Science · 5 min read · Last updated September 2026
Worked example: 12 V battery, 4 Ω resistor. Current I = 12 ÷ 4 = 3 A. Power P = V × I = 12 × 3 = 36 W — the resistor turns 36 joules per second into heat, which is why resistor power ratings matter.
Series: resistances add. Two 5 Ω resistors in series = 10 Ω, so the same 12 V drives only 12 ÷ 10 = 1.2 A.
Parallel: 1/R = 1/R₁ + 1/R₂. Two 5 Ω in parallel = 2.5 Ω, so 12 ÷ 2.5 = 4.8 A — parallel paths increase total current because each path sees the full voltage.
| Configuration | Total R | Current at 12 V | Power |
|---|---|---|---|
| Single 4 Ω | 4 Ω | 3.0 A | 36 W |
| Two 5 Ω in series | 10 Ω | 1.2 A | 14.4 W |
| Two 5 Ω in parallel | 2.5 Ω | 4.8 A | 57.6 W |
Choosing an LED series resistor (supply voltage minus LED drop, divided by desired current), checking whether a wire gauge can carry a motor's inrush current, or verifying a power-supply spec. The Ohm's Law Calculator solves any missing variable; the Resistor Color Code tool decodes the bands on the part itself.
Current through a resistor equals voltage across it divided by its resistance. Double the voltage and current doubles; double the resistance and current halves.
Divide voltage by resistance: I = V ÷ R. A 12 V supply across 4 Ω gives 3 amperes.
P = V × I, which also equals I²R or V²/R. On the 12 V, 3 A example: 36 watts of heat.
Each parallel path carries current independently at the full voltage, so total current adds while voltage stays the same — which means the effective resistance must be lower. Two equal resistors in parallel give exactly half the resistance.