Question

Applying Vector Differential Equations to Electrical Networks question! Will posted below.

Answer 1

A 5V source has been connected in parallel with the capacitor in a series RLC circuit (R= 10 Ohms, C=2F, L=3H) for a long time. At t=0, the source is disconnected. Find a vector differential equation in matrix form which characterises the circuit (without the DC source) and determine the initial conditions.

Answer 2

So I know that:

\[i_c = C \frac{ dv_C }{ dt } + i_c(0^-)\]

\[v_L = L\frac{di_l}{dt} + v_L(0^-)\]

Answer 3

Through Kirchoff's Voltage Law, I know that
\[v_c +iR + L\frac{di}{dt}=0\]

Answer 4

And the relationship in matrix form should look something like
\[\vec{x} = A\vec{x}\]

Answer 5

sorry meant to put: \[\vec{x}'\]

Answer 6

And from that point on, I'm not entirely sure what to do. :( Any help would be great, thanks!

Answer 7

\(v_c +iR + L\frac{di}{dt}=0\)

replace \(i\) by \(dq/dt\) and \(v_c\) by \(q/C\) and you will see a DE in one variable \(q\) :
\[L\dfrac{d^2q}{dt^2} + R\dfrac{dq}{dt}+\dfrac{1}{C}q=0\]

Answer 8

you can split that as a system of two differential equations :

\(\dfrac{dq}{dt} = w\)

\(\dfrac{dw}{dt} = -\dfrac{1}{LC}q-\dfrac{R}{L}w\)

Answer 9

see if you can write that in matrix form

Answer 10

i don't think she is here @ganeshie8

Answer 11

@ILovePuppiesLol Still here!
@ganeshie8 so what I did was I let
\[\vec{x} = \left(\begin{matrix}v_c \\ i_L\end{matrix}\right)\]
so
\[i_c = x_2 = C\frac{dv_c}{dt}\]
\[x_1 + x_2R + L\frac{dx_2}{dt}=0\]
\[\frac{-x_1}L - \frac{x_2R}{L} = \frac{dx_2}{dt}\]
\[\frac{dx_1}{dt} = \frac{x_2}{C}\]
\[\frac{dx_2}{dt} = \frac{-x_1}{L}-\frac{x_2R}{L}\]

Answer 12

So then the matrix is:
\[\left[\begin{matrix}0 & 1/C \\ \frac{-1}{L} & \frac{-v(0^-)}{L}\end{matrix}\right]\left(\begin{matrix}v_c \\ i_L\end{matrix}\right)=\left(\begin{matrix}\frac{dv_c}{dt} \\ \frac{di_L}{dt}\end{matrix}\right)\]

Answer 13

Then substitute the values in.

Answer 14

Does this make sense? @ganeshie8

Answer 15

Looks good to me !