Finding current across capacitor

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SUMMARY

The discussion focuses on calculating the maximum current in an AC circuit with a 30.0 Hz generator and a 45.5 µF capacitor. The maximum current (Io) is determined to be 0.146 A, calculated using the formula Io = Irms × √2, where Irms is derived from the RMS voltage and capacitive reactance (Xc). The current when the voltage across the capacitor is 5.25 V is 138 mA, while the current when the voltage is decreasing is -138 mA, indicating a phase shift in the current direction.

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Homework Statement


An ac generator with a frequency of 30.0Hz and an rms voltage of 12.0V is connected to a 45.5uF capacitor. a) what is the maximum current in this circuit? b) what is the current in the circuit when the voltage across the capacitor is 5.25V and increasing? c) what is the current in the circuit when the voltage across the capacitor is 5.25V and decreasing?


Homework Equations


Irms=Vrms/Xc Xc=1/2pi(frequency)(capacitor) Io=Irms X square root of 2


The Attempt at a Solution



a) I first solved for Xc and got 116.6Ohms and Irms=12V/116.6Ohms=0.102A
Io=0.102X square root of 2=0.146A
I think i have this part right, but check it for me. thanks.

b) V=5.25V my attempt was to just divide by my Xc, but that does not come out right. the answer is the book is 138mA.

c) -138mA

thanks for the help
 
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Part a) is good. I think you did parts b and c right too, is your book known to have errors?
 
very very rarely will it have a wrong answer...
 

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