What Capacitor Values Solve This AC Circuit Problem?

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Homework Help Overview

The discussion revolves around determining the values of two capacitors based on their behavior in an AC circuit. The original poster presents a scenario where the capacitors are connected in both parallel and series configurations across a 10.0 volt rms, 1.0 kHz oscillator, with specified currents for each configuration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use formulas for current in both parallel and series configurations to derive equations for the capacitor values. Some participants question the accuracy of the calculated capacitance values and suggest that they seem too small given the current observed. Others inquire about the inclusion of capacitive reactance and the meaning of the variable 'w' in the equations.

Discussion Status

Participants are actively engaging with the problem, offering insights and questioning assumptions. There is a recognition of potential errors in the original poster's approach, and suggestions for alternative methods, such as plotting results with arbitrary capacitor values, have been raised. No consensus has been reached regarding the correct values or methods.

Contextual Notes

Participants note that the frequency and current values may imply larger capacitance values than those calculated by the original poster. There is also a suggestion that some formulas may be missing from the original poster's approach.

bemigh
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Hey guys,
I have this problem.. When two capacitors.. (different) are in parallel across a 10.0 volt rms, 1.0 khz oscillator, the currrent is 545 mA . When the same capactiors are in series.. its 126 mA. What are the values of the two capacitors...

Alright.. now... The formula when its in parallel is : I=wCV and C= C1+C2
When in series: its I=wCV and 1/C = 1/C1 + 1/C2

Using these formulas.. i plug in the known values or voltage, omega ( which i found) and the current... I have two equations with two unknowns... and after a lot of greasy quadratics.. i have two answers.. but it says they are both wrong.. 4..628e-6 and 3.522e-6

Did i stray anywhere here??
Cheers
Brentski
 
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Brentski,

It seems to me that the capacitance values you got are way too small. They should add up to something like .0545 F, and should add in inverse to .0126 F. Maybe I'm off by a factor of 2 pi, but even then my values are WAY larger than yours.

1KHz is a very low frequency. 10V is a low voltage. But 126mA is a huge current. It's going to take a big capacitor for that kind of current.

Carl
 
Seems like there are some formulas missing? Are you looking to include capacitive reactance? 1 over the square root of 2 pi fc ? (or something like that) . In your equation, what does 'w' represent?

What happens if you plug in a range of arbitrary cap values? then plot the results and approximate whether to increase/decrease values? Id try but don't know what 'w' refers to
 
bemigh said:
Hey guys,
I have this problem.. When two capacitors.. (different) are in parallel across a 10.0 volt rms, 1.0 khz oscillator, the currrent is 545 mA . When the same capactiors are in series.. its 126 mA. What are the values of the two capacitors...

Alright.. now... The formula when its in parallel is : I=wCV and C= C1+C2
When in series: its I=wCV and 1/C = 1/C1 + 1/C2

Using these formulas.. i plug in the known values or voltage, omega ( which i found) and the current... I have two equations with two unknowns... and after a lot of greasy quadratics.. i have two answers.. but it says they are both wrong.. 4..628e-6 and 3.522e-6

Did i stray anywhere here??
Cheers
Brentski
Your expression for current is wrong. It should be: [itex]I = V/X_c = V/\omega C[/itex]

So for the series configuration:

[tex]I_s = V(1/C_1 + 1/C_2)/\omega[/tex]

and in parallel:

[tex]I_p = V/\omega(C_1 + C_2)[/tex]

Plug in values for the two currents and V and work out C1 and C2

AM
 

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