simulate this circuit – Schematic created using CircuitLab

I have a series RLC circuit and I calculated the resonance frequency. Say its w (omega).

Now in second experiment, i use the resistor of half the value used in above ciruit. What effect will it have on my resonant frequency now?


Experiment 1: C = 100nF, L = 100mH, R = 100ohms

Experiment 2: C = 100nF, L = 100mH, R = 50ohms

Only resistor is changed. Nothing else.


1 Answer 1


What effect will it have on my resonant frequency

Theoretically it will have no effect on the resonant frequency. The resonant frequency is purely determined by the capacitor having exactly the opposite reactance of the inductor at a particular frequency and the two reactances cancel leaving the series tuned circuit having only resistance at resonance.

However, with lower values of resistance the peak shape of the resonance will change but the centre point of the peak will remain as previous.

enter image description here

  • \$\begingroup\$ Thank you for your answer, Andy. Can you specify any formula or a relation relating to this? \$\endgroup\$ Commented Nov 18, 2016 at 16:39
  • 1
    \$\begingroup\$ I think you know the resonant frequency formula from what you say in your question. I've added a pretty picture to confirm this. \$\endgroup\$
    – Andy aka
    Commented Nov 18, 2016 at 16:40
  • \$\begingroup\$ Thanks a lot! Can I say that since the equation: Omega=1/(underroot of LC) has no resistance therefore omega is not dependent on R? \$\endgroup\$ Commented Nov 18, 2016 at 16:43
  • \$\begingroup\$ Yes you can but this is only true of some RLC resonant circuits and this happens to be one of them. \$\endgroup\$
    – Andy aka
    Commented Nov 18, 2016 at 16:44
  • \$\begingroup\$ Okay this helps a lot, thanks for your help Andy! Cheers!! \$\endgroup\$ Commented Nov 18, 2016 at 16:46

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