# simple RC circuit corresponding differential equation [closed]

I have an RC circuit as shown below unfortunately I have no knowledge about RC solving but I want to obtain corresponding differential equations for V1, V2 and V3. Having in hand differential equation for V1 is enough for now, if anyone can tell me it!!!

• s-domain analysis is one way to go. – K. Rmth Nov 7 '15 at 17:32
• I'll be grateful if you can give me differential equation for only V1 – Mahmoud Nov 7 '15 at 18:04
• I'm voting to close this question as off-topic because homework without effort. – Brian Carlton Nov 7 '15 at 22:46
• it's not homework. actually it is a circuit model from a paper which I need to use it for my analysis in my dissertion – Mahmoud Nov 8 '15 at 6:34
• Possible duplicate of Deriving 2nd order passive low pass filter cutoff frequency – jippie Nov 8 '15 at 8:03 simulate this circuit – Schematic created using CircuitLab

In box 1, we're converting our current source into a voltage source and further converting it to its s-domain equivalent.

Then in box 2 we're taking the s-domain source together with the s-domain equivalent of the impedances of the given circuit. (see this)

Assuming that the capacitances had no voltage across them initially, (i.e. $V_{0_{C_1}}=V_{0_{C_2}}=0$, the s-domain circuit is equivalent to that in box 3 where $V=-V_1$ and $R_{\text{equiv}}=\left[\left[\left(R_2+\frac{1}{sC_2}\right) \parallel\ R_L \right]\parallel \frac{1}{sC_1}\right]+R_1=R_1+\left[\frac{1}{R_2+sC_2}+\frac{1}{R_L}+\frac{1}{sC_1}\right]^{-1}$

Hence $$V_1=-V=-\left(\frac{R_\text{equiv}}{R_s+R_\text{equiv}}\right)\cdot\frac{iR_s}{s}$$

The expression for $V_1$ will be a function of $s$; applying inverse laplace transform (using tables is easier) to the equation will give you the differential equation you require.