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https://www.eeeguide.com/voltage-and-current-in-series-resonant-circuit/


Here is the graph b/w V vs f. Why at some particular frequency (fc), the capacitor voltage goes beyond supply voltage (Vs) value? Explain for L too. ][1]

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    \$\begingroup\$ The article seems to explain it, with math. \$\endgroup\$ Nov 12 at 8:00
  • \$\begingroup\$ I'd assume that the OP has read the article, and that they need a different approach to understand the phenomena. \$\endgroup\$ Nov 12 at 8:35
  • \$\begingroup\$ Thanks Marko Gulin for your explanation. But I'll explain why I can't get this graph clearly again. When freq = 0 , capacitor will be a open circuit and max voltage (Vs) will appear across it. When freq = infinity, Capacitor will be short circuit. So Vc = 0V. But why the graph does not reduce directly from Vs to 0. Why there is a peak at fc. In resonance condition, it is understandable that, as Vc and VL are in out of phase, they would cancel out. So when f = fr (resonant freq) , Vc and VL can be higher . But when f = fc ( < fr), why capacitor voltage goes maximum? \$\endgroup\$
    – Vishwa
    Nov 12 at 10:35
  • \$\begingroup\$ The graph you posted is correct but exaggerated for most practical scenarios. For R=0 you have fC=fL=fr. For any other finite resistance R>0 the statement fC<fr<fL is correct, but not relevant for practical applications. Do the mathematics yourself and you will see why. \$\endgroup\$ Nov 12 at 17:08
  • \$\begingroup\$ @MarkoGulin I'd like to think OP has enough self-motivation and curiosity to plug in those formulas (even in a spreadsheet) and see exactly what happens, without the need for waiting hours or days for an answer. If the graph is poorly drawn then that should be all the more incentive. But maybe I'm just getting old and grumpy. \$\endgroup\$ Nov 12 at 18:23
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Why at some particular frequency (fc), the capacitor voltage goes beyond supply voltage (Vs) value? Explain for L too.

That graph is just plain wrong. The scaling is terrible and grossly non-linear.

For a series tuned circuit, the frequency that causes the voltage across the capacitor to become maximum is the same frequency that causes the inductor voltage to be maximum. Technically there is a very small difference but, the picture totally misguides the reader in the extent to which this difference appears to be.

This is just plain wrong: -

enter image description here

You'll never find any value of R, L and C that produces a graph that looks anything like this.

Here's a series tuned circuit at resonance: -

enter image description here

All three voltages peak at the same frequency. So, once you forget the incorrect diagram on the website you linked, the answer to your question is simple.

Why at some particular frequency (fc), the capacitor voltage goes beyond supply voltage (Vs) value?

At series resonance, inductive reactance and capacitive reactance values cancel out leaving just the resistance across the source. This means maximum current at this frequency and, the value of the current is Vs/R. You can then use that value of current to determine what the voltages are across either L or C. And what you find is that with small values of R, there is voltage magnification at resonance. Here's a web site (by me) that does the same: -

enter image description here

Note the capacitor voltage magnification of 20 dB (in red). This uses 1 Ω 100 μH and 1 μF as per the microcap simulation higher up. If I change the resistor from 1 Ω to 10 Ω, the magnification is less obvious: -

enter image description here

This is a fundamental aspect of tuned circuits in electronics.

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  • \$\begingroup\$ With resistance in the circuit, maximum voltage on the capacitor appears before and on the inductor after the resonant frequency \$\omega_r = \frac{1}{2\pi \sqrt{LC}}\$. I know I might be embarrassing myself, but I am posting this again because I double-checked the math and did some simulations. Considering your authority, I will triple-check the whole thing. \$\endgroup\$ Nov 12 at 16:43
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    \$\begingroup\$ @MarkoGulin technically, there is a tiny area in the spectrum where there'll be a small difference - but look at the graph in the picture; look at the shapes of the responses - the picture is giving the wrong impression entirely. On my picture above with R = 1 ohm, the frequency difference is hertz i.e. trivial and totally blown out of proportion by the picture in the question. \$\endgroup\$
    – Andy aka
    Nov 12 at 16:50
  • \$\begingroup\$ That is true, although in the OP's question/graph there is no mention of the frequency range.. It might well be 2 Hz as far as we know. Although in most practical applications your WRONG/WRONG/PLAIN WRONG is OK, strictly (mathematically) speaking the OP's graph is not wrong. \$\endgroup\$ Nov 12 at 17:04
  • \$\begingroup\$ The op's graph is wrong - it gives the totally wrong impression of scale - compared to a Picasso drawing of a guitar (which is bizarre), that graph is way off centre and unrecognizable regards scale and impression. It imparts nothing useful. \$\endgroup\$
    – Andy aka
    Nov 12 at 17:17
  • \$\begingroup\$ I've seen videos and lecture materials with the same exaggerated graph. For some reason they teach this at universities. Although I cannot find any practical example where it would be relevant, I still feel that saying it is wrong just adds confusion. You have every right to say it's wrong, but I look at it just as an artwork to make people remember that max does not appear at fr which is mathematically correct. Cheers! \$\endgroup\$ Nov 12 at 17:23
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I am going to split this answer into two parts: (1) how does the voltage and current depend on frequency, and (2) how can voltage on one component be higher than the source voltage.


PART 1.

The Ohm's law states that \$V = I \cdot R\$. This equation would be the same if voltage and current change with time: \$ v(t) = i(t) \cdot R\$. In other words, instantaneous voltage equals instantaneous current amplified by resistance. It does not matter if voltage or current are sine waves, square waves etc.

However, the above equation does not hold true for capacitors and inductors because they are energy storage devices. Now think for a moment in a discrete world - total energy in a battery equals energy from a previous step plus whatever we put into it now

$$ x(k) = x(k-1) + u(k) \Delta T $$

where \$ x \$ is energy, \$ u \$ is input, \$ k \$ is discrete time instant, and \$ \Delta T \$ is time step. In continuous world \$ \Delta T \rightarrow 0 \$ the above equation would be

$$ \frac{d}{dt}x(t) = u(t) $$

You are ready now for equations that describe voltage-current relationship for inductors and capacitors:

$$ v_L(t) = L \frac{d}{dt} i_L(t), \qquad i_C(t) = C \frac{d}{dt} v_C(t), $$

where L and C subscripts denote inductor and capacitor. Please note that although I am using here energy storage equivalence with simple mathematics, the above two equations actually come from physics.

Now try to imagine what happens with the inductor voltage when inductor current is a sine-wave:

$$ v_L(t) = L \frac{d}{dt} I_L \sin(\omega t) = \omega L \cdot I_L \cos(\omega t) $$

You can read this as: at some fixed frequency, inductor voltage equals inductor current phase shifted by \$90^\circ\$ and amplified by \$ Z_L = \omega L \$. If you compare this to Ohm's law, you see that impedance \$ Z_L \$ is equivalent to resistance \$ R \$, the only difference is that unlike resistance the impedance varies with frequency \$ \omega \$. Additionally, Ohm's law does not take into account the phase shift. You can use the same reasoning for capacitors.

We know now that voltage-current characteristics for inductors and capacitors is also linear, the only difference compared to simple resistors is that amplification depends on the frequency and phase shift should also be considered.


PART 2.

When you have an RLC circuit, you actually have two energy tanks (L and C) and one load (R). The energy tanks do not consume energy, ie the energy is still in the circuit and does not go away. The load consumes energy, and when I say consumes I mean that they convert electrical energy to some other form (heat etc.) and the energy is lost from the circuit point of view.

Imagine the two energy tanks as two connected water tanks, where water level (height) equals voltage. When you bounce some water from one tank to the other, water level changes in the tanks. If one tank is empty, the other one is at maximum. But! What if water level can go negative? This is exactly what happens with L and C:

  • there is some voltage source that generates sine-wave voltage of amplitude \$ V_S \$ at a fixed frequency \$ \omega_0 \$
  • there are two energy tanks, namely an inductor and a capacitor, and the voltage on them is also a sine-wave with amplitude \$ V_L \$ and \$ V_C \$
  • the energy bounces between these two energy tanks at a rate that depends on the frequency
  • unlike water level, voltages can go negative!

The graph that you posted shows how voltage amplitude on inductor and capacitor depends on the frequency. At one particular frequency, also known as the resonant frequency, this graph is at minimum or maximum. But, to have a resonant frequency, you need to have two energy tanks.

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The graph is poorly drawn, I agree, and, for my part, I don't know why they would bother adding properly formatted math equations but Paint-level drawings, instead of actual simulations. Oh well.

The point that the picture is trying to make is that the frequencies are never equal except at resonance. Their derivations show the formulas for both the capacitor and the inductor, and if you equate L=C=1 and R=x, then both formulas reduce to the following:

$$\begin{align} \omega_L&=\dfrac{1}{\sqrt{LC}}\sqrt{\dfrac{1}{1-\dfrac{R^2C}{2L}}}=\dfrac{1}{\sqrt{1\cdot 1}}\sqrt{\dfrac{1}{1-\dfrac{x^2\cdot 1}{2\cdot 1}}}=\sqrt{\dfrac{1}{1-\dfrac{x^2}{2}}} \tag{1} \\ \omega_C&=\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{2L}}=\sqrt{\dfrac{1}{1\cdot 1}-\dfrac{x^2}{2\cdot 1}}=\sqrt{1-\dfrac{x^2}{2}} \tag{2} \end{align}$$

If you plot these two on a log Y axis (representing the frequency, X axis being R), you'll get this graph:

test

Since x represents R, the negative side can be omitted. What's left tells you that when R=0 the two frequencies coincide, and when R!=0 the two differ, and they can differ by more than a decade; in fact, they can be at ±∞, if you take the limits. But their amplitudes will drop dramatically, in fact, even with R=1 the peaks are only about 1.25 dB. For R<1 the effect becomes more and more pronounced, but they also coincide more and more, converging towards resonance. The peaks, themselves, are inversely proportional to R, which represents the damping. No damping means infinite peak (resonance).

This can be easily tested by setting L=C=0.159 (1/2/π). Then, with R=1, a 1 V sine at the input will result in a 1 V across all elements. With R=0.5 the peak will be 2 V for L and C, and with R=0.1 there will be 10 V. The voltage across R will remain constant. What will take time, though, will be the time to reach the maximum voltage: the lower the R, the higher the quality factor, thus the longer the settling time.

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