Interesting connection between a word problem and a physics formula

I came across a math word problem about the time it takes workers to finish a job when they’re together vs. separately and I found its connection to a formula for the equivalent resistance of parallel resistors. I wanted to share it here.

I also wanted to ask for which types of problems and for what assumptions the type of logic detailed below can work. I think this assumes the time it takes the workers to finish a job varies linearly with the size of the job, and also that the voltage a resistor produces when a current is flowing through it varies linearly with the current (Ohm’s law basically which actually doesn’t hold for too high temperatures).

Problem for Workers

It takes $T$ days for $1$ job for $n$ workers together. Each worker $i$ does $r_i$ of a job in $1$ day ($r_i$ job / $1$ day).

Since it takes $T$ days for $1$ job when they’re working together, then in $1$ day together the workers can finish $frac{1}{T}$ of a job.

So we have $sum_{i=0}^{n} r_i = frac{1}{T}$

Now worker $i$ takes ($1$ day / $r_i$ job)

So to finish $1$ job it takes worker $i$, $t_i = frac{1}{r_i}$ days to do it.

So we have $sum_{i=0}^{n} r_i = sum_{i=0}^{n} frac{1}{t_i} = frac{1}{T}$

Problem for Resistors

($G$ is conductance and $R$ is resistance)

$R_{eq}$ volts are produced for $1$ amp of current with $1$ big equivalent resistor. Each small resistor $i$, when $G_i$ amps pass through it produces $1$ volt ($G_i$ amps / $1$ volt).

Since one large equivalent resistor produces $R_{eq}$ volts for $1$ amp of current then one large equivalent resistor also produces $1$ volt for $frac{1}{R_{eq}}$ amps of current.

So we have $sum_{i=0}^{n} G_i = frac{1}{R_{eq}}$

Now small resistor $i$ produces ($1$ volt / $G_i$ amps)

So when $1$ amp passes through it small resistor $i$ produces $R_i = frac{1}{G_i}$ volts

So we have $sum_{i=0}^{n} G_i = sum_{i=0}^{n} frac{1}{R_i} = frac{1}{R_{eq}}$