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Question
For a given a.c., i = im sin ωt, show that the average power dissipated in a resistor R over a complete cycle is `1/2I_m^2` R .
Solution
The average power dissipated `overlineP =<<i^2 R>> =<<i_m^2 sin^2 ωt>> = i_m^2R<<sin^2ωt>> `
`because sin^2 ωt = 1/2 (1- cos 2ωt)`
`because <<sin^2 ωt>> = 1/2 (1- <<cos 2ωt>>)= 1/2 (because <<cos 2ωt>> =0)`
`thereforeoverlineP = 1/2i_m^2 R`
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