Engineering Coupled RC circuits with AC current source

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The discussion focuses on analyzing a coupled RC circuit with an AC current source, where the user seeks to determine the voltages across resistors R1 and R2. The user has made initial progress using differential equations but is unsure how to proceed. A suggestion is made to utilize impedance and phasor analysis for steady-state solutions, which can simplify the process without solving differential equations. Additionally, it is emphasized that voltage and current terms should not be mixed in the same equation for clarity. Overall, the conversation highlights different approaches to solving the circuit analysis problem.
no_alone
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Homework Statement


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

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C_1 = C_2 \\<br /> 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

VR1 = VC1
VR1 + VRc - VC2 = 0
VR2 = VC2
IR1 + IC1 + VRC = Iinj
IRc = IR2 + VC2
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} \\<br /> \frac{V_{Rc}}{R_c} = \frac{V_2}{R_2} + C_2*\frac{dV_2}{dt}\\<br />\frac{dV_1}{dt}*C_1 = I_{\omega} -\frac{V_1-V_2}{R_c} - \frac{V_1}{R_1} \\<br /> \frac{dV_2}{dt}*C_2 = -\frac{V_2}{R_2} +\frac{V_1-V_2}{R_c} \\<br /> --&gt; C_1 == C_2 , R_1 == R_2 \\<br /> \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} \\<br /> \frac{dV_2}{dt} = -\frac{V_2}{R_1*C_1} +\frac{V_1-V_2}{R_c*C_1} \\<br /> <br />
 
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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.
 
no_alone said:

The Attempt at a Solution


IR1 + IC1 + VRC = Iinj
IRc = IR2 + VC2
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..
 

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