Question

A battery-powered global positioning system (GPS) receiver operating 9.0 V on draws a current of 0.13 A. How much electrical energy does it consume during 30 minutes?

Short answer

The electric energy consumed by GPS in 30 min is 2100 J.

Step-by-step solution

Concept/Significance of electric current.

The passage of an electric charge is referred to as an electric current. This charge is frequently carried by moving electrons in a wire, but it can also be carried by ions or both electrons and ions, as in an ionized gas.

Determination of the electrical energy does GPS consume during minutes

The electric energy consumed by GPS is given by,

$$\begin{gathered} E=Pt \\ =\left(l.V\right)t \end{gathered}$$

Here, I is the current drawn by GPS, V is the potential of the battery and t is the time taken.

Substitute all the values in the above,

$$\begin{gathered} E=(9.0\mathrm{V}\times 0.13\mathrm{AS})(30\mathrm{m})\left[\frac{60\mathrm{s}}{\mathrm{m}}\right] \\ =2100\mathrm{J} \end{gathered}$$

Thus, the electric energy consumed by GPS in 30 min is 2100 J.