Coupled RC circuits with AC current source

[no_alone]

Homework Statement

Hi, I already asked a question close to this, but now I have different conditions.
This is the circuit

$C_1 = C_2 \\ R_1= R_2$

The current is an AC , and I would like to know the voltage at R_1 and at R_2
I made some progress but I do not really know hot to continue.

The Attempt at a Solution

V_R1 = V_C1
V_R1 + V_Rc - V_C2 = 0
V_R2 = V_C2
I_R1 + I_C1 + V_RC = I_inj
I_Rc = I_R2 + V_C2
$$V_1 = V_{C1} = V_{R1} \\ V_2 = V_{R2} = V_{C2}$$

$$\frac{V_1}{R_1} + \frac{dV_1}{dt}*C_1 + \frac{V_{Rc}}{R_c} = I_{inj} \\ \frac{V_{Rc}}{R_c} = \frac{V_2}{R_2} + C_2*\frac{dV_2}{dt}\\$$$$\frac{dV_1}{dt}*C_1 = I_{\omega} -\frac{V_1-V_2}{R_c} - \frac{V_1}{R_1} \\ \frac{dV_2}{dt}*C_2 = -\frac{V_2}{R_2} +\frac{V_1-V_2}{R_c} \\ --> C_1 == C_2 , R_1 == R_2 \\ \frac{dV_1}{dt} = \frac{I_{\omega}}{C_1} -\frac{V_1-V_2}{R_c*C_1} - \frac{V_1}{R_1*C_1} \\ \frac{dV_2}{dt} = -\frac{V_2}{R_1*C_1} +\frac{V_1-V_2}{R_c*C_1} \\$$

 

[gneill]

With AC sources you can obtain steady-state solutions for such circuits using impedances for the components and phasor values for the sources. No need to write or solve differential equations, the usual DC circuit methods and techniques of analysis will work fine.

 

[rude man]

The Attempt at a Solution

I_R1 + I_C1 + V_RC = I_inj
I_Rc = I_R2 + V_C2

You can't have voltage and current terms in the same equation. Straightening this out is your first task.

It is perectly OK to use differential equations as you have done, especially if you never had phasors..