Question

Camille takes her pocket calculator out of her book bag. In the calculator, a 0.007 C charge encounters 15 Ω of resistance every 1.5 seconds. What is the potential difference of the battery? (6 decimal places)

Answer 1

Given:

Charge, Q = 0.007 C

Resistance = 15 Ω

Time, t = 1.5 seconds

Let's find the potential difference of the battery.

To find the potential difference, apply the Ohm's Law:

`V=IR`

Where:

V is the potential difference

I is the current.

R is the resistance.

To find the current, I, apply the formula:

`\begin{gathered} I=(Q)/(t) \\ \\ I=(0.007)/(1.5) \\ \\ I=0.004667\text{ A} \end{gathered}`

The current passing through the battery is 0.004667 Amperes.

Hence, to find the otential difference, V, we have:

`\begin{gathered} V=IR \\ V=0.004667*15 \\ \\ V=0.070005\text{ V} \end{gathered}`

Therefore, the potential difference of the battery is 0.070005 volts.

ANSWER:

0.070005 Volts