Suppose the circuit in Exercise 7.18 has a resistance of $15\Omega.$ Obtain the average power transferred to each element of the circuit, and the total power absorbed
Exercise
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Reistance, $\text{R}=15\Omega$
$\therefore\ \text{Impedance},\ \text{Z}=\sqrt{\text{R}^2+\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)^2}$
$\text{Z}=\sqrt{15^2+\Big(2\pi\times50\times80\times10^{-3}-\frac{1}{2\pi\times50\times60\times10^{-6}}\Big)^2}$
$=\sqrt{225+779.5}\Omega=31.7\Omega$
$\therefore\ \text{I}_{\text{rms}}=\frac{\text{V}_{\text{rms}}}{\text{Z}}=\frac{230}{31.7}=7.255\text{A}$
Average power transferred to inductor, $\text{L}=\text{E}_{\text{v}}\text{I}_{\text{v}}\cos\frac{\pi}{2}=0$
Average power transferred to capacitor $=\text{E}_{\text{v}}\text{I}_{\text{v}}\cos\Big(\frac{-\pi}{2}\Big)=0$
Average power transferred to $R = I^2rms \times R$
$= (7.255)^2 \times 15$
$= 789.5W.$
Reistance, $\text{R}=15\Omega$
$\therefore\ \text{Impedance},\ \text{Z}=\sqrt{\text{R}^2+\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)^2}$
$\text{Z}=\sqrt{15^2+\Big(2\pi\times50\times80\times10^{-3}-\frac{1}{2\pi\times50\times60\times10^{-6}}\Big)^2}$
$=\sqrt{225+779.5}\Omega=31.7\Omega$
$\therefore\ \text{I}_{\text{rms}}=\frac{\text{V}_{\text{rms}}}{\text{Z}}=\frac{230}{31.7}=7.255\text{A}$
Average power transferred to inductor, $\text{L}=\text{E}_{\text{v}}\text{I}_{\text{v}}\cos\frac{\pi}{2}=0$
Average power transferred to capacitor $=\text{E}_{\text{v}}\text{I}_{\text{v}}\cos\Big(\frac{-\pi}{2}\Big)=0$
Average power transferred to $R = I^2rms \times R$
$= (7.255)^2 \times 15$
$= 789.5W.$
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