Question

Find the total capacitance C tot of the combination of capacitors shown in the figure, where C 1 = 5.15 μF , C 2 = 3.55 μF , C 3 = 7.75 μF , C 4 = 1.25 μF , C 5 = 0.750 μF , and C 6 = 15.0 μF . A complex circuit containing 6 capacitors. The lead wire at the top of the diagram splits into three parallel branches that recombine at the bottom of the diagram. The left branch contains capacitors C subscript 1 and C subscript 2 connected in series. The central branch has the single capacitor C subscript 3. The right branch begins with capacitor C subscript 4, then the branch itself splits into two parallel branches, with capacitor C subscript 5 in one branch and capacitor C subscript 6 in the other. Therefore, the right branch consists of C sub 4 connected in series with the parallel combination of C sub 5 and C sub 6. C tot = 11.926 μF Question Credit: OpenStax College Physics

Answer 1

- 9 = 5. Isufi = 0.750 uF = 3.55uF; C2 = 775 uf; y = 1.25 uf. C = 15.out sol Complex circuit is like this: Remember! In parallel combination: C, ta In series combination GG Gey So, for Total Capicatance: 42 = GG = 5.15X355 = 18.8825 Cita 5.1583.55 8.7 an - 2.10 uF 923 - G2+9= 2.10 +7.75= 9.85 MP 66= 84% = 15+0.75= 15.75 MF Cuso = cuxor = 1,25x15.75 - 19.6875 = 1.158 - Cyto 1.25 + 1575 17 hence Gotal = 9,23+ Cygo 9.85 +1.158= 11.008 uf