C = 1.50µF C2 = 2.65µF%3DC3 = 3. 00µFC4 = 4.50µFE- 14.00 VElectrical QuantitiesTotal21.Capacitance(F)22.23.25.27.29.Charge(C)24.26.28.30.Voltage(V)14.00 V

Answered: Image /qna-images/answer/2afd5c9a-8515-489a-b082-aa93f28cd641.jpg

Answered: C = 1.50µF C2 = 2.65µF HE C3 = 3.00LF…

C = 1.50µF C2 = 2.65µF HE C3 = 3.00LF C= 4.50µF E 14.00 Electrical Quantities Total 21. Сарacitance (F) 22. 23. 25. 27. 29. Charge (C) 24. 26. 28. 30. Voltage (V) 14.00 V

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter21: Electric Fields

Section: Chapter Questions

Problem 82A

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Concept explainers

Combination of Capacitors

The connection of capacitors can be established in a circuit in two ways.

Question

C = 1.50µF C2 = 2.65µF
%3D
C3 = 3. 00µF
C4 = 4.50µF
E- 14.00 V
Electrical Quantities
Total
21.
Capacitance
(F)
22.
23.
25.
27.
29.
Charge
(C)
24.
26.
28.
30.
Voltage
(V)
14.00 V

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