Calculating Current Split in Parallel Circuit with Frequency of 100 sec-1

AI Thread Summary
A capacitor is connected in parallel with a resistive load, and the values given are C = 0.829 µF and R = 229 ohms, with a frequency of 100 sec-1. The user attempts to calculate the current through the capacitor and resistor using the formulas I(resistor) = V/R and I(capacitor) = ω*C*V. The derived fraction of current through the capacitor is approximately 0.106569, but there is confusion regarding whether this represents the correct fraction and how to calculate total effective current in an AC circuit. The user notes that total instantaneous current does not equal the sum of the individual currents in AC circuits, indicating a need for clarification on calculating effective current in parallel configurations. The discussion highlights the complexities of analyzing AC circuits, particularly in determining effective current contributions from capacitive and resistive components.
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Homework Statement



A capacitor is placed parallel across a resistive load in a circuit. C= 0.829*10-6 F and R = 229 ohms. What fraction of the current goes through the capacitor when the frequency is 100 sec-1?

Homework Equations


I(resistor) = V/R
I(capacitor) = C * (dV/dt) = ω*C*V (?)

The Attempt at a Solution


Ic / (Ir + Ic) = [ω*C*V] / [(V/R) + (ω*C*V)] reduces to [ω*C*R] / [1 + ω*C*R]
The answer I got was 0.106569. I don't know if the work is wrong or if the format of my answer wrong, like is that the "fraction" or do I have to do something else to it? Thanks.
 
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The problem here is that I_tot does not equal Ir+Ic. The instantaneous current at any point in time does obey I_tot=Ir+Ic, but not the effective current in an AC circuit. You have to calculate the total effective current in a different way (what way?)
 
The only thing I can find is for an RC circuit in series. is it the same for parallel?
 
do you mean I_tot (sqrt 2)? can someone please help me it's almost due :/
 
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