Hoping this isn't too homework-like, but I'm missing some sort of concept here. Doing a lab, and I have to use previous data to predict the voltage across the 2nd and 3rd elements (e.g. order is LCR, predict voltage across CR). We've measured all the data using an oscilloscope. We put the order of the circuit in series and in all 6 different arrangements and measured the voltages across the second and third elements ($ \triangle V_{23}$).

Basically, what we are given (in other words, what we measured) is:

$ \triangle V_R$, $ \triangle V_L$, $ \triangle V_C$, $ \triangle V_{max}$ as well as $R, L, C$ values

These were measured with the oscilloscope by looking at the amplitude of the waves of each element.

The thing is that we have to find the predicted measurement of $ \triangle V_{23}$ in all the different orders as well. This is the part I understand the least.

My main question is: How do I find the formulas to find $ \triangle V_{23}$? I would have assumed that it would just be the current (rms or max, not sure) multiplied by the addition of the resistance/reactances (e.g. $V = I_{rms}*(X _L + X_C)$ for RLC or RCL), but clearly this is not true because of the fact that my measurements indicate that the order RLC gives 0.3V but RCL gives 0.5V, despite the last two elements being the same.

Why exactly does that difference matter? Must be something basic I'm missing. Has to do with the phasor diagram, right? I just don't get the concept. Sorry if this is too long and homework-like but this is just weird to me.

  • $\begingroup$ With order LC and CL there should be no difference. What was the accuracy of you voltage readings of 0.3 V and 0.5 V? On the oscilloscope input and the signal generator output was there a terminal which was grounded/earthed? $\endgroup$ – Farcher May 9 '17 at 5:52
  • $\begingroup$ That's strange then, but it's reassuring since I was completely lost before. I'm not sure what the problem is then. It's quite possible I got some results mixed up somehow. I'm pretty sure there was a terminal that was grounded. Considering we did something wrong, I think the only thing I can do to fix this is find what I did wrong using the formulas I need to find V across elements 2 and 3. Does it just come down to subtracting the the voltage across the 1st element from the total voltage or something? $\endgroup$ – hhh May 9 '17 at 6:32

The phasor diagram does not depend on the order of components in the circuit even if the inductor has resistance so measuring different voltages when the order of two of the components is reversed is unexpected.

enter image description here

There can be a problem with trying to interpret experimental results if one does not have access to the experimental set up and data.

If the resistance of the inductor is much less than the reactance of the inductor then the difference between the voltages across the inductor and the capacitor should equal the voltage across the inductor and the capacitor.

That is where you may have introduced a large source of error particularly if you are subtracting two large numbers and obtaining a small number?

As an example suppose that $V_{\rm C} = 4.3 \pm 0.1\, \rm V$ which is a $2\%$ error and $V_{\rm L} = 4.8 \pm 0.1\, \rm V$ which again is approximately a $2\%$ error.

When you take the difference $V_{\rm L} - V_{\rm C} = 4.8-4.3 = 0.5 \,\rm V$, the difference will have a "maximum" error of approximately $\pm 0.2 \,\rm v$ which is a much larger error of $40\%$.

The problem with an output from the signal generator and an input to the oscilloscope being grounded/earthed is illustrated below.

enter image description here

In the right hand diagram the capacitor has a short across it and so is no longer part of the circuit.
This will change the impedance of the whole circuit and hence the current passing though the components and the voltages across the components.

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