I was trying to find capacitance of two capacitors in the following way.

  • Connect each capacitor individually to a same inductor and find resonant frequency in each case.
  • Connect the same capacitors in series , to the same inductor and find the resonant frequency.
  • Then calculate the capacitance using following way

we have

\$f_{1} = \frac{1}{2 \pi \sqrt{L c_{1}}}\$ , \$f_{2} = \frac{1}{2 \pi \sqrt{L c_{2}}}\$, \$f_{3} = \frac{1}{2 \pi \sqrt{\frac{L \left(c_{1} + c_{2}\right)}{c_{1} c_{2}}}}\$

Solving these equations, we get

\$\left\{ L : \frac{1}{4 \pi^{2} f_{3} \sqrt{f_{1}^{2} + f_{2}^{2}}}, \ c_{1} : \frac{f_{3} \sqrt{f_{1}^{2} + f_{2}^{2}}}{f_{1}^{2}}, \ c_{2} : \frac{f_{3} \sqrt{f_{1}^{2} + f_{2}^{2}}}{f_{2}^{2}}\right\}\$

I tried to substitute the values for f1, f2 & f3 ( f1 = 2500, f2 = 2030, f3 = 3200 ) . Interestingly , I got a result which is numerically correct , but with a difference of \$1e^{-6}\$. That is, instead of micro farad, I get values in farad, Also, instead of millihenry, I get value in nanohenry

I was trying to figure out why this difference in occurred .

The completed IPython notebook can be found at https://gist.github.com/harish2704/5fe08c80c96307973a11f724a218950d

it will be a great help if someone can help me

  • 1
    \$\begingroup\$ What were the actual measured frequencies? \$\endgroup\$
    – user4574
    Jul 9, 2019 at 20:50
  • \$\begingroup\$ @user4574: I edited the question with this info. \$\endgroup\$
    – harish2704
    Jul 9, 2019 at 20:55
  • \$\begingroup\$ Is the inductance known or are you trying to calculate it too? \$\endgroup\$ Jul 9, 2019 at 20:59
  • \$\begingroup\$ @HarrySvensson: I know capacitance of one of the capacitance. Thus, we can say inductance is also a known value \$\endgroup\$
    – harish2704
    Jul 9, 2019 at 21:02

1 Answer 1


The formula for the series capacitors should be C1 * C2 / (C1 + C2). You have it upside down.


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