Radio Amateur Quiz Canadian License

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Question B-005-006-004

When two 500 ohm 1 watt resistors are connected in series, the maximum total power that can be dissipated by the resistors is:

Correct Answer

2 watts

Explanation

Total resistance = 500 + 500 = 1000 Ω. Max current is limited by the 1W rating of *each* resistor (P=I^2*R => I = sqrt(P/R) = sqrt(1/500)). Since current is the same through both in series, the total max power is simply the sum of individual max powers: 1W + 1W = 2W.

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