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.

Type the three elements into the box, one to a line:

ev,a,0,v
r1,a,c,r1
r2,c,0,r2

Leave Type of analysis on Solve circuit, set Analysis to DC — direct current, and press Run Symbulator.

The answer comes back almost at once — symbolic circuits this small cost nothing. To find the current i, we look at the block for r1 or for r2 and read the current through line. To find the voltage drop on r2, we read the voltage drop line in the r2 block.

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 read the power consumed line of the r2 block. To find the power delivered by the source, we take the opposite of the source's power consumed line — type -pev into Evaluate.

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

Run it in DC, as before.

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 the current through line of the r3 block, ir3, the current through resistor r3. Look at it and you will see an expression rather than a number — \(e/(r1 + 10000)\) — 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 the resistance seen line, re, which is \(r1 + 10000\). We can then write two new equations and solve them for e and r1.

That is what the Solve card is for. You will find it under the results, beside Evaluate. Open it and fill in two boxes.

In the first, headed Equation(s) to solve in terms of the results, write one equation per line, using the same names the results use:

re = 12'k
ir3 = 6'm

In the second, headed Unknown(s) to solve for, name what you want found, separated by commas:

e, r1

Then press Solve equations.

The calculator's restriction does not carry over: the shorthand works in these boxes too, so re = 12'k reads as 12000.

An instant later we get the answers: the card lists e = 72.0 and r1 = 2000.0, that is a source of 72 V and a resistor of 2000 Ω. These are the right answers. Not many other circuit simulators allow this flexibility.

Solution

We use the very same circuit description as before, r1 as the symbolic value included — version 9 has no objection to it; the calculator needed a different name here, and the note below says why.

The statement of the problem gives us the information we need to write the two additional equations: the source sees 12 kΩ, and the current through R3 is 6 mA. Both are written in the same names the answers come back under, and the calculator's SI shorthand works here too:

e,1,0,e
r1,1,2,r1
r2,2,3,4'k
r3,3,0,6'k

re = 12'k
ir3 = 6'm

e, r1

Run it in DC as usual. This time nothing comes back symbolic: Symbulator has enough to pin every value down, so the whole circuit arrives in numbers. The node voltages read v₁ = 72 V, v₂ = 60 V and v₃ = 36 V, and at the foot of Results by element you will find the two unknowns you asked for, listed like any other answer: e = 72 and r1 = 2000.

Both are right, and they arrived alongside everything else rather than in a separate step — the whole circuit is solved, not just the two unknowns.

The speed advantage of the expert mode is not necessarily evident in this simple problem. It does give you an idea of what the expert mode is all about: you get to hand Symbulator extra information before it solves, rather than after. Had this circuit been larger, the benefit in computation time would be clear.