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**Hint:**So as to answer this question, we have to know about the concept of power so that we can proceed in this question. We know that power,

$P = \dfrac{W}{t}$, where $W = $Work done, $t = $Time taken

Then we have to convert this formula into electrical power and then will proceed.

**Complete step-by-step solution:**

We know that:

$ \Rightarrow P = \dfrac{W}{t} - - (i)$ where $W = $Work done, $t = $Time taken

$ \Rightarrow W = Vq - - (ii)$

We know that:

$ \Rightarrow q = It$

Substituting the above equation within the equation\[(ii)\]we will get:

$ \Rightarrow W = VIt - - (iii)$

Then substituting the above equation within equation$(i)$we get:

$ \Rightarrow P = \dfrac{{VIt}}{t}$

$ \Rightarrow P = VI$

\[\therefore V = \dfrac{P}{I} - - (iv)\]

In the above question, the values of $P$and $i$ are $1watt$ and $1A$.

Putting these values within equation $(iv)$we get:

\[ \Rightarrow V = \dfrac{{1watt}}{{1A}}\]

\[ \Rightarrow V = 1watt/A\]

$\therefore V = 1volt$

The potential difference when one watt of power is consumed and when $1A$of current flows is $1V$.

**Note:**The above obtained formula for power standard formula for power i.e.,$P = Vi$ but there are two more formulas we can use according to the values given:

$P = VI$

We know that:

$ \Rightarrow V = IR$

Putting this formula in $P = Vi$ we get:

$ \Rightarrow P = {I^2}R$

$I = \dfrac{V}{R}$

Putting this in formula $P = VI$:

$ \Rightarrow P = V \times \dfrac{V}{R} \\

\Rightarrow P = \dfrac{{{V^2}}}{R} \\ $

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