Subelement A: — Topic :
Question 6A64
Element 6 (Radiotelegraph)The minimum power dissipation rating of a resistor of 20,000 ohms across a potential of 500 V shouldbe:
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.
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