MIPT e-library captcha solution

Following a discussion in Reasonable Deviations, I spend some three hours solving a (quoting) “little school-level problem” posed in a weird Russian CAPTCHA system. Not that it is extra-difficult, but some people may find it interesting to see the solution of its generalized formulation, so here it is.

Problem: we have an electric circuit as depicted in the following figure:

The task is to find the altogether resistance from A to B.

The answer is (numerator and denominator split by “-”):
(n2+n4)*(n1*y+n5*x)+(n1+n2)*n4*x
—————————————
(n1+n2)*y

, where:

x = n3*n2-n1*n4
y = (n1+n2)*(n3+n4)+(n1+n2+n3+n4)*n5

The solution is after the jump.

Let’s depict the currents in this circuit:

We marked each current as a proportion to the one coming from A (and to B), e.g. I3=i*i3. Using Kirchhoff’s circuit laws, let’s write down the system of equations for this circuit:

i1 + i2 = 1
i3 + i4 = 1
i5 = i1 – i3
n1*i1 + n5*i5 – n2*i2 = 0
n3*i3 – n4*i4 – n5*i5 = 0

In a matrix view, the system looks as follows (we take i1 to i5 as the unknown values):

1	1	0	0	0	1
0	0	1	1	0	1
1	0	-1	0	-1	0
n1	-n2	0	0	n5	0
0	0	n3	-n4	-n5	0

Using linear combinations of rows, we rearrange this matrix to the triangular view to get the following (I skip the arithmetical :-) calculations):

1	1	0	0	0	1
0	1	1	0	1	1
0	0	1	1	0	1
0	0	0	1	-(1+n5/(n1+n2))	n1/(n1+n2)
0	0	0	0	1+n5*(1/(n3+n4)+1/(n1+n2))	n3/(n3+n4)-n1(n1+n2)

Solving the above,

i5 = x/y
i2 =(n1+n5*i5)/(n1+n2)
i4 = i2 + i5

where:

x = n3*n2 – n1*n4
y = (n1+n2)*(n3+n4) + (n1+n2+n3+n4)*n5

The desired resistance of the whole circuit is as follows:

Rc = (i2*n2 + i4*n4)*R

After replacing i2 and i4 with the found values, we get the final answer (numerator and denominator split by “-”):

(n2+n4)*(n1*y+n5*x)+(n1+n2)*n4*x
—————————————
(n1+n2)*y

, where:

x = n3*n2 – n1*n4
y = (n1+n2)*(n3+n4) + (n1+n2+n3+n4)*n5