What are the state variable equations of this RLC circuit? How do I write the final matrix? I tried to write the equations for the capacitor and inductor:

$$L\frac{di_3}{dt}=v_1-v_C \leftrightarrow \frac{di_3}{dt}=\frac{v_1}{L}-\frac{v_C}{L}$$

$$C\frac{dv_C}{dt}=i_3\leftrightarrow \frac{dv_C}{dt}=\frac{1}{C}i_3$$

I tried to apply the Kirchoff's law:

(1) $$i_1R_1+L\frac{di_3}{dt}+v_C=v(t)\leftrightarrow i_1=\frac{v(t)}{R_1}-\frac{L}{R_1}\frac{di_3}{dt}-\frac{v_C}{R_1}$$



From these equations I am unable to reach the final matrix. How could I go on?

enter image description here

  • \$\begingroup\$ The purpose is to find the voltage across the capacitor? And it is an AC-circuit? You can use complex analysis. \$\endgroup\$ Oct 28, 2019 at 11:26
  • \$\begingroup\$ She did not specify steady state voltage applied. The voltage source may have a transient component \$\endgroup\$ Oct 28, 2019 at 11:49
  • 2
    \$\begingroup\$ Carmen, this should help you. \$\endgroup\$ Oct 28, 2019 at 11:55

1 Answer 1


The inductor equation: $$L i_3'=v_1-v_C$$

The capacitor equation: $$c v_c'=i_3\ \ \ (1)$$

Kirchoff's current law: $$\frac{u-v_1}{R_1}=i_3+\frac{v_1}{R_2}$$

which can be solved for \$v_1\$ to get $$v_1=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}$$

Substitute this in the inductor equation $$L i_3'=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}-v_c \ (2)$$

The equations (2) and (1) can then be put in matrix form as

$$\left( \begin{array}{c} i_3' \\ v_c' \\ \end{array} \right)=\left( \begin{array}{cc} -\frac{R_1 R_2}{L \left(R_1+R_2\right)} & -\frac{1}{L} \\ \frac{1}{c} & 0 \\ \end{array} \right).\left( \begin{array}{c} i_3 \\ v_c \\ \end{array} \right)+ \left( \begin{array}{c} \frac{R_2}{L \left(R_1+R_2\right)} \\ 0 \\ \end{array} \right) u$$

Verifying using Mathematica. enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.