2
\$\begingroup\$

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

\$\endgroup\$
4
  • 1
    \$\begingroup\$ What were the actual measured frequencies? \$\endgroup\$ – user4574 Jul 9 '19 at 20:50
  • \$\begingroup\$ @user4574: I edited the question with this info. \$\endgroup\$ – harish2704 Jul 9 '19 at 20:55
  • \$\begingroup\$ Is the inductance known or are you trying to calculate it too? \$\endgroup\$ – Harry Svensson Jul 9 '19 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 '19 at 21:02
2
\$\begingroup\$

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

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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