Heat dissipated in an RLC circuit

AI Thread Summary
The discussion focuses on calculating the heat dissipated in a series RLC circuit consisting of a 4 Ω resistor, a 14 mH inductor, and a 99 µF capacitor connected to a 65 V (rms) source at twice the resonance frequency. The average power formula used is Vrms * R * ω^2/(Rω^2 + L^2(ω^2 -ω(0)^2)^2). An attempt to integrate the average power over one period resulted in a value of 8447.31 J, which the user suspects is incorrect. Clarifications are sought regarding the integration limits and the proper application of the average power formula. The discussion emphasizes the need for accurate calculations in determining the heat dissipation in the circuit.
nautola
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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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