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Asked in GUJCET 2018 · Average power and heating
Given: R=10 Ω and peak current I₀=2 A.
Idea: the average power in a resistor is P=Iᵣₘₛ²R, because the heating depends on the mean of I².
Iᵣₘₛ=(I₀)/(√2)=√2 A, so Iᵣₘₛ²=2 A².
P=2×10=20 W.
So the power dissipated is 20 W.
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