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Question

A resistor R, inductor L, and capacitor C are connected in series to an AC source of rms voltage ΔV\Delta V and variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

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Answered 1 year ago
Answered 1 year ago
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The solution is the same as in the previous problem only without introducing the numerical values. The energy supplied in one period is determined from the power relation and the relations of the resonant and regular frequency:

E=PT=Vrms2Rω2R2ω2+L2(ω2ω02)22πω=VrmsR4ω02R24ω02+L29ω042π2ω0=4πVrms2Rω0(4R2+9L2ω02)=4πVrms2R1LC(4R2+9L2LC)=4πVrms2RCLC4R2C+9L\begin{align*} E&=PT\\ &=\dfrac{V_{\text{rms}}^{2}R\omega^{2}}{R^{2}\omega^{2}+L^{2}(\omega^{2}-\omega_{0}^{2})^{2}}\dfrac{2\pi}{\omega}\\ &=\dfrac{V_{\text{rms}}R\cdot4\omega_{0}^{2}}{R^{2}\cdot4\omega_{0}^{2}+L^{2}\cdot9\omega_{0}^{4}}\dfrac{2\pi}{2\omega_{0}}\\ &=\dfrac{4\pi V_{\text{rms}}^{2}R}{\omega_{0}(4R^{2}+9L^{2}\omega_{0}^{2})}\\ &=\dfrac{4\pi V_{\text{rms}}^{2}R}{\dfrac{1}{\sqrt{LC}}\bigg(4R^{2}+\dfrac{9L^{2}}{LC}\bigg)}\\ &=\boxed{\dfrac{4\pi V_{\text{rms}}^{2}RC\sqrt{LC}}{4R^{2}C+9L}} \end{align*}

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