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\$C = 1 \mu F\$

\$L = 2.2 mH\$

\$r = 1.7 \Omega\$

\$f = 4000 Hz\$

\$V_{IN} = 2.6 V\$

I have this RLC circuit where I need to find \$V_{OUT}\$. I have tried it by finding the \$Q\$ factor and then multiplying it with \$V_{IN}\$ which gave me the value \$84.56427 V\$. However, when putting that value in as an answer in the system I'm using, I don't get the right answer, even if I round it up to \$84.6 V\$ or \$85 V\$.

What am I doing wrong?

  • \$\begingroup\$ is Vin a DC source? \$\endgroup\$
    – Big6
    May 5, 2016 at 22:39
  • \$\begingroup\$ So what do you estimate to be the correct answer? Also, importantly, what is your question? \$\endgroup\$
    – Andy aka
    May 5, 2016 at 23:16
  • \$\begingroup\$ Remember units for calculations (Farads, Henry, Ohm) , you're probably forgetting to multiply the values by their factors of 10. \$\endgroup\$ May 6, 2016 at 9:42

1 Answer 1


You have a fixed frequency (4kHz), series RLC circuit with a constant peak Vin of 2.6V, so you can think of it as a voltage divider.

Work out your reactances at 4kHz and Vout will be proportional to Vin.

$$ X_C = \frac{1}{2\pi fC}\Omega = \frac{1}{2\pi(4000)(10^{-6})}\Omega$$ $$ X_L = {2\pi fL}\Omega = {2\pi(4000)(2.2\times10^{-3})}\Omega$$ $$ V_{OUT} = V_{IN}\frac{(X_L+R)}{(X_L+X_C+R)} $$

Hope this helps!

  • \$\begingroup\$ Hi, Thank you for answering. I got the value 1.11459 V, but that doesn't seem right either.. \$\endgroup\$
    – Vetenskap
    May 6, 2016 at 11:37
  • \$\begingroup\$ I mean 1.53108, either way it isn't right \$\endgroup\$
    – Vetenskap
    May 6, 2016 at 12:07
  • 1
    \$\begingroup\$ I understood what I did wrong, Thank you for your help! :) \$\endgroup\$
    – Vetenskap
    May 6, 2016 at 14:14
  • \$\begingroup\$ Gday mate, my equations are correct I crunched the numbers and got 1.1146. Why doesn't that seem correct? You have to think about it without any numbers. This is a passive circuit without any switching of any kind. A fixed frequency, we can only assume its a sine wave. The output cant be higher than the input in this situation. It has to be a ratio of the input and your input is only 2.6V. \$\endgroup\$
    – crowie
    May 6, 2016 at 14:20
  • \$\begingroup\$ @Vetenskap good on you for admitting your mistake. I personally think the best quality in an engineer is knowing when (and why) you are wrong. Its taken me many years of being wrong to know when I am right. Don't have any fear in making mistakes just as long as you admit when you are wrong. \$\endgroup\$
    – crowie
    May 6, 2016 at 14:27

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