I have the following circuit where I would like to obtain an analytic expressions for the value of \$Cs\$ and \$Rs\$ only kowning the phasors \$\bar Vina\$ and \$\bar Vinb\$ there displayed, I don't know the value of current \$ \bar I\$

enter image description here

Therefore, applying KNL on the circuit I obtained the following equations

$$\ \bar I = \bar Vina * \left(\frac{1}{Y}+\frac{1}{Y_{ss}} \right)^{-1} $$ $$\ \bar I = \bar Vinb * \left(Y_{ss} \right) $$

where $$\ Y_{ss} = \frac{1}{R_s}+C_ss $$ $$\ Y = \frac{1}{R} $$ $$\ \bar Vina = Vina *e^{-j * \phi_a} $$ $$\ \bar Vinb = Vinb *e^{-j * \phi_b} $$

From the current, since it is the same for both nodes, I can get the ratio between \$\bar Vinb \$ and \$\bar Vina \$ as $$ \frac{\bar Vinb}{\bar Vina} = \frac{Vinb *e^{-j * \phi_b}}{Vina *e^{-j * \phi_a}} = \frac{\frac{1}{Y_{ss}}} {\frac{1}{Y}+\frac{1}{Y_{ss}}} \Rightarrow\frac{1}{\frac{Y_{ss}}{Y}+1} = \frac{Vinb}{Vina} e^{j*\left(\phi_a -\phi_b \right)} $$

Given that I am interested in \$Cs\$ and \$Rs\$, using above equation I can replace their admittance as follows

$$ \frac{1}{\frac{Y_{ss}}{Y}+1} = \frac{Vinb}{Vina} e^{j*\left(\phi_a -\phi_b \right)} \Rightarrow $$ $$ \frac{{Y_{ss}}}{Y}+1 = \frac{Vina}{Vinb} e^{-j*\left(\phi_a -\phi_b \right)} \Rightarrow $$ $$ R *\left(\frac{1}{R_s}+C_ss \right) +1= \frac{Vina}{Vinb} e^{-j*\left(\phi_a -\phi_b \right)}$$

Expressing the polar complex into its cartesian value would lead to

$$ s=j*\omega \Rightarrow \frac{R}{R_s}+1 +j\omega RC_s= \frac{Vina}{Vinb} \bigl( {cos\left(\phi_a -\phi_b \right)-j*sin\left(\phi_a -\phi_b \right) } \bigr) $$

And finally equalizing real parts and imaginary parts I come up with

$$\frac{R}{R_s}+1 = \frac{Vina}{Vinb}cos\left(\phi_a -\phi_b \right) \Rightarrow R_s = \frac{R}{\frac{Vina}{Vinb}cos\left(\phi_a -\phi_b \right)-1} $$ $$ \omega RC_s= \frac{Vina}{Vinb}\bigl(-sin\left(\phi_a -\phi_b \right)\bigr) \Rightarrow C_s = \frac{1}{\omega R}\frac{Vina}{Vinb}\bigl(-sin\left(\phi_a -\phi_b \right)\bigr)$$

So my questions are two these two

-Are these expression correct? if so,

-What does mean the negative value ( coming from the term \$-sin\left(\phi_a -\phi_b \right)\$) that affetct to the capacitor, since I know that a capacitor should be positive

Thanks in advance,


1 Answer 1


The method you've used looks tedious.

Do you know the value of R? If so, you can determine the total current by Ohm's law.


So, now you know the voltage across the parallel circuit (\$\small V_{inb}\$) and the total current through it, so you can calculate its impedance by Ohm's law.

You can also find find the impedance of the parallel circuit by simple circuit analysis.

Finally, equate the two (complex) impedance expressions to find the values of Cs and Rs.

  • \$\begingroup\$ Thanks for replying, I cannot do that becuase \$I\$ is unknown \$\endgroup\$
    – ndarkness
    Mar 5, 2019 at 14:15
  • \$\begingroup\$ I didn't ask for that. Do you know the two voltages and R? \$\endgroup\$
    – Chu
    Mar 5, 2019 at 14:44
  • \$\begingroup\$ Yes I do know them \$\endgroup\$
    – ndarkness
    Mar 5, 2019 at 14:50
  • \$\begingroup\$ So,what's the problem? Just calculate \$I\$ as in my first equation. \$\endgroup\$
    – Chu
    Mar 5, 2019 at 14:53
  • 1
    \$\begingroup\$ I can, but eventually I will come up to the same expression for \$Cs\$ and \$Rs\$ and now I have the doubt of the sign of \$C_s \$ \$\endgroup\$
    – ndarkness
    Mar 5, 2019 at 15:05

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