FCC Exam Question: 6A64

The minimum power dissipation rating of a resistor of 20,000 ohms across a potential of 500 V shouldbe:

A. 25 watts
B. 12.5 watts
C. 15 watts
D. None of the above
Correct Answer: A

Explanation: To determine the minimum power dissipation rating, first calculate the actual power dissipated by the resistor. Using the power formula P = V^2 / R, where P is power, V is voltage, and R is resistance: P = (500 V)^2 / 20,000 Ω P = 250,000 / 20,000 P = 12.5 watts While the resistor dissipates 12.5 watts, good engineering practice, especially in amateur radio circuits, dictates that components should be "derated" to ensure reliability, longevity, and to prevent overheating. A common guideline is to select a resistor rated for at least double the calculated operating power. Therefore, a resistor dissipating 12.5 watts should have a minimum power rating of 12.5 W * 2 = 25 watts. Option A (25 watts) provides this necessary safety margin. Option B (12.5 watts) is the actual power dissipated, but a resistor with this exact rating would operate at its maximum limit, leading to excessive heat and potential failure. Options C and D are incorrect values.

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.