How Do You Calculate Capacitance in a Series RC Circuit?

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To calculate capacitance in a series RC circuit, the correct approach involves using the impedance formula, where the total impedance Z combines resistance and reactance in quadrature. The relevant equations include I(rms) = E(rms) / Z and Z = √(R² + (1/(2πfC))²). Given the circuit parameters, the current (I(rms) = 0.310 A), resistance (R = 151 Ω), and frequency (f = 7.70 Hz), the capacitance can be determined by rearranging the equations. The initial attempt to calculate capacitance using a simplified approach was incorrect due to not accounting for the frequency-dependent reactance. The final calculation yields a capacitance of C = 0.00244 F.
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
 
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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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