USCG Exam PracticeDC circuits and electronic principles
A battery with an open-circuit EMF of 12 V and an internal resistance of 0.5 Ω supplies a load resistance of 4.5 Ω. A technician then replaces the load with one of 1.5 Ω. What is the terminal voltage with the 1.5 Ω load connected?
- A. 11.4 V
- B. 10.5 V
- ✓ 9.0 VCorrect
- D. 6.0 V
Why C is correct
NEETS Mod. 1 §2-1 and §3-4: with the 1.5 Ω load, total circuit resistance = 0.5 + 1.5 = 2.0 Ω. Current I = 12/2 = 6 A. Internal resistance drop = 6 × 0.5 = 3 V. Terminal voltage = 12 − 3 = 9 V. With the original 4.5 Ω load, I = 12/5 = 2.4 A and terminal voltage = 12 − 1.2 = 10.8 V, confirming that heavier loads cause greater terminal voltage sag.