Syrup Problem

[Thomas Dean]

I am getting up-to-speed, again.

I have a problem from a text book schematic that syrup cannot solve.

restart;with(Syrup);
ckt:="
e1 3 1 4 0 4
h1 1 0 r3 10
i1 2 5 2
r1 2 1 4
r2 3 2 5
r3 3 4 8
r4 4 0 2
r5 5 4 2
v1 3 0 15
.end";
volts:=syrup(ckt,dc,'current');
evalf(volts); evalf(current);

I am not certain of the value for r5, but, as long as it is reasonable, it
should not prevent solution.

I get maple cannot solve the equations.

I1 is node2 --> node5 so, I choose node 5 as +
e1 is node3 +- node1 4*V4
h1 is node1 +- node0 10*Ir3

What am I doing wrong?

tomdean

[Joe Riel]

The problem is that the circuit has no solution.
Note that there is a new version of syrup that has a slightly
different calling convention. I released it on MaplePrimes;
it is also in the Maple Application Center. However, the
versions there are not the very latest and have a few bugs.

To analyze this circuit with the new version I changed
the v1 line to

v1 3 0 V1
.VARS V1

and made V1 a variable (this is a new feature):

Syrup could still not solve the circuit---actually it did, however, it
recognized that the result was not unique and so rejected it.

Using the modified deck, I did

sol := Solve(ckt,'dc'):
Solve: Could not solve equations; circuit may not be
well-defined. Returning equations and variables as [eqs,vars].

solve(op(sol));
{V1 = -11, i[e1] = 1/6 - i[v1], i[h1] = - 1/2 - i[v1], i[i1] = 2, i[r1] = -2/3,
i[r2] = 4/3, i[r3] = -3/2, i[r4] = 1/2, i[r5] = 2, i[v1] = i[v1],
v[1] = -15, v[2] = -53/3, v[3] = -11, v[4] = 1, v[5] = 5, v[e1] = 4,
v[h1] = -15, v[i1] = -68/3, v[r1] = -8/3, v[r2] = 20/3, v[r3] = -12,
v[r4] = 1, v[r5] = 4, v[v1] = -11
}

Note that i[i1] is arbitrary and that V1 must equal -11 for there to be
a solution.

I'll email you the latest version of Syrup.


--
Joe Riel