Quality factor and freq bandwidth in a series RLC circuit

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SUMMARY

The quality factor (Q) in a series RLC circuit is inversely related to the resistance (R). In an experiment with an inductance of 100 mH and capacitance of 47 nF, a resistance of 470 Ω yielded a quality factor of Q1 = 2.88, while increasing the resistance to 1 kΩ resulted in a quality factor of Q2 = 1.58. The frequency bandwidth also increased from Δf = 844 Hz to Δf = 1528 Hz as resistance increased. This demonstrates that higher resistance leads to greater energy loss, thus reducing the quality factor.

PREREQUISITES
  • Understanding of series RLC circuit components (resistor, inductor, capacitor)
  • Familiarity with resonant frequency calculations
  • Knowledge of quality factor (Q) and its significance in circuit performance
  • Basic principles of energy loss in electrical circuits
NEXT STEPS
  • Study the impact of resistance on the quality factor in various RLC configurations
  • Learn about the mathematical derivation of the quality factor formula Q = wL/R
  • Explore the effects of different inductance and capacitance values on resonant frequency and bandwidth
  • Investigate practical applications of RLC circuits in filtering and tuning systems
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Electrical engineers, physics students, and hobbyists interested in circuit design and analysis, particularly those focusing on resonance and quality factor in RLC circuits.

Alex Farraday
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Homework Statement


How and why does quality factor change in a series RLC circuit when the resistance in the circuit increases? I've made a small experiment circuit which looks like one in the picture:

ehir5v.jpg


The current frequency is 1 kHz and the voltmeter should show 2V throughout the whole experiment. Inductance L = 100 mH and capacitance C = 47nF. Now, in the first part of the experiment, I use R = 470 Ω. I get that the max current at the resonant frequency is Imax = 4.13 mA and the resonant frequency is fr = 2436 Hz. Then I calculate boundary frequencies ( left and right of the resonant frequency, where the current Ig = Imax/sqrt(2)) and boundary frequencies are: fg1 = 2076 Hz and fg2 = 2920 Hz, which makes the frequency bandwidth Δf = fg2 - fg1 = 844 Hz.
Then, the quality factor is: Q1 = 2.88

Now, for the second part of the experiment, I use R = 1 kΩ. I get that Imax = 1.97 mA at the resonant frequency fr = 2420 Hz (it's probably the same as the resonant frequency in the first case, but the instruments aren't as precise). Boundary frequencies are: fg1 = 1706 Hz and fg2 = 3234 Hz, which makes the frequency bandwidth Δf = 1528 Hz.
Then, the quality factor is: Q2 = 1.58

My question is, why is the quality factor dependent upon the resistance? Why does the resistance affect the frequency bandwidth and quality factor (for lower R, quality factor is higher and vice versa)?

Homework Equations


Just theoretical explanation.

The Attempt at a Solution


The experiment explained above.
 
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The only mechansm for energy loss in the system is the resistance, since pure reactances are lossless. So with no resistance, the system would be 100% efficient and all energy added to it from the source would be stored and returned.

Q factor is directly proportional to the ratio: (energy stored)/( energy lost) per cycle. So if the losses approach zero, then Q approaches infinity.
 
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Alex Farraday said:

Homework Statement


How and why does quality factor change in a series RLC circuit when the resistance in the circuit increases? I've made a small experiment circuit which looks like one in the picture:

ehir5v.jpg


The current frequency is 1 kHz and the voltmeter should show 2V throughout the whole experiment. Inductance L = 100 mH and capacitance C = 47nF. Now, in the first part of the experiment, I use R = 470 Ω. I get that the max current at the resonant frequency is Imax = 4.13 mA and the resonant frequency is fr = 2436 Hz. Then I calculate boundary frequencies ( left and right of the resonant frequency, where the current Ig = Imax/sqrt(2)) and boundary frequencies are: fg1 = 2076 Hz and fg2 = 2920 Hz, which makes the frequency bandwidth Δf = fg2 - fg1 = 844 Hz.
Then, the quality factor is: Q1 = 2.88

Now, for the second part of the experiment, I use R = 1 kΩ. I get that Imax = 1.97 mA at the resonant frequency fr = 2420 Hz (it's probably the same as the resonant frequency in the first case, but the instruments aren't as precise). Boundary frequencies are: fg1 = 1706 Hz and fg2 = 3234 Hz, which makes the frequency bandwidth Δf = 1528 Hz.
Then, the quality factor is: Q2 = 1.58

My question is, why is the quality factor dependent upon the resistance? Why does the resistance affect the frequency bandwidth and quality factor (for lower R, quality factor is higher and vice versa)?

Homework Equations


Just theoretical explanation.

The Attempt at a Solution


The experiment explained above.
Q = wL/R or 1/wRC.
 

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