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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 the AC source, we will follow these steps: ### Step 1: Identify the given values - The voltage source is given by \( E = 10 \sin(100 \pi t - \frac{\pi}{6}) \). - The current is given by \( I = 2 \sin(100 \pi t + \frac{\pi}{12}) \). ### Step 2: Determine the phase angles From the voltage equation, we can identify the phase angle of the voltage: ...
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