FCC Exam Question: 3-61I1

When an emergency transmitter uses 325 watts and a receiver uses 50 watts, how many hours can a 12.6 volt, 55 ampere-hour battery supply full power to both units?

A. 1.8 hours.
B. 6 hours.
C. 3 hours.
D. 1.2 hours.
Correct Answer: A

Explanation: To determine how long the battery can supply power, first calculate the total power consumption of both units. The transmitter uses 325 watts and the receiver uses 50 watts, for a total of 325 W + 50 W = 375 W. Next, calculate the total current drawn from the battery using the formula Power (P) = Voltage (V) * Current (I). Rearranging for current: I = P / V. I = 375 W / 12.6 V ≈ 29.76 Amperes. Finally, calculate the battery's run time by dividing its ampere-hour rating by the total current drawn: Run Time (hours) = Battery Capacity (Ah) / Current (A) Run Time = 55 Ah / 29.76 A ≈ 1.848 hours. Rounding this value gives approximately 1.8 hours, which corresponds to option A. The other options are incorrect as they do not result from these calculations.

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Includes Elements 1, 3, 6, 7R, 8, and 9.