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schematic

simulate this circuit – Schematic created using CircuitLab

Edit: Impedance value below has been found incorrect and would apply to a purely parallel circuit instead. Question has been answered in below comment. Deleting unrelevent information from this post to give the below answers more relevance.

Question : How do I find Voltage across R at 1000Hz. I built/measured VR in this circuit and VR should be about 2.2V at 1000Hz.

I have calculated:

XL = 20.734 Ohms

XC = 15915.494 Ohms

Z = 18.9908 Ohms < Z found to be incorrect. Now given correctly in comment below

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  • \$\begingroup\$ Your calculation are wrong. For F = 1kHz we have Xc = 16KOhms and XL= 20.7Ohms , and since Xc and XL are connected in parallel so we have Z = (Xc * XL)/(Xc - XL) = 20.7 Ohms and Ztotal is \$Ztot = \sqrt{Z^2 +R^2} = \sqrt{20.7^2+47^2} = 51.35\$ and Vout = 2.5V/51.35 Ohms * 47 Ohms = 2.28V \$\endgroup\$
    – G36
    Commented Jun 5, 2016 at 14:38
  • \$\begingroup\$ Yes that sounds correct. I believe I originally had those values but got all mixed up. Values I posted are for a purely parallel setup where the resistor has its own parallel path next to inductor and capacitor. This clears up my error. If you wish to post above as answer I will mark as solved. \$\endgroup\$ Commented Jun 5, 2016 at 15:30

1 Answer 1

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Let's take this slowly:

Ok, we know the that the complex impedance of a capacitor is

\$Z_C = \frac1{j\omega C}\$

and the of an inductor

\$Z_L = {j\omega L}\$,

with \$\omega=2\pi f\$.

Inserting values: \begin{align} Z_C &= \frac1{j2\pi f C}\\ Z_L &= {j2\pi f L}\\ Z_{L||C} &= \frac1{\frac1{Z_C}+\frac1{Z_L}}\\ &= \frac1{j2\pi f C+\frac1{j2\pi f L}}&\text{extending elegantly yields}\\ &= \frac{j2\pi f C-\frac1{j2\pi f L}}{\left({j2\pi f C+\frac1{j2\pi f L}}\right)\left({j2\pi f C-\frac1{j2\pi f L}}\right)}\\ &= \frac{j2\pi f C-\frac1{j2\pi f L}}{\left({j2\pi f C}\right)^2-\left({\frac1{j2\pi f L}}\right)^2}\\ &= \frac{j2\pi f C+\frac j{2\pi f L}}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} } \\ &= \frac{j\left(2\pi f C+\frac 1{2\pi f L}\right)}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} }\\ &= j\frac{2\pi f C+\frac 1{2\pi f L}}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} } \end{align}

As you can see, the complex value of that sub-circuit is purely imaginary!

Now, do the usual voltage divider calculation for the voltage across R1, and you will find the voltage drop as a function of frequency \$f\$.

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  • \$\begingroup\$ Thanks for effort, but WAY over my head. I need VR at 1000hz. I then need to calculate VR at about 20 differet frequencies. Is there no way to calculate VR? I measure VR at about 2.2V. I know some like to prove formula, but I need to see correct forumla = VR @ 1000Hz. Then usually I will understand why it is so after seeing actual value. \$\endgroup\$ Commented Jun 5, 2016 at 14:47
  • \$\begingroup\$ if you need to calculate that, it can't be over your head, because, well, that's what you'll need to understand about circuit analysis if you want to analyze circuits. \$\endgroup\$ Commented Jun 5, 2016 at 14:49
  • \$\begingroup\$ Also, the formula for the resistance of the parallel C||L is given above. Just plug in your values, and then do a complex voltage divider with R; I don't see your point :) \$\endgroup\$ Commented Jun 5, 2016 at 14:50

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