Question

A battery-powered global positioning system (GPS) receiver operating on 9.0 V draws a current of 0.13 A.
How much electrical energy does it consume during 1.5 h?

Answer 1

Concepts and reason

The concepts required to solve this problem are the electric energy and the power.

Initially, calculate the power using the voltage and the current. Then, using the calculated power and the time find the electrical energy consumed.

Fundamentals

The expression of the electrical power is,

$P = VI$

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

The expression of the electrical energy is,

$E = Pt$

Here, E is the electrical energy, P is the power, and t is the time.

The expression of the electrical power is,

$P = VI$

Substitute 9.0 V for V and 0.13 A for I.

$\begin{array}{c}\\P = \left( {9.0{\rm{ V}}} \right)\left( {0.13{\rm{ A}}} \right)\\\\ = 1.17{\rm{ W}}\\\end{array}$

The expression of the electrical energy is,

$E = Pt$

Substitute 1.17 W for P and 1.5 h for t.

$\begin{array}{c}\\E = \left( {1.17{\rm{ W}}} \right)\left( {1.5{\rm{ h}}} \right)\left( {\frac{{60{\rm{ min}}}}{{1{\rm{ h}}}}} \right)\left( {\frac{{60{\rm{ s}}}}{{1{\rm{ min}}}}} \right)\\\\ = 6318{\rm{ J}}\\\end{array}$

Ans:

The electrical energy consumed by GPS is ${\rm{6318 J}}$ .