A series RLC circuit has a resonant frequency f0. If I apply a sinusoid at frequency f0 to this series RLC circuit, the output of the RLC should be a oscillation at frequency f0 with an amplitude higher than the sinusoidal input amplitude.

However, a property of a series RLC circuit is that, at resonance, the RLC appears as purely resistive, the reactive components cancel.

So, how is it possible that the output of the RLC can output a higher amplitude than the input sinusoid if the circuit appears purely resistive?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ if you put it that way, it sounds like a good question 😉 I am not yet sure if it really is, or if that wording merely leads to some otherwise very obvious oversight. \$\endgroup\$
    – tobalt
    Jun 11, 2022 at 17:39
  • \$\begingroup\$ Why shouldn't it amplify? Other passive circuits amplify - transformer for example. \$\endgroup\$
    – Chu
    Jun 11, 2022 at 17:50
  • \$\begingroup\$ @Chu At resonance, the reactive components of the inductor and capacitor cancel each other out. If I subject an RLC circuit to a sinusoidal signal with frequency equal to the resonant frequency of the RLC, the input impedance seen by that source is purely resistive. So my question is : If the input impedance is purely resistive at the resonant frequency, how on earth can it amplify? \$\endgroup\$ Jun 11, 2022 at 17:57
  • \$\begingroup\$ Depends where you take the output - draw your circuit to remove ambiguity. \$\endgroup\$
    – Andy aka
    Jun 11, 2022 at 18:10
  • \$\begingroup\$ @Andyaka just added \$\endgroup\$ Jun 11, 2022 at 18:15

5 Answers 5


Suppose a high-quality LC series circuit is driven by a function generator having internal 50 ohm resistance:


simulate this circuit – Schematic created using CircuitLab

At resonance (1000 Hz) capacitive reactance cancels inductive reactance so that the function generator sees a load very near zero ohms. If function generator amplitude was 1V, then only R1 limits current to 20 mA. This current is in phase with V1.

But this 20 mA must flow through the inductor, whose reactance is 159.153 ohms. So a voltage appears across the inductor whose amplitude is 3.183 volts, over three times higher than the open-loop function generator's voltage. Seems like you're getting some kind of gain, but you should recognize that current and voltage at L1 have a 90-degree phase relationship, so the only real power is dissipated in R1.


The essence of the series LC is that they exchange energy with each other. If you don't allow yourself to "look inside the box" then the series LC looks like a low impedance (zero, if ideal) at their resonant frequency. However, if you do look inside the box (the middle node where you pick off your choice to examine in your schematic) then you can see that this particular interior node is "dancing around" a lot, but where the voltages across the inductor and the capacitor are roughly equal and oppositely arranged to each other so that the voltage sum across the two appears to be close to zero.

What this kind of reminds me of is an old "thread & button spinner" toy I have played with from time to time. You can see it operate on this youtube video (I've specified it to start about where you can see it working well.) Only a little energy is being supplied by the slight pulling needed by the operator's hands every cycle, while the button rotates so fast you can hear the wind whistling by. The motion seen from both ends of the toy is modest. But the motion in the center is wild and fast!

Let's set this up in a simple LTspice schematic:

enter image description here

I'll explain the equations shown there and then get to the results.

The KCL is a little bit annoying, at first:

$$\begin{align*} \frac1{L_1}\int V_{_\text{OUT}}\:\text{d}t + C_1\frac{\text{d}}{\text{d}t} V_{_\text{OUT}} &= \frac1{L_1}\int V_{_\text{GND}}\:\text{d}t \\\\ \frac{V_{_\text{GND}}}{R_1} + \frac1{L_1}\int V_{_\text{GND}}\:\text{d}t &= \frac{V_{_\text{IN}}}{R_1} + \frac1{L_1}\int V_{_\text{OUT}}\:\text{d}t \end{align*}$$

But if I take derivatives, solve and substitute a bit, and finally move things around I will wind up with:


By substitution of \$Z=V_{_\text{OUT}}^{\quad'}=\frac{\text{d}}{\text{d}t}V_{_\text{OUT}}\$ I find that the characteristic equation is \$Z^{''}+\frac{R}{L}Z^{'}+\frac1{LC}Z\$ with \$\omega_{_0}=\frac1{\sqrt{LC}}\$ and \$\zeta=\frac{R}2\sqrt{\frac{C}{L}}\$. Knowing that \$Q=\frac1{2\zeta}\$ I can find that if I know \$f_{_0}\$ and a desired \$Q\$, I can solve for the rest as shown in the above schematic.

I've assigned \$\tau_{_0}=\frac1{\omega_{_0}}\$. So \$L=\tau_{_0}\cdot Q\cdot R\$ and \$C=\frac{\tau_{_0}}{Q\,\cdot\, R}\$.

If I assign \$Q=1\$ then this is what I get:

enter image description here

Note that \$V_{_\text{GND}}\$ stays very close to ground. It doesn't move around that much. But now also note that \$V_{_\text{OUT}}\$ (the green trace, which is the capacitor voltage) and that \$V_{_\text{GND}}-V_{_\text{OUT}}\$ (the bluish trace, which is the inductor voltage) are moving almost exactly oppositely to each other and, broadly speaking, sum up close to zero all the time. Finally, note that all the magnitudes, including \$V_{_\text{IN}}\$, are the same.

Let's change to \$Q=2\$:

enter image description here

Note that \$V_{_\text{GND}}\$, while still close to ground, is swinging around a little bit more. Note that the peaks for the capacitor and inductor voltages are now twice as large as the input voltage peaks and still act in a way that their sum remains close to zero.

Let's try \$Q=5\$:

enter image description here

And now you can see that the peaks for the capacitor and inductor voltages are now five times as large as the input voltage peaks and yet still always sum to near zero. (\$V_{_\text{GND}}\$ has somewhat yet-larger swings.)

The key things to notice are that the voltage drop across the resistor remains very close to the source voltage and that the voltage drop across the LC pair remains very close to zero. What's happening "inside the box" is that energy is being transferred back and forth between the inductor and capacitor and that the rate at which the supply voltage itself is changing is just exactly what is needed so that the summed voltage across the inductor and capacitor remains close to zero throughout a cycle. If the source frequency were to be any higher, or any lower, then the timing of the exchange of energy between the inductor and capacitor would no longer match up so well and the effect would be lost.


The L and C cancel one another because the voltages across them are 180 degrees out of phase. The voltages don't both go to 0V.


The key here is to look at the waveforms or the phasor diagram.

enter image description here Image source: Waveforms for Series AC Circuits

In an ac circuit, current (red) I is in phase with the voltage drop across the resistor (blue) \$V_R\$. I lags the voltage drop across the ideal inductor (brown) \$V_L\$. I leads the voltage drop across the capacitor (purple) \$V_C\$. These are shown in the phasor diagram on the left. The \$V_L\$ phasor is the same magnitude and opposite in polarity to the \$V_C\$ phasor.

At resonance, they cancel out and the circuit becomes resistive. Similarily, the purple waveform added to the brown waveform will be 0.

Power waveform is shown in green. At resonance, all the power provided by source goes to the resistor.

When a capacitor AND an inductor are in a circuit, the capacitor supplies leading reactive power to meet the needs of lagging reactive power of the inductor. Essentially, the capacitor acts as a power source for the inductor, decreasing the load on the source. This is the basis of resonance or power factor correction.

Individually, the source would have to supply reactive power to inductors OR capacitors. And reactive power would go back and forth between the devices and the source.

But the collapsing electric field of capacitor creates a magnetic field in inductor and vice versa. \$V_C\$ and \$V_L\$ can be significantly larger than the source voltage \$V_S\$, but you are not getting something for nothing, but rather taking advantage of the fundamental relationships between current and the voltage for the circuit components.

Resonance \$f_R\$ is defined as inductive reactance \$X_L\$ equal to capacitive reactance \$X_C\$. Above \$f_R\$, \$X_L > X_C\$ and the circuit is inductive. Below \$f_R\$, \$X_C > X_L\$ and the circuit is capacitive. The circuit has real power P and net reactive power the source has to provide. The capacitor still supplies the inductor, but extra reactive power goes between source and device. Extra leading reactive power probably will not go to the source but find a closer lagging load.

At resonance \$X_C = X_L\$ and all of the power required for the inductor comes from the capacitor and all the power provided by the source goes to the resistor. Power factor = 1.


The weakness of formal approach

The OP's question shows what absurdities formal methods can lead to when they are used to explain circuit phenomena. They give superficial explanations that are easy to remember and reproduce but do not reveal the essence of the phenomenon; clear and simple intuitive explanations are needed for this purpose.

This answer can be seen as a continuation of my yesterday's answer to almost the same question, but here I have tried to present my ideas more systematically.

The "rechargeable source" concept

In LC electrical circuits, such a powerful concept is a "rechargeable source" ("battery"). From this point of view, a capacitor can be thought of as a "rechargeable voltage source" and a coil as a "rechargeable current source".

What is an LC circuit?

So the LC circuit itself consists of two such "sources" connected to each other. They are recharged constantly switching roles - one is charging and the other charged. If there are no losses in the circuit, this process should go on forever and the energy initially put in will not change. Note that the current through and the voltage across the two elements are the same.


simulate this circuit – Schematic created using CircuitLab

Inserting an input source

If we decide to interfere with the operation of this closed system to help it, we can do it in two ways.

Current source in parallel

First we can help the "current source" (coil) by connecting an external current source in parallel. It must work in sync with the internal "source" to help it, not hinder it; so it must have the same frequency and phase.


simulate this circuit

But if the LC circuit is ideal (lossless), there is nothing to help it - the two elements serve each other and the current flows internally between them. The portions of energy injected by the current source during each cycle are accumulated in the elements and the voltage amplitude continuously grows. Thus the parallel resonance "amplifies".

LC tank resonance current input

Series resonance in RLC circuit frequency domain

Voltage source in series

Then we can help the "voltage source" (capacitor) by connecting an external voltage source in series. As above, it must have the same frequency and phase as the internal "source".


simulate this circuit

Now the portions of energy added by the voltage source during each cycle are accumulated in the elements and the voltage amplitude continuously grows as above. Thus the series resonance "amplifies".

LC tank resonance series input

Added resistor in series

Let's now "unfold" the series LC circuit as it is normally drawn - with the input source on the left and a connected resistor on the right. Thus, the input and output are connected through the LC network. What can we say about this arrangement?


simulate this circuit

This is a network of two "sources" in series but one of them is charging (i.e., it is really a source) while the other is being charged (i.e., it is actually a load). So a source and a load are connected in series, and their equal voltages add up (as above, both the current through and the voltage across the two elements are the same). But one is a voltage (emf) and the other is a drop. So they are subtracted and the voltage between the input (1) and output (3) is zero.

Series resonance in RLC circuit_1

As a result, the input voltage is transferred unchanged to the output resistor.

Series resonance in RLC circuit_2

  • 1
    \$\begingroup\$ It might be worth showing how it does not in fact amplify at all: the average input and output power are equal ("output" being all losses i.e. resistors), thus these can serve as impedance transformers. \$\endgroup\$ Apr 12, 2023 at 9:49
  • \$\begingroup\$ @Tim Williams, It reminds me of a voltage multiplier... \$\endgroup\$ Apr 12, 2023 at 10:06

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