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

  • \$\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


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\$.

  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.