While going through a book, I came across a statement "The magnitude of transfer function is minimum during resonance". What I did not understand is that transfer function means a ration of output to input, so across which terminal is the output taken? Does it not matter, that is, is it minimum irrespective of the component across which we take output?

  • \$\begingroup\$ without a picture who knows \$\endgroup\$
    – Andy aka
    Feb 4, 2017 at 12:22
  • \$\begingroup\$ Also, you might not be aware that on questions receiving good answers, the normal practise is to "formally-accept" what you see is your preferred answer. Call it a fee for getting good information. \$\endgroup\$
    – Andy aka
    Feb 4, 2017 at 14:07
  • \$\begingroup\$ Oh, I am new and wasn't aware of it. How do we do it? Is there an accept button or is it just a formal acknowledgement through commenting? \$\endgroup\$ Feb 5, 2017 at 20:11

2 Answers 2


It depends if RLC series resonator is put in series or shunt in a circuit.

At series resonance \$X_L = -X_C\$

This means that those impedances cancel leaving just the series resistance, R at resonance. Then it becomes an R ratio divider with whatever else is at source or load.

If RLC is a series resonator across Rs source impedance, it becomes a notch filter with R ratio defining depth of notch.


simulate this circuit – Schematic created using CircuitLab

The quality of L or C components is usually rated by ratio of X/R ratio = Q at some high frequency where Q= 100 is high, <10 is low.


In real world terms the statement means that the impedance of the series RLC network as a whole becomes minimum at the resonant frequency. Therefore, if you were to measure the current this series RLC draws from a signal generator, you would find it to be quite large relative to other lower and higher frequencies.

If you look across each element of the RLC you would find that the L and C have quite large voltage drops, and are operating 180 degrees out-of-phase. The voltage across L lags the phase of the input driving voltage by 90 degrees, the capacitor leads it by 90, so the net difference is 180 degrees between the these two elements.

If the R is significant in the circuit,(which it usually is NOT because it merely represents incidental heat losses in the C & L), and it is actually a discreet element so you can measure its voltage drop, you would find its voltage and current to be in-phase with the driving voltage and driving current.

I hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.