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A series L-C-R circuit is connected acro...

A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?

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To find the average power dissipated in the series L-C-R circuit connected to an AC source, we will follow these steps: ### Step 1: Identify the given values The given values are: - The EMF of the source: \( E = 10 \sin(100 \pi t - \frac{\pi}{6}) \) - The current: \( I = 2 \sin(100 \pi t + \frac{\pi}{12}) \) ### Step 2: Determine the peak values of voltage and current ...
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