Hi again,
(Note: Here R1=Rp,R2=Rc,R3=Rk, and Rs=R1 from the original circuit.
In other words, here R1,R2,R3 are the resistors in parallel with C1,C2,C3
and Rs is the resistor in series with V in the original circuit.
The solution is the same, with different component names.)
Here is a slightly different form for the time solution for the
voltage across R2:
v2(t,V)=V*(A*(e^(a*t)-1)+B*(e^(b*t)-1)+C*(e^(c*t)-1))
Note that this isnt a particularly complicated solution, but we do
need to find all the constants A,B,C,a,b, and c, so what we do
here is first find these constants A,B,C, and a,b,c and then we have the
solution for all time t.
First we define some sundry constants, all of which depend on the circuit components
and the voltage of the battery:
Code:
B3=C1*C2*C3*R1*R2*R3*Rs
B2=C1*C2*R1*R2*R3+C1*C2*R1*R2*Rs+C1*C3*R1*R2*R3+C1*C3*R1*R3*Rs+C2*C3*R1*R2*R3+C2*C3*R2*R3*Rs
B1=C1*R1*R2+C1*R1*R3+C1*R1*Rs+C2*R1*R2+C2*R2*R3+C2*R2*Rs+C3*R1*R3+C3*R2*R3+C3*R3*Rs
B0=R1+R2+R3+Rs
A2=V*R2*C1*C3*R1*R3
A1=V*R2*(C1*R1+C3*R3)
A0=V*R2
Next we find the three real roots of this equation:
B3*x^3+B2*x^2+B1*x+B0=0
using whatever method we have available to find numerical roots such as:
(a,b,c)=roots("B3*x^3+B2*x^2+B1*x+B0=0")
This yields three real roots, a,b, and c, and so we now have three of the
required constants. The roots will always be real with this circuit because
there are no oscillatory components. They are also negative.
Next, since we now know a,b, and c we can calculate the other three
constants A,B, and C from these equations:
V=1
A=(A2*a*a+A1*a+A0)/(B3*a*(a-b)*(a-c))
B=(A2*b*b+A1*b+A0)/(B3*b*(b-a)*(b-c))
C=(A2*c*c+A1*c+A0)/(B3*c*(c-a)*(c-b))
and now all we do is insert all of the constants into the time solution and
we have the solution for v2(t) which is the voltage across the resistor R2 at
any given time t:
v2(t)=A*(e^(a*t)-1)+B*(e^(b*t)-1)+C*(e^(c*t)-1)
We can now insert any time t and get the solution. If we change any of the
circuit values however we need to go back and recalculate all of the constants
again first.
Also,
v2(t,V)=A*(e^(a*t)-1)+B*(e^(b*t)-1)+C*(e^(c*t)-1) with V=1,
and with another V we get:
v2(t,V)=V*(A*(e^(a*t)-1)+B*(e^(b*t)-1)+C*(e^(c*t)-1))
and that's it.
Of course we could also use the state vector differential equation in the
previous post too putting it into the form
x'=
Ax+
Bu and then solving
after calculating the state transition matrix.