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A sinusoidal A.C. current flows through a resistor of resistance 10 Ω. If the peak current through the resistor is 2 A, then the power dissipated is ________.

Asked in GUJCET 2018 · Average power and heating

Answer: (2) 20 W

Step-by-step solution

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.

Why the other options are wrong

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