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Asked in GUJCET 2023 · RMS and mean values
Given: the measured output 24 V is an r.m.s. value, and the bulb takes P=12 W.
Idea: for a resistive load P=VᵣₘₛIᵣₘₛ.
Iᵣₘₛ=P/(Vᵣₘₛ)=(12)/(24)=0.5 A.
Peak current: Iₘ=√2 Iᵣₘₛ=1.414×0.5=0.707 A.
So Iₘ≈0.71 A, that is 1/(√2) A.
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