# To derive differential equation from Kcl Kvl for RC circuit

I have a RC circuit named called active realization of compensator, which is used in control systems, and is given in the following link.

https://www.dropbox.com/s/zpmq5ql6qr58gjb/Circuit.png?dl=0

I want to derive its differential equation by applying KCL and KVL. So, I did apply KCL at node A and B, which results into the following equations.

$$\frac{V_i -V_A}{R_1} + \frac{V_B - V_A}{R_2} = 0$$

and

$$\frac{V_A -V_B}{R_2} + C\frac{d(0 - V_B)}{dt} = 0$$

I considered KCL as sum of the current flowing away from node is equal to zero.

Now, I want to solve these equations by eliminating the voltages $V_A$ and $V_B$ to obtain a differential equation in term of $V_i$ and $V_o$.

Can anybody help me out in this regards.

Thanks in advance.

## 1 Answer

I think that reason your struggling is because you did not quite realise that: $$V_A = V_o$$ With that in mind, and after a little algebra with the first equation, we have the following identity: $$V_B = V_o (\frac{R_2 +R_1}{R_1}) - V_i \frac{R_2}{R_1}$$ Taking the derivative of both sides: $$\frac{dV_B}{dt} = \frac{dV_o}{dt}(\frac{R_2 +R_1}{R_1}) - \frac{dV_i}{dt} \frac{R_2}{R_1}$$ So the final step is plugging this new equations into your second one to get ride off $V_B$ and its derivative: $$V_o \frac{1-R_2^2 + R_1 R_2 }{R_2^2 R_1} + V_i \frac{1}{R_1} = \frac{dV_o}{dt}(\frac{R_2 +R_1}{R_1}) - \frac{dV_i}{dt} \frac{R_2}{R_1}$$