Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 757: 51

Answer

(a) $I_{7.1\Omega}=0.29A,I_{3.2\Omega}=1.1A$ (b) $1.4A$ (c) $11.3V$

Work Step by Step

(a) We know that the expression of the voltage drops in loop $ABCF$ can be written as $(4.5\Omega) i_1+(1\Omega) i_1+(3.2\Omega)(i_1-i_2)+(0.50)i_1-12.0V=0$ $\implies (9.2\Omega) i_1-(3.2\Omega) i_2=12.0V$.....eq(1) This simplifies to: $i_2=0.19875i_1$....eq(2) We plug in this value in eq(1) $(9.2\Omega) i_1-(3.2\Omega)(0.19875i_1)=12.0V$ $\implies i_1=1.40A$ We plug in this value in eq(2) to obtain: $i_2=0.19875(1.40A)$ $i_2=0.29A$ The current flowing through the resistance $3.2\Omega$ can be determined as $i_{3.2\Omega}=i_1-i_2$ $\implies i_{3.2\Omega}=1.40A-0.28A=1.1A$ (b) We know that the current flowing through the battery is $i_1$ -- that is, the current flowing through the battery is $1.40A$. (c) We know that $V=12.0V-i_1r$ We plug in the known values to obtain: $V=12.0V-(1.40A)(0.50\Omega)$ $\implies V=11.30V$
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