Find power when resistor, capacitor, and inductor are connected in a series

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SUMMARY

The discussion focuses on calculating the power dissipated in a series circuit containing a resistor, capacitor, and inductor. The average power delivered to the resistor alone is 0.952 W, while adding a capacitor reduces the power to 0.477 W, and adding an inductor further reduces it to 0.255 W. The final power dissipated when both the capacitor and inductor are included in series with the resistor is determined to be 0.666 W. The relevant equations used include P(1) = (V^2)/(R), P(2) = (V^2*R)/Z^2, and P(3) = (V^2*R)/(R^2+(X(L)-X(C))).

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  • Familiarity with impedance calculations
  • Knowledge of power formulas in electrical circuits
  • Basic concepts of reactance for capacitors and inductors
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Homework Statement


When a resistor is connected by itself to an ac generator, the average power delivered to the resistor is 0.952 W. When a capacitor is added in series with the resistor, the power delivered is 0.477 W. When an inductor is added in series with the resistor (without the capacitor), the power delivered is 0.255 W. Determine the power dissipated when both the capacitor and the inductor are added in series with the resistor. Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.


Homework Equations



P(1)= (V^2)/(R)
P(2)= (V^2*R)/Z^2)
P(3)= (V^2*R)/(R^2+(X(L)-X(C))

The Attempt at a Solution



I am having trouble setting up the equation for P(4). I know the answer is .666W but cannot manipulate the equations to get the right answer.
 
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Express the impedances in terms of R, Xc and XL for all cases and use the "relevant equations" to relate R, XC, XL to the powers P1, P2, P3. ehild
 

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