Question

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce 3.15 x 10⁴ J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Answer 1

Concepts and reason

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

Initially, calculate the power using the electrical energy relation. Then, using the calculated power and the time find the current in the circuit.

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 energy is,

$E = Pt$

Rearrange the above expression for power.

$P = \frac{E}{t}$

Substitute $3.15 \times {10^4}{\rm{ J}}$ for E and 5.25 h for t.

$\begin{array}{c}\\P = \frac{{3.15 \times {{10}^4}{\rm{ J}}}}{{5.25{\rm{ h}}}}\left( {\frac{{1{\rm{ h}}}}{{3600{\rm{ s}}}}} \right)\\\\ = 1.667{\rm{ W}}\\\end{array}$

The expression of the electrical power is,

$P = VI$

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

Rearrange the above expression for I.

$I = \frac{P}{V}$

Substitute 1.667 W for P and 3.70 V for V in the above expression.

$\begin{array}{c}\\I = \frac{{1.667{\rm{ W}}}}{{3.70{\rm{ V}}}}\\\\ = 0.45{\rm{ A}}\\\end{array}$

Ans:

The current is 0.45 A.