Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Stop to Think 26.5 - Page 728: 1


$C_b > C_a=C_d>C_c$

Work Step by Step

We can find the capacitance of parallel capacitors by using $C_{eq} = \displaystyle \Sigma_i^n$ $C_i = C_1+C_2+...+C_n$ and the capacitance of series capacitors using $C_{eq} = \Sigma_i^n$ $\displaystyle \frac{1}{C_{i}} = \frac{1}{C_1} +\frac{1}{C_2} +...+\frac{1}{C_n}$ ---- The capacitance of A is just $5\mu$F since there's only one capacitor The capacitance of B is $6\mu$F since they're connected in parallel and $C_b = 3 + 3 = 6\mu$F The capacitance of C is $1.5\mu$F since they're connected in series and $\displaystyle \frac{1}{C_c} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ $\displaystyle C_c = \frac{3}{2}\mu$F The capacitance of D is $5\mu$F. We must first find the capacitance of the capacitors on the right side which are in series. $\displaystyle \frac{1}{C_{right}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$ $C_{right} = 2\mu$F Now D is in parallel: $C_d = C_{left} + C_{right} = 3+2 = 5\mu$F From largest to smallest capacitance, we can rank these as $C_b > C_a=C_d>C_c$
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