FCC Exam Question: 6A73

Given: Input power to a receiver is 75 watts. How much power does the receiver consume in 24 hours ofcontinuous operation?

A. 1800 watthours
B. 1.80 kilowatthours
C. A & B
D. None of the above
Correct Answer: C

Explanation: The question asks for the total energy consumed by the receiver over a period, not its instantaneous power. Energy consumed is calculated by multiplying power by time. Given: Input Power (P) = 75 watts, Time (t) = 24 hours. Energy (E) = P × t = 75 watts × 24 hours = 1800 watthours (Wh). This calculation matches option A. To express this energy in kilowatthours (kWh), we convert watthours to kilowatthours by dividing by 1000 (since 1 kilowatt = 1000 watts). E = 1800 Wh / 1000 = 1.80 kilowatthours (kWh). This conversion matches option B. Both 1800 watthours and 1.80 kilowatthours represent the exact same amount of electrical energy consumed, just expressed using different units. Therefore, both A and B are correct statements, making C the comprehensive correct answer. This demonstrates a fundamental understanding of electrical energy calculation and unit conversion, crucial for evaluating power usage in amateur radio stations.

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