Finding Phase Shift in RLC Circuits

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The discussion revolves around calculating the phase shift in an RLC circuit with specified resistance, capacitance, and inductance at a given frequency. The initial attempts to find the phase shift using the inverse tangent formula led to results around -1.5 radians, but these were deemed incorrect. A participant confirmed that the current lags the voltage by approximately -1.54 radians, suggesting that the textbook might have rounded the answer prematurely. Another participant recalculated the phase shift to be around -1.44 radians, indicating potential discrepancies in the provided answers. The conversation highlights the complexities of phase shift calculations in RLC circuits and the importance of precise values.
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Homework Statement



An RLC circuit has a resistance of 2.0 kΩ, a capacitance of 8.0 µF, and an inductance of 9.0 H. If the frequency of the alternating current is 4.0/π kHz, what is the phase shift between the current and the voltage?
A) -1.6 rad
B) -1.5 rad
C) 36 rad
D) 3.1 rad



Homework Equations



Tan-1 (XL-Xc/R)


The Attempt at a Solution



at first i thought that since current phasors lag by 90 degrees that it may be -1.5 rads. but that isn't correct.
then i used the the inverse tangent equation above in hopes of finding the angle which came out to be 88.4 degrees (which is about 1.5 rads)
i then took the 88.4 degrees and subtracted it from 90 degrees, because I thought it might turn out a correct answer because the phasors are perpendicular to each other. but that gives you 1.6 which is not correct either.
I'm not sure what equation to use ...
 
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Your work is correct, the current lags the voltage by 1.54 rad. The book might have rounded off too early, so it can be A or B. ehild
 
ehild said:
Your work is correct, the current lags the voltage by 1.54 rad. The book might have rounded off too early, so it can be A or B.


ehild

yeah that's what i thought, but i submitted both of those and neither is correct but i think that may be a mistake! thanks!
 
I calculated -1.44 rad:
w = 8 rad/s
angle = tan-1{(8*9 - 1/64e-6)/2e3} = tan-1(-7.78) = -82.7 deg = -1.443 rad.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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