(RLC in series) What is the R Voltage when at resonance?

In summary: You are correct that the impedance of the circuit at resonance is equal to R. It sounds as if you measured the supply voltage when you started, with nothing connected. As R will have a low value, I expect it is loading the signal generator and you are noticing the volt drop from the internal resistance. I suggest measuring VRLC with the circuit connected. The wider bandwidth which you are noticing may also be caused by the internal resistance of the signal generator being in series with the tuned circuit. Also notice that L will have some resistance. If you want to find the total resistance in the circuit, you might be able to measure both voltage and current and use Ohm's Law.The value of R was measured 996 Ω, also C=
  • #1
Gjagur
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We made a RLC circuit in the lab and took some values of R and LC Voltage while we changed the frequency.

So the experimental data seem to suggest that at resonance (VLC=18 mV : min) the Voltage of the resistor is 834 mV. But the initial voltage given was measured 1.426 V (All Values rms).

Since at resonance it is Xim=0 so Xall=R shouldn't all the Voltage drop in the resistance, why is it less?

It's been bothering me for a while and by using the relevant equations I can't seem to figure out why does that happen. Why isn't even the total voltage V2=(VR2+VLC2)1/2 the same as the initial one. I also calculate bandwidth (R/L) wrong (by about 50%).

Did something went wrong in the experiment or do I not understand something?
 
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  • #2
Gjagur said:
We made a RLC circuit in the lab and took some values of R and LC Voltage while we changed the frequency.

So the experimental data seem to suggest that at resonance (VLC=18 mV : min) the Voltage of the resistor is 834 mV. But the initial voltage given was measured 1.426 V (All Values rms).

Since at resonance it is Xim=0 so Xall=R shouldn't all the Voltage drop in the resistance, why is it less?

It's been bothering me for a while and by using the relevant equations I can't seem to figure out why does that happen. Why isn't even the total voltage V2=(VR2+VLC2)1/2 the same as the initial one. I also calculate bandwidth (R/L) wrong (by about 50%).

Did something went wrong in the experiment or do I not understand something?
Is the value of R small? The source resistance might be adding to the circuit resistance.
 
  • #3
You are correct that the impedance of the circuit at resonance is equal to R. It sounds as if you measured the supply voltage when you started, with nothing connected. As R will have a low value, I expect it is loading the signal generator and you are noticing the volt drop from the internal resistance. I suggest measuring VRLC with the circuit connected. The wider bandwidth which you are noticing may also be caused by the internal resistance of the signal generator being in series with the tuned circuit. Also notice that L will have some resistance. If you want to find the total resistance in the circuit, you might be able to measure both voltage and current and use Ohm's Law.
 
  • #4
The value of R was measured 996 Ω, also C=33nF and L = 10mH (nominal values). I had tried to find if the internal resistances were the cause but I calculated that there needed to be about 700 Ω of internal resistance and I find that pretty big. Also the internal resistance of L should use at resonant frequency 18mV at most (no?).

VLC was calculated at 1kHz as 1.316 V (VR=287mV) So is the total Voltage changing (assuming that the phase difference of R and LC is π/2)? Also I remember in the oscilloscope (connected to R and the source) that at resonant frequency the two graphs were completely identical, so the VRLC was almost equal to the VR.
 
  • #5
Can you adjust the circuit for resonance than measure Vs, Vr, Vl and Vc and send us the figures.
 
  • #6
I can't right now buy I had taken notes of VL=464 mV, VC=462 mV. As for Vs I guess according to the oscilloscope should be close to the VR. I might recreate the circuit and take more precise readings when I get the chance because I am really wandering what could go wrong.
 
  • #7
These numbers do not make sense, as Vl and Vc should be much greater than Vs. You may be having a problem using a CRO because one side is grounded, which might short circuit part of the circuit. I suggest using a multimeter to obtain the four voltages, assuming the frequency is low.
 
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  • #8
All the measurements were done with multimeter (although a bit hastily) the oscilloscope was only used for verification. I guess I have to look onto it again to make sure it wasn't a mistake. Thanks anyway for the answers I just wanted to know if it was something I didn't understand correctly/overlooked by me.

By the way should really the VC and VL be always greater than Vs? For example in the exact circuit that I am referring to, it is resonant frequency f=8761Hz therefore XL=Lω=550.5 and since Xall=R=103Ω, I=Ir=IL=V/R=1.426mA and so VL=I*XL=785mV.

Am I doing something wrong?
 
  • #9
I did not realize you were using such high R value, so that Q is only about 0.5. The circuit barely oscillates.
 
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1. What is RLC in series?

RLC in series refers to a circuit that contains a resistor (R), an inductor (L), and a capacitor (C) connected in a series. In this type of circuit, the components are connected end to end, with the same current flowing through each component.

2. What is resonance in RLC circuit?

Resonance in RLC circuit occurs when the inductive reactance (XL) and capacitive reactance (XC) are equal, resulting in a frequency at which the circuit has the highest impedance. This means that the voltage across the circuit is at its maximum and the current is at its minimum.

3. How do you calculate the R Voltage at resonance?

The R Voltage at resonance can be calculated using the formula VR = IZ, where VR is the voltage across the resistor, I is the current in the circuit, and Z is the total impedance of the circuit. The total impedance can be calculated using the formula Z = √(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance.

4. What happens to the R Voltage at resonance?

At resonance, the R Voltage is at its maximum value because the circuit has the highest impedance. This means that the voltage across the resistor is at its peak and the current is at its minimum, resulting in a high R Voltage.

5. How is R Voltage affected by changing the frequency in an RLC circuit?

The R Voltage in an RLC circuit is directly affected by the frequency. As the frequency increases, the inductive reactance (XL) also increases, while the capacitive reactance (XC) decreases. This results in a decrease in the total impedance (Z) and a decrease in the R Voltage. Conversely, as the frequency decreases, the R Voltage increases due to an increase in total impedance.

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