If your cell phone has a battery that can hold 1859 milliampere-hours of charge, how long will...

Note: According to the research conducted by Lawerence Berkeley National Lab, a cell phone uses approximately {eq}\bf{3\text{ W}} {/eq} of power when charged.

The expression for cell phone charge {eq}Q {/eq} in terms of supply current {eq}I {/eq} and operation time {eq}t {/eq} is given by,

{eq}Q=I\times t\text{ }\ \ \ \ \quad \left( 1 \right) {/eq}

Now, contemplating the provided data, the cell phone is capable of holding a charge of {eq}1859\text{ mAh} {/eq}. Consequently, substitute the value of the charge {eq}Q=1859\text{ mAh} {/eq} in equation 1 to obtain an expression for operation time {eq}t {/eq}.

{eq}\begin{aligned} 1859\text{ mAh}&=I\times t \\ t&=\dfrac{1859\text{ mAh}}{I}\text{ }\ \ \ \ \quad \left( 2 \right) \\ \end{aligned} {/eq}

Further, determine the supply current {eq}I {/eq} required to draw a {eq}\text{3}\text{.5 V} {/eq} voltage continuously from the expression for the power {eq}P {/eq} dissipated by the cell phone in terms of drawn voltage {eq}V {/eq} and supplied current {eq}I {/eq}. Consequently, the expression can be given by,

{eq}P=VI\text{ }\ \ \ \ \quad \left( 3 \right) {/eq}

Now, substitute the values of power {eq}P=3\text{ W} {/eq}, voltage {eq}V=3.5\text{ V} {/eq} in equation 3 to obtain the supply current {eq}I {/eq}.

{eq}\begin{aligned} 3\text{ W}&=\left( 3.5\text{ V} \right)I \\ I&=\dfrac{3\text{ W}}{3.5\text{ V}} \\ &\approx 0.86\text{ A} \\ \end{aligned} {/eq}

Finally, substitute the value of supply current {eq}I=0.86\text{ A} {/eq} in equation 2 to obtain the operation time {eq}t {/eq} of the cell phone.

{eq}\begin{aligned} t&=\dfrac{1859\text{ mAh}}{0.86\text{ A}} \\ &=\dfrac{1.859\text{ Ah}}{0.86\text{ A}} \\ &\approx 2.2\text{ h} \\ \end{aligned} {/eq}

Therefore, considering the provided specifications, the cell phone can operate for {eq}\bf{2.2 \text{ hours}} {/eq}.