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 at once. To find the current i, we read ir1 or ir2. To find the voltage drop on r2, we read 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 read pr2. To find the power delivered by the source, we read the power delivered row of the source's card, -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 R_T 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 ir3, the current through resistor r3. Look at it and you will see an expression rather than a number, in terms of the two unknowns, e and r1:

This is what we call a symbolic answer. The problem also says that R_T is 12 kΩ; the equivalent resistance as seen by the source e is re, which is:

We can then write two new equations and solve them for e and r1.

That is what the Solve card, under the results beside Evaluate, is for. It has two boxes.

In the first box, headed Equation(s) to solve in terms of the results, write the equations, using the same names the results use.

The shorthand works in these boxes too, so re = 12'k reads as 12000.

You can write the equations separated by and, like this:

re = 12'k and ir3 = 6'm

Or one equation per line, like this:

re = 12'k
ir3 = 6'm

In the second box, headed Unknown(s) to solve for, name what you want found. You can write the unknowns separated by commas, like this:

e, r1

Or one unknown per line, like this:

e
r1

Then press Solve equations.

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

Solution

We use the same circuit description as before, r1 included.

The problem gives us the two extra equations: the source sees 12 kΩ, and the current through R₃ is 6 mA. They use the names the answers come back under, and the 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. This time everything comes back numerical: the node voltages read v₁ = 72 V, v₂ = 60 V and v₃ = 36 V, and at the foot of Results by element are the two unknowns, listed like any other answer: e = 72 and r1 = 2000. Both are right, and the whole circuit was solved with them in one step.

The speed advantage of Expert Mode is not necessarily evident in this simple problem. But it does give you an idea of what 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.