# solving Impedance divider

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)

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

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)]


gives

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

Or use this

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

FullSimplify[z/(z + z)]


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

Edit

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

• Use I instead j to represent $\sqrt{-1}$ Dec 17, 2019 at 20:14
• Thanks a lot ... I was using " I " instead of j and mathematica was simplifying it in other form. Dec 18, 2019 at 8:28
• @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) Dec 18, 2019 at 8:34