Heat dissipated in an RLC circuit

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SUMMARY

The discussion focuses on calculating the heat dissipated in an RLC circuit consisting of a 4 Ω resistor, a 14 mH inductor, and a 99 µF capacitor connected in series to a 65 V (rms) source at twice the resonance frequency. The average power formula used is Vrms * R * ω² / (Rω² + L²(ω² - ω(0)²)²). The calculated heat dissipated during one period is 8447.31 J, indicating a potential miscalculation in the integration process or formula application.

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Homework Statement


A 4 Ω resistor, 14 mH inductor, and 99 µF
capacitor are connected in series to a 65 V
(rms) source having variable frequency.
Find the heat dissipated in the circuit during one period if the operating frequency is
twice the resonance frequency.
Answer in units of J


Homework Equations


Vrms * R * ω^2/(Rω^2 + L^2(ω^2 -ω(0)^2)^2) = avg power


The Attempt at a Solution


I integrated the average power from 0 to ∏√(LC) because that was the time for one period. Then I got 8447.31 J. This is a massive number, which I'm sure is wrong. But I don't really know what I'm doing wrong.
 
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Your Relevant Equation yields the real component of the complex current (RMS complex current, since you're starting with Vrms). Multiply it by R to find the average power (or, perhaps you meant to have R2 in the numerator to begin with...)

A full cycle is ##2\pi## radians, so that T = ##2\pi/\omega##.
 

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