How Do You Calculate Capacitance in a Series RC Circuit?

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Homework Help Overview

The discussion revolves around calculating the capacitance in a series RC circuit, given specific values for resistance, voltage, frequency, and current. Participants are exploring the relationships between these variables using relevant equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants attempt to apply equations for current in the circuit but express difficulty in obtaining the correct capacitance value. Others question the handling of resistance and reactance, suggesting that they should be combined in quadrature. There is also a focus on the importance of frequency in determining reactance.

Discussion Status

Participants are actively engaging with the problem, offering corrections and alternative approaches. Some have pointed out the need to consider the electric potential and the proper formulation of Ohm's law in the context of the circuit. There is no explicit consensus yet, but various interpretations and methods are being explored.

Contextual Notes

Participants are discussing the implications of using incorrect equations and the necessity of incorporating frequency into their calculations. The original poster's attempts at a solution are noted, but there is an acknowledgment of potential misunderstandings in the application of the equations.

SereneKi
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Homework Statement



Consider a series RC circuit with R = 151 Ω and an unknown capacitor C as shown below.


http://www.flickr.com/photos/67342906@N0…


The circuit is driven by an alternating electromotive force with εRMS = 174 V at a frequency of f = 7.70 Hz. The current in the circuit is IRMS = 0.310 A. Calculate the capacitance C.

Homework Equations



(1) I(rms)=V(source) / ( R+ (1/C))
(2) I(rms)=C x V(source) / ( 1+ RC)

The Attempt at a Solution



tried to plug them into equation (1) to get C but it didn't work

0.31 =174 / (151+ 1/C )

C=0.00244 F
 
Last edited by a moderator:
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SereneKi said:

Homework Statement



Consider a series RC circuit with R = 151 Ω and an unknown capacitor C as shown below.


http://www.flickr.com/photos/67342906@N0…


The circuit is driven by an alternating electromotive force with εRMS = 174 V at a frequency of f = 7.70 Hz. The current in the circuit is IRMS = 0.310 A. Calculate the capacitance C.

Homework Equations



(1) I(rms)=V(source) / ( R+ (1/C))
(2) I(rms)=C x V(source) / ( 1+ RC)

The Attempt at a Solution



tried to plug them into equation (1) to get C but it didn't work

0.31 =174 / (151+ 1/C )

C=0.00244 F

Your Relevant Equations are not correct in that they are not handling the summation of resistance (of the resistor) and reactance (of the capacitor) properly; they should add in quadrature (square root of sum of squares, just like vector components). Also, the reactance should depend upon the frequency of the driving signal, which in this case is given as 7.70 Hz.
 
Last edited by a moderator:
gneill said:
Your Relevant Equations are not correct in that they are not handling the summation of resistance (of the resistor) and reactance (of the capacitor) properly; they should add in quadrature (square root of sum of squares, just like vector components). Also, the reactance should depend upon the frequency of the driving signal, which in this case is given as 7.70 Hz.

So

Irms = sqrt( R^2 + (1/2(pi)fC)^2)
 
SereneKi said:
So

Irms = sqrt( R^2 + (1/2(pi)fC)^2)

Now you're missing the electric potential :smile: Remember you're writing Ohm's law, so I = E/Z, where here Z is the sum of resistance and reactance as you've calculated above.
 
gneill said:
Now you're missing the electric potential :smile: Remember you're writing Ohm's law, so I = E/Z, where here Z is the sum of resistance and reactance as you've calculated above.

So

Erms/Irms = sqrt( R^2 + (1/2(pi)fC)^2)
 
SereneKi said:
So

Erms/Irms = sqrt( R^2 + (1/2(pi)fC)^2)

Yes. Solve for C.
 

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