Determine the effective current value in the circuit with C and R

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The discussion revolves around calculating the effective current in a circuit with a resistor and capacitor. The initial calculations yield an equivalent resistance (R_eq) of 12.28 Ohm and a current (I) of 17.9 A, which contradicts the textbook answer of 13 A. Participants highlight the importance of considering the phase difference between the capacitor and resistor when calculating current. A schematic diagram is requested to clarify the circuit configuration, as it may aid in identifying errors in the calculations. The conversation emphasizes the need for accurate phase analysis in AC circuits.
Leonid92
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Homework Statement
The rheostat and capacitor bank are connected in parallel and connected to a sinusoidal voltage source of 220 V with a frequency of 50 Hz. The resistance of the rheostat is 20 Ohm, the capacitance of the capacitor bank is 100 μF. Determine the effective current value in the unbranched part of the circuit.
Relevant Equations
1) Xc = 1/(ω*C) = 1/(2*π*f*C)
2) I = U/resistance
Given:
U = 220 V
f = 50 Hz
r = 20 Ohm
C = 100 μF
Find: I
Solution:
1) Xc = 1/(ω*C) = 1/(2*π*f*C) = 1/(2*π*50*10^-4) = 31.83 Ohm
2) R_eq - equivalent resistance
R_eq = (r*Xc)/(r+Xc) = (20*31.83)/(20+31.83) = 12.28 Ohm
3) I = U/R_eq = 220/12.28 = 17.9 A

True answer given in the textbook is 13 A.

Could you please tell, what is wrong in the solution?
 

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Posting a schematic diagram would help to avoid guesswork on our part. Thanks.
 
kuruman said:
Posting a schematic diagram would help to avoid guesswork on our part. Thanks.
A schematic diagram is not given in the problem statement. I attached a sheme drawn on my own.
 
Thank you for the schematic. I am not sure how you got R_eq. I would calculate the current in the capacitor and rheostat branches separately and then add them noting that they are not necessarily in phase with the voltage.
 
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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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