**Problem Statement:**What is the resistance of a 1.00×10²Ω, a 2.5×10³Ω, and a 4.00×10³Ω resistor connected in parallel?**Answer Units: [Ω]****Explanation:**To find the equivalent resistance of resistors connected in parallel, use the formula:\[\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]Where \( R_1 = 100 \, \Omega \), \( R_2 = 2500 \, \Omega \), and \( R_3 = 4000 \, \Omega \). Calculate the reciprocal of the equivalent resistance and then take the inverse to find \( R_{\text{eq}} \). **Question:**What is the resistance of a \(1.00 \times 10^2 \, \Omega\), a \(2.5 \times 10^3 \, \Omega\), and a \(4.00 \times 10^3 \, \Omega\) resistor connected in series?**Answer Units:** \([k\Omega]\)**Explanation:**To find the total resistance of resistors connected in series, simply add their resistances together. The formula for total resistance \( R_{\text{total}} \) in series is:\[ R_{\text{total}} = R_1 + R_2 + R_3 \]Where \( R_1 = 1.00 \times 10^2 \, \Omega \), \( R_2 = 2.5 \times 10^3 \, \Omega \), and \( R_3 = 4.00 \times 10^3 \, \Omega \).Convert each resistance to kilo-ohms:- \( R_1 = 1.00 \times 10^2 \, \Omega = 0.10 \, k\Omega \)- \( R_2 = 2.5 \times 10^3 \, \Omega = 2.50 \, k\Omega \)- \( R_3 = 4.00 \times 10^3 \, \Omega = 4.00 \, k\Omega \)Adding these values gives:\[ R_{\text{total}} = 0.10 \, k\Omega + 2.50 \, k\Omega + 4.00 \, k\Omega = 6.60 \, k\Omega \]Thus, the total resistance of the resistors connected in series is \( 6.60 \, k\Omega \).

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Answered: What is the resistance of a 1.00x10², a…

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What is the resistance of a 1.00x10², a 2.5x10³, and a 4.00x10³ resistor connected in parallel?

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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