Answered: 29. (I) What is the maximum voltage…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Can you please type out all of your solution so that they are easy for me to read I have bad eyesight...

$### Question 29 **Question:** What is the maximum voltage that can be applied across a 3.9-kΩ resistor rated at 1/4 watt? --- ### Explanation In this question, we are given a resistor with a resistance value of $ 3.9 \, \text{kΩ} $ and a power rating of $ \frac{1}{4} \, \text{watt} $. We are required to determine the maximum voltage that can be applied across it. To solve this problem, we can use the power formula for resistors: \[ P = \frac{V^2}{R} \] Where: - $ P $ is the power rating of the resistor (in watts). - $ V $ is the voltage across the resistor (in volts). - $ R $ is the resistance of the resistor (in ohms). Given: - $ P = \frac{1}{4} \, \text{watt} $ - $ R = 3.9 \, \text{kΩ} = 3900 \, \text{Ω} $ Rearranging the formula to solve for $ V $: \[ V = \sqrt{P \times R} \] Now, substitute the given values: \[ V = \sqrt{\frac{1}{4} \, \text{watt} \times 3900 \, \text{Ω}} \] \[ V = \sqrt{0.25 \times 3900} \] \[ V = \sqrt{975} \] \[ V \approx 31.2 \, \text{volts} \] ### Answer The maximum voltage that can be applied across a 3.9-kΩ resistor rated at 1/4 watt is approximately **31.2 volts**.$

Transcribed Image Text:### Question 29 **Question:** What is the maximum voltage that can be applied across a 3.9-kΩ resistor rated at 1/4 watt? --- ### Explanation In this question, we are given a resistor with a resistance value of $ 3.9 \, \text{kΩ} $ and a power rating of $ \frac{1}{4} \, \text{watt} $. We are required to determine the maximum voltage that can be applied across it. To solve this problem, we can use the power formula for resistors: \[ P = \frac{V^2}{R} \] Where: - $ P $ is the power rating of the resistor (in watts). - $ V $ is the voltage across the resistor (in volts). - $ R $ is the resistance of the resistor (in ohms). Given: - $ P = \frac{1}{4} \, \text{watt} $ - $ R = 3.9 \, \text{kΩ} = 3900 \, \text{Ω} $ Rearranging the formula to solve for $ V $: \[ V = \sqrt{P \times R} \] Now, substitute the given values: \[ V = \sqrt{\frac{1}{4} \, \text{watt} \times 3900 \, \text{Ω}} \] \[ V = \sqrt{0.25 \times 3900} \] \[ V = \sqrt{975} \] \[ V \approx 31.2 \, \text{volts} \] ### Answer The maximum voltage that can be applied across a 3.9-kΩ resistor rated at 1/4 watt is approximately **31.2 volts**.

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