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Question
The average power dissipated in a resistor when an AC current of peak value im flows through it is ______.
Options
P = \[i_m^2R\]
P = \[\frac{1}{2}i_m^2R\]
P = 2\[i_m^2\]R
P = 0
MCQ
Fill in the Blanks
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Solution
The average power dissipated in a resistor when an AC current of peak value im flows through it is P = \[\frac{1}{2}i_m^2R\].
Explanation:
The instantaneous power is p = \[i_m^2R\sin^2\omega t\]. Averaging over a full cycle using (sin2ωt) = \[\frac {1}{2}\] gives P = \[\frac{1}{2}i_m^2R\] = I2R, where (I) is the rms current. Power dissipation is always positive regardless of current direction.
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