Circuit Theory/Simultaneous Equations/Example 6

Find currents and voltages. Assume the sources and resistance values are knows. Solve symbolically.

Solving these equations symbolically might fail. The goal is to begin entering the differential equations in symbolic form accurately and methodically .. maybe a solution will pop out!

Label

There is one current source in series with a resistor and inductor that all share the same current: $i_{S}$ .

Loops

There are two loops. The direction is CW this time because the current source is pointing down.

The + and - added to the current source is the reverse of the rule of moving around the circle with your finger. The rule was violated because there is only one source in the circuit and it is going to have to pump energy into the circuit. Thus current has to be going into the - and coming out the +.

Ultimately, the polarity added has to impacts:

the direction of the current
the sign of the voltage term in the loop equations

As long as the equation for the loop is written with a negative sign for the current source's voltage, then violating the rule is ok.

Remember that the + and - right now are not guessing the polarity of the answer, but are capturing the layout of the circuit.

There are no trivial loops (components in parallel).

Junctions

There are two trivial junctions where series components share the same current. One is between the current voltage source and $R_{1}$ . The other is between current source and the inductor.

There are two non-trivial junctions, see the areas constant EMF that are shaded in different colors.

This means that just one non-trivial junction can generate an equation.

Count Eq

We are being told that the resistors and voltage sources have values, and are thus to be treated as knowns. We just don't know what the specific values are so we have to work symbolically.

$R_{1}=\surd$
$R_{2}=\surd$
$I_{S}=\surd$
$i_{1}=?$
$i_{2}=?$
$v_{1}=?$
$v_{2}=?$
$v_{3}=?$
$v_{4}=?$
$v_{5}=?$
$v_{6}=?$

There are 8 unknowns. There are 2 equations from the resistors, 1 from the capacitor, 2 from the inductors, 2 from the loops and 1 from the junction. So the problem can be solved explicitly.

Terminal Eq

The current source symbol chosen implies a DC current source. After a long time, the capacitor is going to open and the two inductors are going to short. The circuit reduces to a 2 resistor series circuit. Obviously, we need to assume that the current source is continuously changing ... for example alternating current. Then it makes sense to write the differential equations.

$v_{1}(t)=i_{1}(t)*R_{1}$
$v_{2}(t)=i_{2}(t)*R_{2}$
$i_{1}(t)=C_{4}{d \over dt}v_{4}(t)$
$v_{5}(t)=L_{5}{d \over dt}i_{2}(t)$
$v_{6}(t)=L_{6}{d \over dt}i_{S}(t)$

If the source current $i_{S}(t)$ is a function of time, so are all the other voltages and currents.

Loop Eq

$L_{1}:v_{4}(t)+v_{1}(t)-V_{3}(t)+v_{6}(t)=0$
$L_{2}:v_{5}(t)+v_{2}(t)-v_{4}(t)=0$

Junction Eq

$J_{1}:i_{1}(t)-i_{2}(t)-i_{S}(t)=0$

Solve the Equations

Algebra

We are going to solve these by hand before the semester is out. This circuit borders on the verge of being to complex to solve by hand in any case. The goal here is to see what the symbolic computation systems come up with.

Differential Equations

The differential equations exist! They exist in the form of the terminal equations. Obviously to solve them we are going to need to know the initial conditions. However, no initial conditions or values were given so we can assume they are zero.

The traditional "by hand" solution requires knowing the resistor, inductor and capacitor values because they point to three alternative solutions. It is going to be interesting to see how to enter the differential equations into MuPad and to see how it responds symbolically.

Looking at the equations above, it is interesting that the differential nature comes from the terminal relations. This makes sense. The voltages and currents, even as a function of time have to add to zero.

Symbolic Computations

MuPad

File:Ex6-6.png

MathWorks MuPAD solution .. click here for the text to cut and paste .. MuPad Solution contains variable solution

A symbolic solution is impossible with differential equations. We are going to work with symbols as much as possible. This is why we are practicing entering the symbols into MuPad.

To really solve, we are going to need values for the resistors, and an equation for $i_{S}$ .

The next chapter begins the process of explaining what needs to be done. This was an exercise in walking up to the edge of the unknown and figuring out where to stop.

Numeric Solution

There is no numeric solution because no numbers were given.

Simulate

Simulation is impossible with out numbers.

Build Intuition

Differential equations require known voltages and/or currents that are function of time and capacitors and/or inductors.
If the circuit just has resistors in it, then there are no differentials in the terminal relationships.
If no voltage or current is changing with time, then the capacitors and inductors turn into opens and shorts ... and disappear. So then there are no differential equations.
But if a switch is thrown to turn on a DC circuit, things are changing in the short amount of time after the switch is thrown. Differential equations would be appropriate. So how do we write an equation for the this situation? ... on to the next chapter.