Chapter 10 (answers)

Chapter 10 capacitors and electric energy storage

22-19 Solution

(dl)0(a)

A00AQCV(dl)dlVE(dl)23CC0(b) If ld, 0A/(dd)230A/d3

22-26 (a)They are conneted in parallel. (b) CC1C20A0Ad1d2(d1d2)0A

d1d2 (c) 22-48

if d1(or d2)0, Cmaxif d1= d2d, Cmin20A d1(a)we express the energy as UCV22

0lAV212120A0AWUfUiV(CfCi)V()22ddl2d(dl)Q2(b)we express the energy as U2CQ2Q2Q2ddlQ2lQ2lWUfUi()2Cf2Ci20A0A20A20Adl0AV2lW2(dl)2QCiV0AV

22-56

They are conneted in parallel.KAKA(KK2)0A

CC1C2102012d2d2d22-57

They are conneted in seriesdd11122 CC1C2K10AK20AC2K1K20A(K1K2)d22-85

Q2x WFdEQdQ(2xx)20A20AQWe express U in terms of CQ2Q2Q22xxxQ2

UUfUi()2Cf2Ci20A0A20ASo,WU