Calculating Impedance Using Capacitive Reactance

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Homework Help Overview

The discussion revolves around calculating the impedance of an RL circuit consisting of a resistor and an inductor connected in series with an AC source. The original poster seeks to understand how to incorporate capacitive reactance into the impedance formula, despite the circuit lacking a capacitor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between impedance, resistance, and reactance in RL circuits. Questions arise regarding the derivation of the impedance formula and its application to circuits without capacitors. There is also discussion about the role of phasor diagrams in understanding voltage relationships.

Discussion Status

Participants are actively engaging with the concepts, questioning assumptions about the circuit's components and the application of formulas. Some guidance is offered regarding the use of phasor diagrams and Kirchhoff's voltage law, but no consensus on a final approach has been reached.

Contextual Notes

The original poster expresses confusion about the absence of a capacitive element in the circuit while attempting to apply a formula that includes capacitive reactance. This highlights a potential misunderstanding of the circuit's configuration and the relevant equations.

PeachBanana
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Homework Statement



A 29kΩ resistor is in series with a 62mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is 80Hz .

Homework Equations



XL = ωL
Z = [(R^2 + (XL - XC)^2)] ^ 1/2


The Attempt at a Solution



R = 29,000 Ω

XL = 0.062 H (2∏ * 80 Hz)
XL = 31.16 Ω

I'm having difficulty figuring out how to find XC. XC = 1 / ωC but what is "C?"
 
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PeachBanana said:

Homework Statement



A 29kΩ resistor is in series with a 62mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is 80Hz .

Homework Equations



XL = ωL
Z = [(R^2 + (XL - XC)^2)] ^ 1/2

The Attempt at a Solution



R = 29,000 Ω

XL = 0.062 H (2∏ * 80 Hz)
XL = 31.16 Ω

I'm having difficulty figuring out how to find XC. XC = 1 / ωC but what is "C?"
Your formula for Z applies to RLC circuits. You have an RL circuit. Do you know how that formula for Z was derived? It was probably using a phasor diagram. If you understand the derivation, it's pretty easy to see what you need to do.
 
Was the formula for "Z" derived using the Pythagorean theorem? If so, I still do not see the connection to capacitance.
 
Describe how the formula was derived.
 
It looks as if it uses the following relations:

Vrms = Irms * Z
Vnaught = InaughtZ

The book also shows a phasor diagram where Vnaught is the sum vector and acts as the hypotenuse of the triangle.
 
And where does the triangle come from?
 
The triangle comes from a phasor diagram. I'm attaching an image made in paint of the one I see in the textbook. It's a tad messy, but it is readable.
 

Attachments

  • phasor.png
    phasor.png
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In your picture, I should actually point in the same direction as VR=IR.

Do you understand why the one leg is VL-VC and why you add all the voltages?
 
Are the voltages being added together to find the peak voltage source, Vnaught?
Is Vc subtracted from Vl because Vc lags the current by 90°?
 
  • #10
PeachBanana said:
Are the voltages being added together to find the peak voltage source, Vnaught?
Yes, it's Kirchoff's voltage law applied to the elements in series.

Is Vc subtracted from Vl because Vc lags the current by 90°?
And because VL leads the current by 90°. The two phasors point in opposite directions, so when you add them vector-wise, you subtract their magnitudes.

So back to your original problem, if you follow the same analysis, what do you get for the impedance?
 
  • #11
Since this is an RL circuit then

Z = [(R^2 + XL^2)]^1/2
 
  • #12
Yup, you have it.
 

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