# Resonance in RLC Circuits

theuniverse

## Homework Statement

I did an experiment using an RLC circuit (in series) and an oscilloscope to determine the resonance frequency and amplitude (in volts) of my components (R total = 2300ohm, L = 0.081H, C = 1.5E-8F). I plotted all the data I collected (V vs. f) and got a nice response curve which looks like a bell curve with max amplitude of 21.5V @ 16kHz where my initial input voltage was 8V.

The problem:
I am asked to compare my experimental data to a theoretical calculation of both the resonance frequency and amplitude. But I have no idea what equations to use or how to find the values otherwise. Are there any equations that can help me with that?

Thanks!

Mentor
http://en.wikipedia.org/wiki/RLC_circuit" [Broken]

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theuniverse
Hey gneill!
Finding the resonance frequency looks easy enough 1/sqrt(LC)...

I found another formula meanwhile: A=(v0/C)/sqrt((R^2)(w^2)+L^2(w^2-(1/LC))^2). I tried differentiating it with respect to w (resonance frequency) after substituting all my variables in and substituting it into the original equation of A to find the amplitude...

However the results I'm getting for the amplitude do no produce anything that looks like a response curve and the values of amplitude range between ~8-10 as oppose to my experimental values which range from 5-23 (I've collected my data by repeating the experiment 5 times so I'm sure that my values are rather accurate). Any ideas?

Mentor
Are you familiar with the concept of impedance? (Sort of an "up scale" version of resistance).

Which component(s) were you taking the voltage measurements across for plotting?

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theuniverse
I saw impedance show up in my reading from time to time but I am not too familiar with it.
I attached the circuit diagram that I used... The measurements were taken across the capacitor.

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• circuit.jpg
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Mentor
Impedance is a sort of resistance that takes into account the effects that reactive components (like capacitors and resistors) have on the voltage and current in a circuit. It is measured in Ohms, just like resistance, but it is a complex number rather than a simple real number. This allows the impedance to reflect the fact that in a circuit driven by a sinusoidal source, the current will lead or lag the voltage waveform by some amount (phase shift) -- complex numbers allow the angular offset of the waveforms to be represented.

The impedance of a resistor is just the resistance value -- a straightforward real number. The impedance of capacitors and inductors is purely imaginary:

$$Z_C = \frac{1}{j \omega C}$$

$$Z_L = j \omega L$$

The impedance values can be treated just like resistances when analyzing a circuit. Of course, the rules of complex arithmetic apply, and often you are interested in the magnitude of a result rather than a complex value.

In your case you have a circuit that is essentially a voltage divider, with a resistor and inductor "on top" of a capacitor across which you're measuring an output voltage.

If your driving voltage is V (a sinusoid) then the voltage across the capacitor will be

$$V_o = V_{in} \frac{Z_C}{R + Z_L + Z_C}$$

Of course the result will be complex, reflecting the phase shift being produced by the reactive components.

When you measure with a meter or oscilloscope you're typically looking at the magnitude of the output voltage. So take the absolute value, and obtain:

$$|V_o| = V_{in} \frac{1}{\sqrt{R^2 \omega^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C + 1}}$$

This isn't quite the expression that you found.

Now, you specify that the input voltage was 8V, but presumably that was 8V RMS? That would give a peak-to-peak value of about 22.6V, which is close to your reported maximum. Were you measuring peak-to-peak on the oscilloscope for your results?

I don't see the "bell curve" you report coming out of this expression. A peak, yes, but I'd expect a relatively large flat response at low frequencies (since the inductor has a low impedance at low frequencies and the capacitor a high impedance there), followed by a peak and then a plunge in output as the capacitor "shorts out" higher frequencies and the inductor blocks them.

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theuniverse
Thanks for the explanation about the impedance!
Yea, I measured peak to peak to get my results. I tried plotting a theoretical curve using the formula for amplitude you provided but I am getting an exponential curve (20Hz-25000Hz) which gives me a much higher peak value than the one you found (22.6). I used excel with the following formula using the variables I listed originally.
8.6/SQRT((R*(w^2)*(C^2))+((w^4)*(L^2)*(C^2))-(2*(w^2)*(L)*(C))+1)
What can be the problem?

Edit: Re: the bell curve, I managed to fix it and the curve now looks like you described it. The theoretical curve is still an exponential though...

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Mentor
Thanks for the explanation about the impedance!
Yea, I measured peak to peak to get my results. I tried plotting a theoretical curve using the formula for amplitude you provided but I am getting an exponential curve (20Hz-25000Hz) which gives me a much higher peak value than the one you found (22.6). I used excel with the following formula using the variables I listed originally.
8.6/SQRT((R*(w^2)*(C^2))+((w^4)*(L^2)*(C^2))-(2*(w^2)*(L)*(C))+1)
What can be the problem?

Edit: Re: the bell curve, I managed to fix it and the curve now looks like you described it. The theoretical curve is still an exponential though...

I must apologize. When I "LaTexed" the formula I missed a square on the first R in the denominator. The formula should have read:

$$|V_o| = V_{in} \frac{1}{\sqrt{R^2 \omega^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C + 1}}$$

I've fixed it in the original post, too.

Again, sorry for the typo.

theuniverse
I tried the new formula and although the curve starts to resemble my experimental one, my peak value is only 9.9V while the lowest value is 8.6V. For which frequency did you find the peak value to be 22.6V?

Mentor
I tried the new formula and although the curve starts to resemble my experimental one, my peak value is only 9.9V while the lowest value is 8.6V. For which frequency did you find the peak value to be 22.6V?

Actually, my curve's peak was a bit higher than that, around 26.3V. Since the supply was 8V RMS (how accurate is that number?), that makes the peak-to-peak driving voltage 2√2 times that, or about 23 V.

The peak occurred around f = 3260 Hz, or ω = 20490 rad/sec.

theuniverse
What program did you use to find the values and plot them?
I tried maple and excel but I can't get the anything above 10V for the peak...

Mentor
What program did you use to find the values and plot them?
I tried maple and excel but I can't get the anything above 10V for the peak...

I use Mathcad, a commercial program. Yes, I'm spoiled.

But Excel should be able to handle it fine. You can do semilog plots with Excel easily enough.

theuniverse
I noticed that when I put the equation and calculate the values the term under the square root always equals to ~1 so that most of my terms end up being very close to the numerator term (ie. Vin). looks like all the other values are too small to affect the 1 under the square root in the equation. I think I tried every variation of the equation by now and it still doesn't work.
There is also barely any difference if I use "w" in Hz or rad/s since the number under the square root still tends to ~1...

Mentor
I noticed that when I put the equation and calculate the values the term under the square root always equals to ~1 so that most of my terms end up being very close to the numerator term (ie. Vin). looks like all the other values are too small to affect the 1 under the square root in the equation. I think I tried every variation of the equation by now and it still doesn't work.

What frequency range are you plotting?

theuniverse
20Hz - 25000Hz since that's pretty much what I used in my experiment.

Mentor
20Hz - 25000Hz since that's pretty much what I used in my experiment.

And that's Hz, not radians per second (ω) ?

theuniverse
I converted the frequency into radians. I attached my excel file, I think it'll be easier to see what I did that way.

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• Book1.xls
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Mentor
I don't see a problem there. You've plotted RMS values based upon your 8V RMS input. You were recording P-P volts though, right?

theuniverse
Yea I was recording peak to peak. Am I missing something here?
Edit: I think I see it now... do I have to convert from RMS to p-p (RMS*2*SQRT(2))?

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Mentor
Yea I was recording peak to peak. Am I missing something here?

8V RMS has a p-p amplitude of about 22.6V.

I'll trade you an excel workbook; attached is one that I put together just before you sent yours!

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• RLC.xls
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theuniverse
Yea it all makes sense now. I was wondering though, why did my experimental curve started from a much lower point (~5V) while the theoretical starts almost near the peak? Both graphs have the same scale and cover almost the same frequencies. Also the resonance frequency is around 4kHz as oppose to the 15kHz in my experimental curve. Looks like a very large error compared to what it should have been.

Mentor
Yea it all makes sense now. I was wondering though, why did my experimental curve started from a much lower point (~5V) while the theoretical starts almost near the peak? Both graphs have the same scale and cover almost the same frequencies. Also the resonance frequency is around 4kHz as oppose to the 15kHz in my experimental curve. Looks like a very large error compared to what it should have been.

I can't say for sure why the difference. It could be that one or more part values were not as they seemed, or that the coil had some internal resistance, or the capacitor had some leakage resistance, or the supply voltage didn't remain steady over the frequency range.

You might note also that the peak doesn't occur right at the natural frequency ωo, but at the damped frequency, ωd (you'll find it on the wiki page).

EDIT: Actually, the peak occurs before the damped frequency! Solving for the peak (differentiating the transfer function w.r.t. ω , setting to zero, blah, blah,... yields

$$\omega_{peak} = \omega_o \sqrt{1 - \frac{1}{2}\omega_o^2 R^2 C^2}$$

in this case about 3.3 kHz.

P.S. Are you sure that the capacitance was 1.5 x 10-8 F and not 1.5 nF or 1500pF or .0015 µF?

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theuniverse
Yea I see what you mean. I appreciate the help gneill! Thanks.

Mentor
theuniverse, I made an edit in my previous post that might be worth looking at.

theuniverse
If it is 0.0015 µF? then how would it affect the result?

Mentor
If it is 0.0015 µF? then how would it affect the result?

Well, then your capacitance would be 1.5 x 10-9 F rather than 1.5 x 10-8 F, and your peak would shift in frequency accordingly, perhaps enough to make sense of your observed data...

theuniverse
When I tried changing the value in the excel file you sent me the curve seemed to get even further from the experimental curve. But yes, the capacitance was supposed to be 0.0015 µF as I found out from a few classmates. However, I measured it to be 0.015 µF... so I am not too sure what to do with the results now.

Mentor
What instrument did you use to measure the output voltage?

theuniverse
Just a regular oscilloscope for the circuit and a volt meter for the capacitance.
However, assuming that I read it incorrectly, why didn't the equation provided before made the curve look closer to the experimental values?

Mentor
Well, it's a puzzler. If the capacitance turned out to be 0.0015 µF then the peak would occur around 14.4 kHz, closer to your observed 16 kHz. But the peak voltage would go through the roof -- the circuit would have a higher Q factor and a higher peak, upwards of 70V peak to peak for an 8V RMS input. Unless, of course, your input voltage was also measured with the scope and it was an 8Vp-p voltage. Then the peak output would be in the neighborhood of 26Vp-p.

theuniverse
I'm trying to add this part to my lab now. I now assume that 0.0015 µF and 8V peak to peak rather than RMS. I don't see however how you find the values, because when I plug the new variable values in the previous equation I don't get what you posted.

Mentor
Hmm. If you go to the Excel workbook that I posted and change the capacitance and voltage, how does the plot look?

theuniverse
It works fine now, was changing the wrong variable. Definetly one of the more confusing labs I did... but I learned a lot from it mainly thanks to you!

Mentor
No problem. I enjoyed it, too.

Good luck.