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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 # 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: $$P{\rm{ }} = {\rm{ }}1.00{\rm{ }}W$$.

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

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

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: $$P{\rm{ }} = {\rm{ }}1.00{\rm{ }}W$$. ### Want to see more solutions like these? 