Series Within Parallel Resistance Networks
If electrical resistors r1 and r2 are connected in series they give
a total resistance of r1 + r2. Likewise connecting r3 and r4 in
series gives a total resistance of r3 + r4. Putting these series
combination in parallel gives a combined resistance of
(r1 + r2)(r3 + r4)
rx = ------------------
r1 + r2 + r3 + r4
Quentin Grady notes that if we connect the node between r1 & r2 to
the node between r3 & r4 the circuit is transformed to a parallel
within series connection, and the resistance becomes
r1 r3 r2 r4
ry = ------- + -------
r1 + r3 r2 + r4
He asks for positive integer solutions of this equation, i.e.,
integers values of r1, r2, r3, r4, rx, and ry, "preferably under
100", that satisfy this equation.
There are a very large number of solutions to this. Even if we
require r1 through r4 to be distinct, and add the restriction that
the values of (r1*r3)/(r1 + r3) and (r2*r4)/(r2+r4) must both be
integers, there are still an over-abundance of solutions. Most of
them have r1*r4 = r2*r3, which implies that rx=ry. In other words,
the overall resistance is unchanged by inserting the jumper. An
example is [r1,r2,r3,r4] = [3,9,6,18]. You can easily define
infinite families of such solutions. For example, with
r1 = 8j+2 r2 = 24j+8k+10 r3 = 24j+6 r4 = 72j+24k+30
we have rx = ry = 24j + 6k + 9. On the other hand, there are also
many solutions with rx > ry. An example is [5,99,95,9].
There are so many solutions to these equations that I think it's
more interesting to restrict the problem a little more by requiring
not only that Rx and Ry are integers but also the components of
Ry, i.e., R1R3/(R1+R3) and R2R4/(R2+R4). In other words, we seek
four integers a,b,c,d such that each of the quantities
ac bd (a+b)(c+d)
M = ----- N = ---- K = ----------
a+c b+d a+b+c+d
is an integer. The expressions for M and N imply
(a-M)(c-M) = M^2 (b-N)(d-N) = N^2
This means there are integers W,X,Y,Z such that WX=M^2, YZ=N^2, and
a = M+W b = N+X c = M+Y d = N+Z
Thus for any integers M,N we need only select factorizations of M^2
and N^2 into two factors to generate values of a,b,c,d that satisfy
the equations for M and N. Now we just need to ensure that the
equation for K is also satisfied.
The equation for K implies the existence of integers m,n such that
mn=K^2 and
(a+b) = K + m (c+d) = K + n
Thus we have
(a+b) - (c+d) = m - n (1)
(a+b) + (c+d) - 2K = m + n (2)
K^2 = m n (3)
Letting Q denote a+b+c+d-2K, equations (2) and (3) imply that m
and n are the roots of
m^2 - Qm + K^2 = 0
so we have
_________ _________
Q + /Q^2 -4K^2 Q - /Q^2 -4K^2
m = -------------- n = --------------
2 2
To make these integers it's clear that the quantity in the radical
must be a square, so there is an integer f such that
f^2 + (2K)^2 = Q^2
By Pythagorean triples this implies the existence of integers q,r,s
such that
f^2 = q(r^2 - s^2) K = qrs Q = q(r^2 + s^2)
and so we can express m and n as
q(r^2 + s^2) + q(r^2 - s^2)
m = --------------------------- = qr^2
2
q(r^2 + s^2) - q(r^2 - s^2)
n = --------------------------- = qs^2
2
Substituting these expressions for m, n, and K back into equations
(1) and (2) gives
q(r^2 - s^2) = (a+b) - (c+d) (4)
q(r + s)^2 = (a+b) + (c+d) (5)
Now recall that we have
a = M+W b = N+Y c = M+X d = N+Z
where WX=M^2 and YZ=N^2. It follows that there must be integers
e,f,w,x,y,z such that
W = ew^2 X = ex^2 Y = fy^2 Z = fz^2
and so M = ewx and N = fyz. Substituting these expressions into
equations (4) and (5) gives
q(r^2 - s^2) = (ewx+ew^2 + fyz+fy^2) - (ewx+ex^2 + fyz+fz^2)
q(r + s)^2 = (ewx+ew^2 + fyz+fy^2) + (ewx+ex^2 + fyz+fz^2)
Cancelling terms, these equations reduce to
q(r+s)(r-s) = e(w^2 - x^2) + f(y^2 - z^2)
q(r+s)(r+s) = e(w + x)^2 + f(y + z)^2
Multiplying the first by (w+x) and the second by (w-x), and then
subtracting one form the other gives
q(r+s)[(w+x)(r-s) - (w-x)(r+s)] = f(y+z)[(w+x)(y-z) - (w-x)(y+z)]
which reduces to
q(r+s)(xr-ws) = f(y+z)(xy-wz)
Similarly we can eliminate f from the two preceeding equations
to give the relation
q(r+s)(zr-ys) = e(w+x)(wz-xy)
This shows that there are two general classes of solutions to the
overall problem. The degenerate case is when
xr-ws = xy-wz = zr-ys = 0
which is true iff (r/s) = (w/x) = (y/z). In this case we can set
w=jy and x=jz for any rational j, and we have the parametric set of
solutions
a = ewx+ew^2 = ey(y+z)j^2 M = y z e j^2
b = fyz+fy^2 = fy(y+z) N = y z f
c = ewx+ex^2 = ez(y+z)j^2 K = y z (ej^2 + f)
d = fyz+fz^2 = fz(y+z)
On the other hand, if the quantities xr-ws, xy-wz, and zr-ys are
not zero, we can still have solutions by solving the equations
q(r+s)(xr-ws) = f(y+z)(xy-wz)
q(r+s)(zr-ys) = e(w+x)(wz-xy)
for e, f, and q. Clearly the first equation is satisfied if we set
q = (y+z)(xy-wz) and f = (r+s)(xr-ws). Substituting this value of q
into the second equation gives
(y+z)(r+s)(zr-ys)(xy-wz) = e(w+x)(wz-xy)
or
(y+z)(r+s)(ys-zr)
e = -------------------
(w+x)
To ensure that e is an integer we can multiply e, f, and q by (w+x)
to give the solution
e = (r+s)(y+z)(ys-zr)
f = (w+x)(r+s)(xr-ws)
q = (y+z)(w+x)(xy-wz)
Any common factors can be divided out of these three numbers. Then
the final form of the solution (in the non-degenerate case, and
without reducing to lowest terms) is
a = w(ys-zr)J M = (r+s)(y+z)(ys-zr)wx
b = y(xr-ws)J N = (r+s)(x+w)(xr-ws)yz
c = x(ys-zr)J K = (x+w)(y+z)(xy-wz)rs
d = z(xr-ws)J
where J = (x+w)(y+z)(r+s). By the way, notice that
Rx - Ry = K - N - M = (ys-zr)(xr-ws)(xy-wz)
so this is the amount by which the two circuits differ. The smallest
solutions are those for which each of the three values ys-zr, xr-ws,
and xy-wz have their minimum possible value, which is +1, so most of
the small solutions give Rx - Ry = +1.
There's also a connection with what are called "Farey sequences".
Remember that in order to give all positive resistances we require
each of the quantities ys-zr, xr-ws, and xy-wz to be positive, which
is the case if and only if
(y/z) > (r/s) > (w/x)
Now, observe that the four basic resistors a,b,c,d will be proportional
to w,y,x,z if we can ensure that the "commutators" are all +1. All we
really need to do is take any four integers w,x,y,z such that xy-wz=1
and then set r=w+y and s=x+z. This automatically ensures not only that
r/s is between y/z and w/x, but that the quantities xr-ws and ys-zr are
both equal to 1. This automatically gives a solution of the simple
form
a = wJ M = (r+s)(y+z)wx
b = yJ N = (r+s)(x+w)yz
c = xJ K = (x+w)(y+z)rs
d = zJ
where J = (x+w)(y+z)(r+s) and the difference Rx-Ry equals 1. Of course,
the difference between Rx and Ry is also equal to
(ad-bc)^2
------------------
(a+c)(b+d)(a+b+c+d)
which shows (again) why the difference cannot be negative.
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