Journeyman Electrician Exam (NEC)General Electrical Knowledge and Plan ReadingHard
A client reports that a 120V circuit designed for general lighting always trips its 15A overcurrent protective device shortly after being turned on, even though the total connected load (sum of lamp wattages) is only 1200W. All lamps are incandescent. What is the MOST likely cause for the tripping?
- AExcessive reactive power.
- BInsufficient voltage.
- CHigh circuit resistance.
- DInrush current from incandescent lamps.
Show answer & explanationAnswer & explanation
Correct answer: D. Inrush current from incandescent lamps.
The calculated continuous current for 1200W at 120V is 10A (1200W/120V). A 15A breaker should easily handle this. However, incandescent lamps, when first energized, draw a momentary inrush current (cold filament resistance is much lower than hot) that can be 10-15 times their steady-state current. This brief surge can trip an OCPD, especially if it's a thermal-magnetic breaker.
Why the other options are wrong
- A. Incandescent lamps are primarily resistive loads and have very little reactive power, so this is not a factor for tripping.
- B. Insufficient voltage would lead to lower current for a resistive load (P=V^2/R), not higher current and tripping.
- C. High circuit resistance would decrease the current (I=V/R), making it less likely to trip the breaker, not more.
Incandescent Lamp Inrush Current
Incandescent lamps draw a momentary high current (inrush current) upon energization due to the low resistance of their cold filament, which can be significantly higher than their steady-state operating current.
- Cold filament resistance is much lower than hot filament resistance.
- Inrush current can be 10-15 times the normal operating current.
- Lasts for a very short duration (milliseconds).
- Can cause nuisance tripping of overcurrent protective devices.
Memory trick: Overload, Short, Ground, or Inrush Surge.