Electrical Resonance: Q-Factor, Active Resistance & Graphs

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SUMMARY

The discussion centers on the calculation of active resistance in an electrical resonance circuit involving a Q-factor. The circuit includes a 100mH inductor, a 0.253 μF capacitor, and a combination of 10 ohm resistors alongside a significant 98000 ohm active resistor. The calculated active resistance using the formula R=Q/sqrt(C/L) yields approximately 8900 ohms, which is notably 11 times less than the active resistor value. Participants are seeking clarification on the underlying reasons for this discrepancy.

PREREQUISITES
  • Understanding of electrical resonance concepts
  • Familiarity with Q-factor calculations
  • Knowledge of circuit components: resistors, inductors, and capacitors
  • Ability to interpret and analyze circuit diagrams
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  • Explore the implications of resistance values on circuit performance
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Electrical engineering students, circuit designers, and hobbyists interested in understanding resonance phenomena and the implications of Q-factor on circuit behavior.

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Homework Statement


The scheme is here: http://img196.imageshack.us/img196/9366/lcre.jpg
G generates soundwaves, Re are all 10 ohm resistors , R is an 98000 active resistor, L=100mH, C=0.253 μF
The assignment was to measure the voltage through different parts of the circuit near resonance frequency and draw some graphs, but that's not the issue here.
When calculating the active resistance from the formula
R=Q/sqrt(C/L),
where Q is the Q-factor, the result is ~8900 ohms. This answer is correct and supposed to be so, but I have to explain WHY is this 11 times less than in the active resistor and this is the problem here. Can anyone give me some good advice?
 
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