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Q42PE

Expert-verifiedFound in: Page 734

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A charge of \(4.00{\rm{ }}C\) charge passes through a pocket calculator’s solar cells in \(4.00{\rm{ }}h\). What is the power output, given the calculator’s voltage output is \(3.00{\rm{ }}V\)? (See Figure \(20.40\).)**

**The power output of the calculator is obtained as: \(P{\rm{ }} = {\rm{ }}0.833{\rm{ }}mW\).**

**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 calculator is: \(q{\rm{ }} = {\rm{ }}4.00{\rm{ }}C\).
- The time interval needed for that charge to pass is:\(\begin{aligned}{}\Delta t{\rm{ }} &= {\rm{ }}(4.00{\rm{ }}h)(\frac{{3600{\rm{ }}s}}{{1{\rm{ }}h}})\\ &= {\rm{}}14400{\rm{}}s\end{aligned}\)

- The voltage output of the calculator is: \(\Delta V{\rm{ }} = {\rm{ }}3.00{\rm{ }}V\).

Current in the calculator 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:

\(\begin{aligned}{}I{\rm{ }} &= {\rm{ }}\frac{{4.00{\rm{ }}C}}{{14400{\rm{ }}s}}\\ &= {\rm{ }}\frac{1}{{3600}}A\end{aligned}\)

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:

\(\begin{aligned}{}P{\rm{ }} &= {\rm{ }}(\frac{1}{{3600}}A)(3.00{\rm{ }}V)\\ &= {\rm{ }}8.33{\rm{ }} \times {\rm{ }}{10^{ - 4}}{\rm{ }}W\\ &= {\rm{ }}(8.33{\rm{ }} \times {\rm{ }}{10^{ - 4}}{\rm{ }}W)(\frac{{1000{\rm{ }}mW}}{{1{\rm{ }}W}})\\ &= {\rm{ }}0.833{\rm{ }}mW\end{aligned}\)

Therefore, the power output is: \(P{\rm{ }} = {\rm{ }}0.833{\rm{ }}mW\).

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