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

AI Thread Summary
The discussion focuses on calculating the average power dissipated when a resistor, capacitor, and inductor are connected in series with an AC generator. The initial power delivered to the resistor alone is 0.952 W, which drops to 0.477 W when a capacitor is added and further decreases to 0.255 W with an inductor. The challenge lies in determining the power when both the capacitor and inductor are included, with a known answer of 0.666 W. Participants emphasize the need to express impedances in terms of resistance and reactance to derive the correct power equations. The conversation highlights difficulties in manipulating the equations to achieve the desired result.
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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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