Question

Solving this electronic problem with impedance ?

There is two of this image in series, two capacitors `C_1` `C_2` and two resistors `R_1` `R_2`

To be more precise this is two real capacitors (capacitor and resistor in parallel) in series.

So i have the differential equation etablished by Kirchhoff's laws :

`i(1/R_1 + 1/R_2) + (C_1 + C_2)(di)/(dt) = C_1C_2(d^2u)/(dt^2) + (C_2/R_1 + C_1/R_2)(du)/(dt) + u/(R_1R_2)`

i tried to do it by impedance because i thought it will take less time.

Impedance of the first real capacitor :

`Z_1 = (1/Z_(C_1) + 1/Z_(R_1))^(-1) = Z_(R_1)/(1+Z_(R_1)/(Z_(C_1))`

Impedance of the second :

`Z_2 = (1/Z_(C_2) + 1/Z_(R_2))^(-1) = Z_(R_2)/(1+Z_(R_2)/(Z_(C_2))`

the equivalent impedance of the entire circuit is :

`Z_(eq) = Z_1 + Z_2`

and then `Z_(eq) = u/i`

but it lead to

`i = 1/R_1(u + R_1C_1jwu) + 1/R_2(u + R_2C_2jwu)`

and :

`i = u(1/R_1 + 1/R_2) + (du)/dt(C_1+C_2)`

`j` is the imaginary unit

Obviously it's not the same, i know i'm good with Kirchkoff's law because i checked the answer, but i'm not good with impedance, why ?

Answer 1

See below.

Explanation:

When you solve using impedances you are assuming that the circuit is submitted to a sinusoid periodic input. You are solving without considering the transient modes. So be careful with this approach.