Answered: R, В D A R, R, R, G H E F

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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Question

The circuit to the right consists of a battery (?0=4.50 V)and five resistors (?1=211 Ω, ?2=182 Ω, ?3=363 Ω, ?4=734 Ω, and ?5=565 Ω). Determine the current passing through each of the specified points.

Expert Solution

Step 1

The given values of the battery emf and resistances are:

V₀ = 4.5 V, R₁ =211 Ω, R₂=182 Ω, R₃ = 363Ω, R₄ =734 Ω, and R₅= 565 Ω

Now from the given circuit the net resistance R:

as the resistances R₂, R₃, R₄ and R₅are in parallel so the net resistance will be $R_{e} = \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} + \frac{1}{R_{5}}$ and these all resistances are in series with R₁the net resistance will be

$R = R_{e} + R 1 R = R_{1} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} + \frac{1}{R_{5}} = 211 + \frac{1}{182} + \frac{1}{363} + \frac{1}{734} + \frac{1}{565} = 211.0113 Ω$

Now the current on the circuit I_E = 4.50V/211.0113Ω

= 0.0213 A

The current in a series connection is same so I_E = I₁ = 0.0213 A

so the voltage drop across the resistance R_1,

$V_{1} = I_{1} R_{1} = 0.0213 \times 211 = 4.494 V$

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