I am tasked with finding the current I through the following circuit at an array of frequencies. I have a solution however I am fairly new to AC systems and just want to make sure I am on the right track.

RLC Circuit

The values of \$ V_R, V_C, \$ and \$ V_L\$ were measured using an oscilloscope, and we can assume for the purpose of this question that they are 0.8 V, 3.8 V, and 5.6 V respectively.

Here is my solution assuming a frequency of 500 Hz and a voltage of 14.1 V peak-to-peak, also there is a correction that the capacitor is \$2.2 \ \mu F \$ not \$0.22 \ \mu F \$:

\$ I = \frac E Z \$

\$ \omega = 2\pi f \$

\$ Z = Z_R + Z_C + Z_L\$

\$ Z_R = 480 + j0 \ \Omega \$

\$ Z_C = 0 - \frac j {\omega C} \ \Omega = 0 - j144.7 \ \Omega \$

\$ Z_L = 88 + j {\omega L} \ \Omega = 88 + j314.2 \ \Omega \$

Summing the impedances we obtain an effective impedance of \$592.8 \ \angle \ 16.6^ \circ \ \Omega\$

Now we need the phase angle which can be found with the vector sum of the measured voltages.

\$ \theta = arctan(\frac {V_L - V_C} {V_R} ) = 66.0^\circ \$

With this the final answer for the current should be

\$ I = \frac {14.1 \ \angle \ 66.0^ \circ} {592.8 \ \angle \ 16.6^ \circ} \$

Which gives a final current of \$ I = 23.8 \ \angle \ 49.4^ \circ \ mA\$

Why might my measurements be different from the calculated values?

Edit: the measured voltages are peak-to-peak not RMS.

  • \$\begingroup\$ Thank you @Andy aka for pointing out that the measured voltages are not RMS \$\endgroup\$ – user104126 Mar 16 '17 at 16:05
  • \$\begingroup\$ You apply \$5V_{RMS}\$ = \$14.14V_{PP}\$ and you only have measured values of 0.8V, 3.8V, and 5.6V. If these scope readings are correct, you have something significantly wrong. \$\endgroup\$ – StainlessSteelRat Mar 16 '17 at 16:27
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    \$\begingroup\$ I’m voting to close this question because it's got fundamental errors in the question with no effort to correct by OP. Question and Answers have no value to this site. \$\endgroup\$ – StainlessSteelRat May 30 at 19:36

Have I made any mistakes here?

Editing your question after I made my answer is a really big mistake. I can see you are ignoring my request so I point out here that changing your mind from RMS to p-p does not alter the problem you have. Current I (as calculated by V/R or V/Xc) does not match what you have calculated and neither do they match each other. Now stop amending your question thus causing me to make amendments to my answer and wasting my time. I'm trying to help you yet all you are doing is wasting my time.

If VR is 0.8 V RMS and R has a resistance value of 480 ohms then the current I has a magnitude of 0.8/480 amps = 1.667 mA. Clearly your calculated current of 23.8 mA is a mile out.

Sanity check: 3.8 volts across Xc and Xc = 144.7 ohms hence I is 26.3 mA.

The only feasible conclusion is that you have made mistakes in posting the detail of the question or the question is misleading.

  • \$\begingroup\$ I apologize for not specifying, but it's 0.8V peak-to-peak not RMS \$\endgroup\$ – user104126 Mar 16 '17 at 15:42
  • \$\begingroup\$ That makes the current even smaller. Now please go and undo your correction to your question or make it absolutely plainly clear that this correction was made AFTER I made my answer. Of course you can be lazy and ignore what I strongly request but that wouldn't be playing ball. \$\endgroup\$ – Andy aka Mar 16 '17 at 15:44
  • \$\begingroup\$ Look, I'm new at this and I'm just trying to learn here. \$\endgroup\$ – user104126 Mar 16 '17 at 15:49
  • \$\begingroup\$ As for the "mistake" in posting the details, it's a lab experiment so the details were "Measure the AC voltage in these three locations, record the data in a table, and calculate the current I." My conclusion then is that I may have made a mistake using the oscilloscope to measure the values, however that is a fairly straight forward process so I'm not entirely sure how that could be. \$\endgroup\$ – user104126 Mar 16 '17 at 15:55
  • \$\begingroup\$ You had your window. \$\endgroup\$ – Andy aka Mar 16 '17 at 15:56

Have I made any mistakes here?

You have problems with your question.

\$5V_{RMS} = 14.14V_{PP}\$

I get effective impedance of \$592.8 \ \angle \ 17.6^ \circ \ \Omega\$.

Just dealing with magnitudes. \$ I_{PP} = \frac {14.14V} {592.8\Omega} = 23.8 mA_{PP}\$

So apply that to \$480\Omega\$, and you get \$4.05V_{RMS}\$ and \$11.45V_{PP}\$. No where close to 0.8 V, 3.8 V, and 5.6 V that you report.

\$V_{C_{RMS}}\$ = 1.22V and \$V_{C_{PP}}\$ = 3.45V. With experimental error, 3.8V is probably fine.

\$V_L \ne 5.6V\$ because of the \$ 88\Omega\$ resistor.

\$V_{L_{RMS}}\$ = 4.05V and \$V_{L_{PP}}\$ = 11.45V. The actual measured value across your real inductor would be 11.6V. Again no where close to 0.8 V, 3.8 V, and 5.6 V that you report.

So Yes, you have made some mistakes. I would suspect the incorrect frequency because:

an array of frequencies.

  • \$\begingroup\$ @StainlessSteelRet I think VL = 5.6V is OK because it already includes the 88Ω resistor. It would have been better to rename this voltage to VLR or VRL. \$\endgroup\$ – Elec1 May 30 at 7:24

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