Symbolic circuits and expert mode

Solution

All the values in this circuit are variables: no element value is known. This is an example of a symbolic circuit. Imagine you are asked to find symbolic expressions for the current i, the voltage drop in R2, and the power consumed by R1 and R2. Since we are being asked to find symbolic answers from a symbolic circuit, this is a purely symbolic problem.

This is how I would describe this circuit. First, the nodes. The node called a in the schematic stays node a. The node called b is my ground, so it becomes node 0. The node between the two resistors I call c.

Then the elements. I name the source ev and the resistors r1 and r2, and give them variables for values: v for the source, r1 and r2 for the resistors. It is no problem that a symbolic value carries the same name as its element.

"ev,a,0,v:r1,a,c,r1:r2,c,0,r2"→cir
s\dc(cir)

This simulation took 15 seconds in my calculator. To find the current i, we ask for ir1 or ir2. To find the voltage drop on r2, we ask for vr2.

Both are expressions rather than numbers:

which is the voltage divider you would have written by hand.

To find the power consumed in r2, we ask for pr2. To find the power delivered by the source, we ask for the negative of pev.

It is this ability to simulate symbolically, as if it were nothing, that puts Symbulator in a league of its own. Symbolic answers from a symbolic circuit take no extra effort.

Solution

This is a very nice numerical-from-symbolic problem. We can solve it into numbers because, although it hides two values from us (the source E and the resistor R1), it gives us in exchange two answers (the equivalent resistance RT and the current I3) to solve for them with.

Since this circuit is structurally identical to B11's Example 5.7 solved in Lesson 1, we will use the same names for the nodes. The description is identical except for the values. Symbulator accepts numbers, variables or even algebraic expressions as values. Here I use e for source e, and r1 for resistor r1.

"e,1,0,e:r1,1,2,r1:r2,2,3,4'k:r3,3,0,6'k"→cir
s\dc(cir)

A moment later Symbulator is done. From its symbolic answers and the answers the problem gives us, we write two equations and solve them for the two unknowns.

The problem says that I3 is 6 mA. In Symbulator, I3 is ir3, the current through resistor r3. If you evaluate it, you will see it produces an algebraic expression in terms of the two unknowns, e and r1. This is what we call a symbolic answer. The problem also says that RT is 12 kΩ; the equivalent resistance as seen by the source e is given by re, which when evaluated gives another algebraic expression in terms of r1. We can then write two new equations and solve them for e and r1:

solve(re=12000 and ir3=.006,{e,r1})

An instant later we get the answers: e = 72 V and r1 = 2000 Ω. These are the right answers. Not many other circuit simulators allow this flexibility.

Solution

We will use the same circuit description as before, with a single change: we will use rx for the value of resistor R1, instead of the r1 value we used before. Like this:

"e,1,0,e:r1,1,2,rx:r2,2,3,4'k:r3,3,0,6'k"→cir

We use rx instead of r1 because expert mode saves the value it finds into that variable, and r1 — like the other r# variables — cannot store anything: trying to results in an error.

To run an expert mode simulation, type this:

s\ex(cir)

When prompted, select DC and press Enter. Now you will see a prompt asking you to add equations, variables and conditions. You may recall from algebra that solving a set of equations into numbers needs as many equations as unknowns. The statement of the problem gives us the information we need to write the two additional equations.

In "Add equations" type:

re=12'k and ir3=6'm

In "Add unknowns" type:

e,rx

Now we have six variables and six equations. Press Enter and wait a few seconds. A few other dialogs appear; in this and every other expert example in this volume, just press OK without changing anything. When Symbulator says "Done", go ahead and retrieve the answers: rx gives 2000 and e gives 72. Both are right.

The speed advantage is not obvious in so simple a problem, but the idea is: you halt the simulation in mid-air and give Symbulator extra information. Had the circuit been larger, the saving would be clear.