Lighting and accessories are the visible loads of the electrical system. A switch may control a relay coil rather than the load current directly, a fuse protects the wiring, and pulse-width modulation can dim a lamp without dissipating the excess energy as heat.
Two 55 W lamps
Problem. Two 55 W halogen lamps operate at 12 V. What total current do they draw?
Solution. P_total = 110 W. I = 110 ÷ 12 = 9.17 A. The wiring, switch or relay and fuse must be rated for that current and for the lamp’s inrush behaviour.
Switching and protection
A relay lets a small switch control a large lamp current; a fuse or circuit breaker protects the feed wiring; a module may control the lamp directly with a semiconductor output. The correct ratings come from the vehicle manufacturer’s data, because lamp inrush, wire length and ambient temperature all affect the choice.
LED replacement lamps
An LED lamp draws less current than the incandescent lamp it replaces. On vehicles that monitor lamp current, that lower current can be interpreted as a failed lamp and trigger a warning. The fix is the correct manufacturer-approved solution, not an arbitrary resistor that overheats.
- Lighting current is P/V for a resistive lamp near its rated voltage.
- Halogen lamps draw a high inrush current when cold.
- Relays separate control current from load current.
- PWM dimming switches the lamp rapidly to control average power.
- LED replacement can change the current monitored by the vehicle.
- I = P/V gives the load current.
- Lamp inrush can exceed the steady current.
- Relays and fuses protect the control and feed circuits.
- LED replacement can confuse lamp monitoring.
Worked example
A 21 W brake lamp at 12 V draws 21 ÷ 12 = 1.75 A. A trailer with two such lamps draws 3.5 A, plus the side markers, so the trailer module and wiring must be rated for the total.