Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 23 - Think and Discuss: 121

Answer

a. 2 A. b. 1 A. c. 3 A. d. $4 \Omega$.

Work Step by Step

a. $I = \frac{V}{R} = \frac{12V}{6 \Omega} = 2 A$. b. A pair of resistors wired in series has an equivalent resistance equal to the sum of the resistances. This is Rule 2 on page 443. The two resistors, considered as a branch, then have an equivalent resistance of $12 \Omega$. $I = \frac{V}{R} = \frac{12V}{12 \Omega} = 1 A$. c. The total current in the circuit, which is supplied by the voltage source, equals the sum of the currents in the branches (Rule 3, page 444). 2 A + 1 A = 3 A. d. We effectively have a $12 \Omega$ resistor in parallel with a $6 \Omega$ resistor. The equation is not given in this textbook, but for 2 resistors in parallel, the following relation holds. $$\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$ Apply the equation to this problem. $$\frac{1}{R_{parallel}} = \frac{1}{12 \Omega} + \frac{1}{6 \Omega } = \frac{3}{12 \Omega} $$ $$R_{parallel} = 4 \Omega$$ As a confirmation, one may also apply Ohm's Law and arrive at the same answer. $$3 A = I = \frac{V}{R} = \frac{12V}{x}$$. $$x = 4 \Omega$$.
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