During the last day or so, I have studied the topic of resonant frequency over a few circuits. Now, I could easily see that every time my circuit operated upon that resonant frequency something special happened. I'm aware that the deal with resonant frequency has to do something with oscillation but I couldn't figure out exactly what.

In a regular RLC circuit in series with an AC voltage source operating in resonant frequency I get that the impedance equals to the impedance of the resistor alone. So?? What does oscillation has to do with it? In the different example where we deal with RLC parallel circuit we get that for AC voltage source operating in resonant frequency I ended up with a pure imaginary resistance, what does it mean? In LC parallel circuit with an AC current source I ended up with infinite impedance, what is that have to do with anything?

So, in general I'm a bit confused with the whole resonant frequency making weird stuff on any different circuit, can someone please help me clarify this a bit..?

  • \$\begingroup\$ user3921 - have you got an answer you are satisfied with? If not then please raise a comment asking for what it is you need clarifying. \$\endgroup\$
    – Andy aka
    Feb 21, 2021 at 22:17

3 Answers 3


Im aware that the deal with Resonance frequency has to do something with oscillation but i couldn't figure out exactly what

Probably it's main use is in filters - because the impedance changes so great as a signal inputted passes through the resonant frequency, you can use this to make radios very selective in what they receive and largely block-out all the other stations. Because radios tend to use sinewaves as their primary oscillator you can also use resonance to help you get a cleaner sinewave. In fact many oscillators use an LC or RLC circuit so that a clean and well-defined (in terms of frequency) sinewave is produced.

An industrial use is power factor correction - you have a lagging power factor due to high power induction motors and the electricity company bills you for reactive power taken - add the right capacitor in parallel with your induction motor and the current reduces by tens of percent usually - what is this miraculous cost saving technique - it's parallel resonant tuning aka power-factor correction.

So you have parallel and series resonant circuits - both exhibit large changes in impedance as the inputted signal passes through resonance - the series circuit reduces its impedance to just a few ohms and the parallel circuit increases its impedance to theoretically infinite and this is because inductors and capacitors take current differently.

In an inductor the current lags the voltage by 90 degrees and in a capacitor it leads by 90 degrees - in effect there is a 180 degree phase difference between the two currents and if the inputting voltage source is connected to a parallel LC circuit, at the resonant frequency the current taken by the inductor is totally cancelled by the current taken by the capacitor - the net effect is that no current is taken from the inputted signal. This means infinite impedance.


simulate this circuit – Schematic created using CircuitLab

The current flowing thru the capacitor is always opposite (but equal in magnitude) to the current in the inductor at resonance so, if you analysed the current flowing from the signal generator it has to be zero. By the way I've chosen values that do work at 159.155 kHz.

With series circuits, the L and C share the same current so the individual voltages are forced to be 180 degrees apart and it's like two 9 volt batteries - put them in series and the voltage is 18 volt but put them in series opposition and the voltage is zero. An L and a C in series at resonance produce no net voltage across them - this means that current is flowing due only to the other component, the series resistor. Impedance = R.


simulate this circuit

And if it's still a little confusing ask yourself what the impedance of two resistors in series is BUT, imagine one was positive 10 ohms and the other was negative 10 ohms - the answer is zero ohms. Now think about them in parallel - the current drawn by one is equal and opposite to the current drawn by the other hence the impedance is infinite.


yes. when your inductor and capaitor reactances equal out depending on if they are in series or shunt your circuit will see them as short(your circuit will only see pure resistance of that rezistor) or open(your circuit will only see infinite imaginary impedance) at that perticular resonance frequencies, respectively. that way thry are used in circuit as a notch reject or notch pass, in other words to filter one perticular frequency or reject/amplify perticular frequencies by using LC circuit in series or shunt. Here is a little excercise for you! try computing impedance of LC circuit when they are in series and shunt and apply the angular frequency(omega) of 1/sqrt (LC) in your formula you will see that you get 0 and infinite in series and shunt respectively.


Regarding the relation between "resonant frequency" and oscillators:

1.) Oscillators need a passive and frequency-dependent circuit as well as an active elemenet (amplifier).

2.) These two parts are connected in a closed loop which must produce a gain of "1" around this loop (unity loop gain) at ONE SINGLE frequency only (which is the desired frequency).

3.) Hence, we need a frequency selective passive circuitry that for all other frequencies (except the desired one) allows only a gain BELOW unity.

4.) For this purpose, an RLC series or parallel combination (bandpass) can be used. (But note, this is only one alternative for designing harmonic oscillators).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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