Question

A $1.5V$flashlight battery is connected to a wire with a resistance of $3.0\Omega$. FIGURE P27.69 shows the battery's potential difference as a function of time. What is the total charge lifted by the charge escalator?

Short answer

The total charge lifted by the charge escalator is $\theta =1800C$.

Step-by-step solution

Given information

A $1.5\mathrm{V}$flashlight battery is connected to a wire with a resistance of $3.0\Omega$. Given picture shows the battery's potential difference as a function of time.

Explanation

Given that, Resistance $\left(R\right)=3\Omega$

The rate of flow of electrical charge $d\theta =1dt$

Hence, the total charge $\theta =\int Idt$

$=\int \frac{V}{R}dt$

$=\frac{1}{R\int Vdt}$

where, $V$indicates voltage$=1.5$

$R$indicates Resistance $=3$

$dt$indicates time

$\int vdt=$Area of triangle$=\frac{bh}{2}$

$\theta =\frac{1}{3}\left[\frac{1.5\times 2\times 60\times 60}{2}\right]$

$=\frac{1}{3}\left[5400\right]$

$\theta =1800C$

The total charge lifted by the charge escalator is $1800c$