Question

One watt of power is consumed when $1A$ of current flows at potential difference $.......V$.

Answer 1

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} \\\