LRC Circuit Problem

  • Thread starter rabcdred
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  • #1
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Homework Statement



An oscillator producing 10 volts (rms) at 200 rad/s is connected in series with a 50 Ω resistor, a 400 mH inductor, and a 200 μF capacitor. The rms voltage (in volts) across the inductor is

Homework Equations


Xc=1/wC, Xl=wL, Vrms= Irms(Z), Z=(R^2 +(Xl-Xc)^2)^(1/2)


The Attempt at a Solution


I know I am supposed to make an attempt here but I really have absolutely no idea what to do. All I could do was solve for Z and got Z= 74.33Ω. Please Help! Thanks.
 

Answers and Replies

  • #2
gneill
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Homework Statement



An oscillator producing 10 volts (rms) at 200 rad/s is connected in series with a 50 Ω resistor, a 400 mH inductor, and a 200 μF capacitor. The rms voltage (in volts) across the inductor is

Homework Equations


Xc=1/wC, Xl=wL, Vrms= Irms(Z), Z=(R^2 +(Xl-Xc)^2)^(1/2)


The Attempt at a Solution


I know I am supposed to make an attempt here but I really have absolutely no idea what to do. All I could do was solve for Z and got Z= 74.33Ω. Please Help! Thanks.

If you know how to deal with complex numbers then an easy approach is to work with the complex impedance, current, and voltages. You can then treat all the components just as you would resistors using Ohm's law and so forth.
 
  • #3
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Thanks for the response. Unfortunately, I do not know how to work with complex numbers. Do you have any other suggestions?
 
  • #4
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You are almost there....you have the equations you need.
I also get Z = 74.3 ohms so now you can calculate the current. It is a series circuit so the current is the same through each component. It should be straightforward to calculate the 3 voltages
 
  • #5
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Thanks. I took your advice but I still cannot get the correct answer. I calculated 6.7 volts being dissipated across the resistor and concluded that the inductor would have to be less than 3.3volts. This answer choice was wrong however. The answer choices are 6.7V, 2.5V, 3.4V, 10V, and 7.6V. 10V and 2.5V are incorrect.
 
  • #6
gneill
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Thanks. I took your advice but I still cannot get the correct answer. I calculated 6.7 volts being dissipated across the resistor and concluded that the inductor would have to be less than 3.3volts. This answer choice was wrong however. The answer choices are 6.7V, 2.5V, 3.4V, 10V, and 7.6V. 10V and 2.5V are incorrect.

Did you multiply your apparent current by each of the component impedances (R, XL, XC)?
 
  • #7
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Yes I did. However, I did not get any of the possible answer choices when I did V=Irms * Xl
 
  • #8
gneill
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Yes I did. However, I did not get any of the possible answer choices when I did V=Irms * Xl

What values did you get for each of the voltages?
 
  • #9
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I got V(resis)= 6.73V, V(ind)= 10.78V, V(capac)=3.36V. Maybe my current is wrong. I got Irms= 0.134589502.
 
  • #10
gneill
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I got V(resis)= 6.73V, V(ind)= 10.78V, V(capac)=3.36V. Maybe my current is wrong. I got Irms= 0.134589502.

Those values all look fine for a series RLC with the parameters that you've specified.

If 10.78 V is not a choice for the inductor voltage and the selection of closest value to that is deemed incorrect, then it is possible that the question has been altered at some point (to make it a "new" question) without updating the answer key.
 
  • #11
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Yep. It seems that way. Thanks so much for all of your help. I really appreciate it.
 
  • #12
gneill
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As a point of interest, if the input voltage of 10 V was a peak value rather than an rms one, then the resulting rms value on the inductor would be 7.61 V.
 

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