11th Edition

ISBN: 9780134159386

Author: Dale Ewen, Neill Schurter, Erik Gundersen

Publisher: PEARSON

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Textbook Question

Chapter 2.1, Problem 22P

R = k L d 2 for L

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Chapter 2.1, Problem 21P

Chapter 2.1, Problem 23P

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Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC

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Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µC

Chapter 2 Solutions

Applied Physics (11th Edition)

Ch. 2.1 - =stforS Ch. 2.1 - a=tforV Ch. 2.1 - w = mg for m Ch. 2.1 - F = ma for a Ch. 2.1 - E = IR for R Ch. 2.1 - V = lwh for w Ch. 2.1 - Ep = mgh for g Ch. 2.1 - Ep = mgh for h Ch. 2.1 - 2 = 2gh for h Ch. 2.1 - XL = 2 f L for f

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