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Q43PE

Expert-verifiedFound in: Page 734

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**How many watts does a flashlight that has \(6.00 \times {10^2}{\rm{ }}C\) pass through it in \(0.500{\rm{ }}h\) use if its voltage is \(3.00{\rm{ }}V\)?**

The power output of the flashlight is obtained as: \(P{\rm{ }} = {\rm{ }}1.00{\rm{ }}W\).

**In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega \({\rm{(\Omega )}}\) is used to represent resistance in ohms.**

Electric current is the magnitude of the physical quantity of electric current \({\rm{I}}\) in a wire which equals the magnitude of the electric charge \(q\) that passes through a cross-section of the wire divided by the time interval \(\Delta t\) needed for that charge to pass:

\(I{\rm{ }} = {\rm{ }}\frac{{|q|}}{{\Delta t}}\)……………..(I)

The unit of current is the ampere \(A\), which is equivalent to one coulomb per second \((C/s)\). A current of \(1A\) (one ampere, or amp) means that \(1{\rm{ }}C\) of charge passes through a cross-section of the wire every second. The direction of the current is from higher to lower potential. So, it is in the direction that positive charges that would move.

If a potential difference \(\Delta V\) is maintained across a circuit element, the power, or the rate at which energy is supplied to the element, is:

\(P{\rm{ }} = {\rm{ }}I\Delta V\)………….(II)

- Charge passing through the flashlight is: \(q{\text{ }} = {\text{ }}6.00{\text{ }} \times {\text{ }}10{\text{ }}C\).

- The time interval needed for that charge to pass is:\(\begin{aligned}{}\Delta t{\rm{ }} &= {\rm{ }}(0.500{\rm{ }}h)(\frac{{3600{\rm{ }}s}}{{1{\rm{ }}h}})\\ &= {\rm{ }}1800{\rm{ }}s\end{aligned}\).
- The voltage output of the flashlight is: \(\Delta V{\rm{ }} = {\rm{ }}3.00{\rm{ }}V\).

Current in the flashlight is found by using Ohm's law with the help of the first equation as:

\(I{\rm{ }} = {\rm{ }}\frac{q}{{\Delta t}}\)

Entering values and we obtain:

\(\begin{aligned} I{\text{ }} = {\text{ }}\frac{{6.00{\text{ }} \times {\text{ }}10{\text{ }}C}}{{1800{\text{ }}s}} \\ = {\text{ }}\frac{1}{3}A \\ \end{aligned} \)

The power output of the calculator is found with the help of the second equation as:

\(P{\rm{ }} = {\rm{ }}I\Delta V\)

Substitute numerical values, and we get:

\(\begin{aligned}{}P{\rm{ }} &= {\rm{ }}(\frac{1}{3}{\rm{ }}A)(3.00{\rm{ }}V)\\ &= {\rm{ }}1.00{\rm{ }}W\end{aligned}\)

Therefore, the power output of the flashlight is: \(P{\rm{ }} = {\rm{ }}1.00{\rm{ }}W\).

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