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

by jamiewilliams
Tags: capacitor, connected, inductor, power, resistor, series
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jamiewilliams
#1
Mar17-12, 03:30 PM
P: 11
1. The problem statement, all variables and given/known data
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.


2. Relevant equations

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

3. 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.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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ehild
#2
Mar18-12, 01:10 AM
HW Helper
Thanks
P: 10,506
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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