Solving this electronic problem with impedance ?

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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 ?

1 Answer
Mar 29, 2018

Answer:

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.