Solving this electronic problem with impedance ?
There is two of this image in series, two capacitors C1 C2 and two resistors R1 R2
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(1R1+1R2)+(C1+C2)didt=C1C2d2udt2+(C2R1+C1R2)dudt+uR1R2
i tried to do it by impedance because i thought it will take less time.
Impedance of the first real capacitor :
Z1=(1ZC1+1ZR1)−1=ZR11+ZR1ZC1
Impedance of the second :
Z2=(1ZC2+1ZR2)−1=ZR21+ZR2ZC2
the equivalent impedance of the entire circuit is :
Zeq=Z1+Z2
and then Zeq=ui
but it lead to
i=1R1(u+R1C1jwu)+1R2(u+R2C2jwu)
and :
i=u(1R1+1R2)+dudt(C1+C2)
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 ?
There is two of this image in series, two capacitors
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 tried to do it by impedance because i thought it will take less time.
Impedance of the first real capacitor :
Impedance of the second :
the equivalent impedance of the entire circuit is :
and then
but it lead to
and :
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
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.