# Using KCL/KVL to derive differential equation of a mixed RLC-ciruit

I have the following LTI sinusoidal circuit, and I am trying to derive its differential equation using KCL and/or KVL. I need this to derive the Power Factor as function of the components, so I can optimize it. Solving the equations is generally no problem.

I already asked a similar question a while ago, but what got me confused this time is that my teacher said that I should end up with two equations, one depending on C and the other depending on L, but that is not what I derived. simulate this circuit – Schematic created using CircuitLab

If I am correct:

$$i(t) = i_{C1} + i_{L1} + i_R$$ With $$i_R = i_{L2} = i_{C2}$$

$$v(t) = v_{C1} + v_{L1} + v_{R} + v_{L2} + v_{C2}$$

But if I use the i(t), and derive the differential equation, then I find the same equation of a simple parallel RLC-circuit. This does not seem correct, and I do not find the two equations my teacher was talking about.

I am allowed to use the identities:

$$v_b(t) = L_b \frac{d i_b (t)}{dt}, i_b (t) = C_b \frac{d v_b (t)}{dt}, v_b (t) = R_b i_b (t)$$ With $$i(t) = I_m \cos(\omega t + \phi_i), v(t) = V_m \cos(\omega t + \phi_v), \phi = \phi_u - \phi_i, etc.$$

I am a math student btw, so unfortunately my circuit-theory knowledge is limited.

The thing you got wrong here is the Kirchhoff's laws : while you're right on the current laws, you made a little mistake on the voltages : they are equivalent in each parallel wire, so that : $$v(t) = v_{C1}(t) = v_{L1}(t) = v_R(t)+v_{C2}(t)+v_{L2}(t)$$ Note that these equations are given without regard to the sign and the conventions. This way should lead you to the proper solution. Always remember : the Kirchhoff's laws says that the sum of all the voltages in a loop (closed network) is equal to zero, and that the sum of all currents in a node are equal to zero.
• Ah yes you are right! So basically I will get two equations, one depending on C1 and the other on L1 if I solve $$1) v(t) = v_{C1}(t) = v_R(t)+v_{C2}(t)+v_{L2}(t),$$ $$2) v(t) = v_{L1}(t) = v_R(t)+v_{C2}(t)+v_{L2}(t)$$ Right? I am trying it now, but if I am not mistaken, this is quite complex because I of the three different currents? If i try to solve 1) I find this equality? $$i_1 (t) = C1R i_3 (t) + C1L2 \frac{d^2 i_3 (t)}{dt^2} + \frac{C1}{C2} i_3 (t)$$ With $$i_1, i_2, i_3$$ being the current through first, second and third loop, and same for voltage $$v_1, v_2, v_3.$$ – user260710 May 29 '16 at 0:02
• Do you know about complex impedances ? That is the most efficient way of solving these, by using : $$L = jL\omega ; C=\frac{1}{jC\omega}$$ And by swapping back in time domain. If you cannot use these, I'm working on a solution now – Albits May 29 '16 at 9:26
• The first equality should give you : $$i_1(t)=RC1\frac{di_3(t)}{dt}+\frac{C1}{C2}i_3(t)+C1L2\frac{d²i_3(t)}{dt}$$ – Albits May 29 '16 at 9:41