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A circuit contains a 1V cell and some identical 1 ohm resistors. A voltage of a/b, where $a\leq b$, is to be made across a voltmeter using the minimum number of resistors in the circuit. The voltage across part of a circuit = resistance across that part/total resistance. The total resistance of n resistors in parallel = 1/n, and resistance of n in series = n.
Pretend that there is no resistance in the wires, and that the voltmeter does not draw any current. $a,b,m,n,p\subset N$

Denote the minimum number of resistors required f(a,b).

I have noticed that $f(mp,mn+1)\leq n+m$ if $m. When is $f(mp,mn+1)< n+m$, if at all?

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    @joriki $a/b$ could be written as the difference of two fractions,e.g., a voltage divider with $12$ resistors will give voltage of $1/12$ but $1/12 = 1/3 - 1/4$ and two voltage dividers with $3$ and $4$ resistors can be used to get $1/12$ volts with just $7$ resistors. So, $f(a,b) \leq \min M+N$ where the minimum is over all $M, N$ for which there exist $m$, $n$ such that $\frac{a}{b} = \frac{m}{M} - \frac{n}{N}$ – 2011-10-18

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Saying anything general about $f(a,b)$ may be difficult. I note the review by Wan Ding Ding of Lou Shapiro, Some open questions about random walks, involutions, limiting distributions, and generating functions, Adv. in Appl. Math. 27 (2001), no. 2-3, 585–596, MR1868982 (2003b:60060), which says, in part,

In this paper, the author systematically presents 12 open questions which had originally been presented at the Foatafest in the fall of 2000.... The first open question presented by the author is to find the simple family of circuits that have resistances $C_{2n}/C_{2nāˆ’1}$, $n\ge1$, where $C_m={{2m} \choose m}/(m+1)$ is the $m$th Catalan number [where all resistors are one ohm resistors].