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Expert-verified Found in: Page 734 ### College Physics (Urone)

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: $$P{\rm{ }} = {\rm{ }}0.833{\rm{ }}mW$$.

See the step by step solution

## Step 1: Define Resistance

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.

## Step 2: Concepts and Principles

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)

## Step 3: The given data and the required data

• 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: $$\Delta V{\rm{ }} = {\rm{ }}3.00{\rm{ }}V$$.

## Step 4: Evaluating the power output of the calculator

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: $$P{\rm{ }} = {\rm{ }}0.833{\rm{ }}mW$$. ### Want to see more solutions like these? 