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Unformatted text preview: 1 1% 1 1& independent equations relating the two sets of capacitances. The set of equations are # ! '1 ! (, C1 C y ' KC y KCx ( ) * 1 1% 1 1& 1 1 # '1 ! ! and # . From these, subbing in the expression for K , we get ' KC y KC x ( ( C2 C x ) C3 KC yC x * C1 # (C xC y 0 C yCz 0 Cz Cx ) Cx , C2 # (CxC y 0 C yC z 0 C zC x ) C y and C3 # (CxC y 0 C yC z 0 C zC x ) C z . (b) Using the transformation of part (a) we have the equivalent networks shown in Figure 24.73b: Figure 24.73b C1 # 126 $ F , C2 # 28 $ F , C3 # 42 $ F , C4 # 42 $ F , C5 # 147 $ F and C6 # 32 $ F . The total equivalent capacitance %1 1 1 1 1& 0 0 0 0 is Ceq # ' ( 72 $ F 126 $ F 34.8 $ F 147 $ F 72 $ F * ) !1 !1 %% 1 %1 1& 1 && 0 0' 0 34.8 $ F # ' ' ( (. ' ) 42 $ F 32 $ F ( * ) 28 $ F 42 $ F * ( ) * !1 # 14.0 $ F, where the 34.8 $ F comes from 24-24 Chapter 24 (c) The circuit diagram can be redrawn as shown in Figure 25.73c. The overall charge is given by Q # CeqV # (14.0 $ F)(36 V) # 5.04 " 10!4 C . And this is also the charge on the 72 $ F capacitors, so V72 # 5.04 " 10!4 C # 7.0 V . 72 " 10!6 F Figure 24.73c Next we will find the voltage over the numbered capacitors, and their associated voltages. Then those voltages will be changed back into voltage of the original capacitors, and then their charges. QC1 # QC5 # Q72 # 5.04 " 10 !4 C . VC5 # 5.04 " 10!4 C 5.04 " 10!4 C # 3.43 V and VC1 # # 4.00 V . Therefore, !6 147 " 10 F 126 " 10!6 F !1 %1 1& VC2 C4 # VC3C8 # (36.0 ! 7.00 ! 7.00 ! 4.00 ! 3.43) V # 14.6 V . But Ceq (C2C4 ) # ' 0 ( # 16.8 $ F and C2 C4 * ) !1 %1 1& Ceq (C3C6 ) # ' 0 ( # 18.2 $ F , so QC2 # QC4 # VC2 C4 Ceq (C2 C4 ) # 2.45 " 10!4 C and ) C3 C6 * QC QC QC QC3 # QC6 # VC3C6 Ceq( C3C6 ) # 2.64 " 10!4 C . Then VC2 # 2 # 8.8 V , VC3 # 3 # 6.3 V , VC4 # 4 # 5.8 V and C3 C2 C4 QC VC6 # 6 # 8.3 V . Vac # VC1 0 VC2 # V18 # 13 V and Q18 # C18V18 # 2.3 " 10!4 C . Vab # VC1 0 VC3 # V27 # 10 V and C6 Q27 # C27V27...
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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