge is lost. Voltage law (mesh law): the sum of voltages around a closed loop is zero — what you raise the voltage with in the source, you lose again across the resistances.

A conductor's resistance depends on the material, the length and the cross-section: R = ρ · L / A, where ρ is the resistivity. Copper has ρ ≈ 0.0175 Ω·mm²/m (Danish table value for the conductivity κ ≈ 56 at 20 °C; calculating 1/0.0175 theoretically gives ≈ 57), aluminium ρ ≈ 0.028 Ω·mm²/m (κ ≈ 35). Aluminium thus conducts only about two thirds as well as copper at the same cross-section — which is why an aluminium cable needs a larger cross-section for the same job.

Resistance increases with temperature. For copper the temperature coefficient α ≈ 0.0039 per K, so a conductor heated from 20 °C to 70 °C gets approx. 20 % higher resistance: R_70 ≈ R_20 · (1 + 0.0039 · 50) ≈ R_20 · 1.20. This is one of the reasons why voltage drop is always calculated at operating temperature, not at 20 °C.

Power times time is energy: E = P · t. If P is in kilowatts and t in hours, you get kilowatt-hours (kWh) — the unit the electricity bill is settled in.