What exactly is happening above?
Note that the $R$ in
$$V = RI$$
is ordinarily understood to be a constant. We say something like
For the ideal resistor, the voltage $V$ is proportional to the
current $I$ where the constant of proportionality is the
resistance $R$
In your post, you wrote an equation involving the work $W$
$$V = \frac{W}{T}\frac{1}{I}$$
and claimed that this implies that the voltage is inversely proportional to the current.
Now, think about it. If that were true, then doubling the current would halve the voltage, correct? That is, if $\frac{W}{T}$ is a constant of proportionality, then doubling the current would not change the factor $\frac{W}{T}$, correct?
But the factor does change when the current is doubled. If $I' = 2I$, then it's easy to show that
$$\frac{W'}{T} = \frac{4W}{T}$$
Thus
$$V' = \frac{W'}{T}\frac{1}{I'} = \frac{4W}{T}\frac{1}{2I} = 2\frac{W}{T}\frac{1}{I} = 2V$$
So, in fact, doubling the current doubles the voltage, i.e., the voltage and current are proportional. The equation you wrote does not imply that the voltage is inversely proportional to the current.