Determine the capacitor value so that this circuit is in resonance

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Homework Help Overview

The discussion revolves around determining the capacitor value required for resonance in an electrical circuit involving inductance and resistance. The participants are analyzing the relationships between inductive reactance, capacitive reactance, and impedance in a parallel and series configuration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate the impedance of the circuit using given values for inductance and frequency. There are questions about the correct application of formulas for reactance and impedance, as well as concerns about rounding errors and the correct signs in calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the circuit's impedance and how to combine various components. Some guidance has been offered regarding the use of complex numbers for impedance representation, but no consensus has been reached on the final approach to finding the capacitor value.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use. There are also discussions about the implications of phase angles in the calculations, indicating a need for careful consideration of vector addition in impedance.

baywa2
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Homework Statement
Need to determine capacitor value so that the circuit is in resonance
Relevant Equations
E=21∠0º
ω=10^3 rad/s
R1=10Ω
R2=40Ω
L=100mh
C=?
246672

L=100mh=0.1H
ω=10^3 rad/s -> f = 159Hz
XL= ωL= 2πfL= 2π*159*0.1= 99.90 Ω
Z parallel = [(XL∠90º)*R2] / [(XL∠90º)-R2]= 37.13∠-21.8º
XC= 1/ωC= 1/(2 π f C)
I don't see how I am supposed to get to C
 
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baywa2 said:
L=100mh=0.1H
ω=10^3 rad/s -> f = 159Hz
XL= ωL= 2πfL= 2π*159*0.1= 99.90 Ω
OK. Here, it is easier to use XL = ωL rather than XL = 2πfL. Your value of 99.90 has a bit of round off error.
Z parallel = [(XL∠90º)*R2] / [(XL∠90º)-R2]
Check the sign of R2 in the denominator.

Are you familiar with using complex numbers to represent impedances? For me, that makes things easier.
I don't see how I am supposed to get to C
Think about how to combine the resistance R1 and the reactance of C with the result for Z_parallel. Can you see that C, R1, and Z_parallel are all in series?
 
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Ok
so Z parallel = [(XL∠90º)*R2] / [(XL∠90º)+R2]
I get Z is some 37 Ω
So do I then use R1 and C for Z? Z=R1+XC∠-90º
Then add the Z and Zpar? 10+37+XC∠-90º -> 47+XC∠-90º
 
baywa2 said:
Ok
so Z parallel = [(XL∠90º)*R2] / [(XL∠90º)+R2]
I get Z is some 37 Ω
There should also be a phase angle for the 37 Ω.

So do I then use R1 and C for Z? Z=R1+XC∠-90º
Then add the Z and Zpar? 10+37+XC∠-90º -> 47+XC∠-90º
You can't combine the 10 and the 37 to get 47 since the 37 has a nonzero phase angle. It's like adding vectors. The 10 Ω corresponds to a vector pointing along the x-axis since it has zero phase angle. The 37 Ω corresponds to a vector pointing at some angle above the x axis.
 

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