Equations from Circuit Diagrams

In this section, we will look at several example circuits and develop the state equations following the procedure outlined in the previous section.

The circuit above consists of two batteries with voltages B1 and B2, one capacitor and two resistors. Note that the wide end of the battery symbol is the + end.

Here are the steps we follow, all stated in further detail in the previous section:

Below, we will solve the problem in this way. But note the following:  OK, here is the solution for this problem, step by step:
First, we draw in the current arrows. Note that I have only drawn one arrow next to b1, and have not drawn separate currents for C1 and b2: I am implicitly noting that these elements are simply in series and will have the same current.That is, iC1=ib1=ib2

I have also drawn in three nodes. Since I have already noted that b1, b2 and C1 are simply in series and have the same current, I have not drawn nodes in between these elements. If you want to draw in those nodes, that is fine -- the node equations you write down will then tell you the currents in each of these elements is the same! 

Finally, I have drawn in plus and minus signs across my elements and batteries - these are consistent with our rules for voltmeters in circuits and I will use these signs when I apply KVL.

 
Node Equations (KCL) 
iC1 = iR1 + iR2 

iR2 = iR3 

iC1 = iR1 + iR3

Loop Rules (KVL) 
VR2 + VR3 - VR1=0 

VR1 - Vb2 - Vb1+ VC1 =0

Constitutive Laws 
VR1 = R1 iR1 

VR2 = R2 iR2 

VR3 = R3 iR3 

VC1 = (1/C1) qC1

Charge-current relation: q'C1 = iC1 

State Variable: qC1 

State Equation will be of the form: q'C1 = function(qC1

Number eqns: 10; Number unknowns: 9 

Hmmm - I see that the 3rd node equation is a linear combination of the first two, and thus can be eliminated, giving me 9 equations and 9 unknowns. OK!
Now I can solve for our single state equation, starting with the charge-current relation for the capacitor:
Phew! Finally - q'C1 as a function of qC1 and known parameters. If you simplify the circuit by using equivalent resistances, RS and RP, as described above, you will find the algebra reduces dramatically! Try it!


Here is another example to try. Follow the same steps as above:

If you do it properly, wou will obtain two state equations, one for each capacitor:

 

Here is a last example, this time with given voltage bus values:

 

First, redraw with current arrows:

This time when we write our loop rules, we will really be using the series and parallel rules for the voltages to take advantage of the named potentials at the top and bottom wires (aka busses, rails).

Node Equations (KCL) 
iR1 = iC1 + iR2 

iC1 + iR3 = iR4

Loop Rules (KVL) 
VR1 + VR2 = 5 - 0 

VR3 + VR4 = 5 - 0 

VR1 + VC1 + VR4 = 5 - 0

Constitutive Laws 
VR1 = R1 iR1 

VR2 = R2 iR2 

VR3 = R3 iR3 

VR4 = R4 iR4 

VC1 = (1/C1) qC1

Charge-current relation: q'C1 = iC1 

State Variable: qC1 

State Equation will be of the form: q'C1 = function(qC1

Number eqns: 11; Number unknowns: 11 

 

solving for VR1 in terms of VC1:

Then substituting this in for VR1,

becomes