How to represent this equation $$ Z1 = \frac{1}{{\frac{1}{R_1} +j \cdot w \cdot C_1}} $$

in this form in mathematica: $$ Z_1 = \frac{R_1}{{1 +j \cdot w \cdot C_1 \cdot R_1}} $$

And then to solve this $$ Z_3 = \frac{Z_1}{Z_1 + Z_2} $$

where $$ Z_2 = \frac{R_2}{{1 +j \cdot w \cdot C_2 \cdot R_2}} $$

to get the equation in this form : (I need to check the dependence of each variable on poles and zeros)

$$ Z_3 = \frac{R_1 + j w (C_2 R_1 R_2)}{(R_1 + R_2) + j w (C_2 R_1 R_2 + C_1 R_1 R_2)} $$

Please it would be very helpful if someone can help me.


You are not solving, just simplifying it. I don't recommend subscript notation.

z1 = r1/(1 + j w c1 r1);
z2 = r2/(1 + j w c2 r2);

FullSimplify[z1/(z1 + z2)]


$z3=\frac{\text{c2} j \text{r1} \text{r2} w+\text{r1}}{j \text{r1} \text{r2} w (\text{c1}+\text{c2})+\text{r1}+\text{r2}}$

Or use this

z[1] = r[1]/(1 + j w c[1] r[1]);
z[2] = r[2]/(1 + j w c[2] r[2]);

FullSimplify[z[1]/(z[1] + z[2])]

$z(3)=\frac{r(1) (c(2) j r(2) w+1)}{(c(1)+c(2)) j r(2) r(1) w+r(1)+r(2)}$


z1 = 1/(1/r1 + j w c1 );
z2 = 1/(1/r2 + j w c2 );
Simplify[z1] /. j -> I
Simplify[z2] /. j -> I
z3 = FullSimplify[z1/(z1 + z2)] /. j -> I
  • 1
    $\begingroup$ Use I instead j to represent $\sqrt{-1}$ $\endgroup$ – Cesareo Dec 17 '19 at 20:14
  • $\begingroup$ Thanks a lot ... I was using " I " instead of j and mathematica was simplifying it in other form. $\endgroup$ – chetan kulkarni Dec 18 '19 at 8:28
  • $\begingroup$ @OkkesDulgerci... Thanks and it works...Can you please help me with the first part of the question: representing without a fraction in denominator. (I have many equations and I cant do it for each impedance) $\endgroup$ – chetan kulkarni Dec 18 '19 at 8:34

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