Question

A flashlight bulb uses 190.1W of power when the current in the bulb is 0.64 A. What is the voltage?

Answer 1

Given:

`\begin{gathered} P=190.1\text{ W} \\ I=0.64\text{ A} \end{gathered}`

The power is given as,

`P=VI`

Here, P is the power, V is voltage and I is the current.

Rearranging equation in order to get voltage,

`V=(P)/(I)`

Putting all known values,

`\begin{gathered} V=\frac{190.1\text{ W}}{0.64\text{ A}} \\ =297.03\text{ V} \end{gathered}`

Therefore, the voltage is 297.03 V.