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?
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.
6A166
6A105
6A408
6A11
6A468
Pass Your FCC Exam!
Study offline, track your progress, and simulate real exams with the GMDSS Trainer app.
Includes Elements 1, 3, 6, 7R, 8, and 9.