# 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 : $i \left(\frac{1}{R} _ 1 + \frac{1}{R} _ 2\right) + \left({C}_{1} + {C}_{2}\right) \frac{\mathrm{di}}{\mathrm{dt}} = {C}_{1} {C}_{2} \frac{{d}^{2} u}{{\mathrm{dt}}^{2}} + \left({C}_{2} / {R}_{1} + {C}_{1} / {R}_{2}\right) \frac{\mathrm{du}}{\mathrm{dt}} + \frac{u}{{R}_{1} {R}_{2}}$ 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 : ${Z}_{e q} = {Z}_{1} + {Z}_{2}$ and then ${Z}_{e q} = \frac{u}{i}$ but it lead to $i = \frac{1}{R} _ 1 \left(u + {R}_{1} {C}_{1} j w u\right) + \frac{1}{R} _ 2 \left(u + {R}_{2} {C}_{2} j w u\right)$ and : $i = u \left(\frac{1}{R} _ 1 + \frac{1}{R} _ 2\right) + \frac{\mathrm{du}}{\mathrm{dt}} \left({C}_{1} + {C}_{2}\right)$ $j$ 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 ?

Mar 29, 2018

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.