Question

Consider the circuit shown in the figure, with C1 = 2.52 uF and C2 = 7.94 uF. 6.00 F 2.00 F C 90.0 V (a) Find the equivalent capacitance (in pF) of the system. 1.77x This is the equivalent capacitance for the upper branch. You need to find the equivalent capacitance for the system. uf (b) Find the charge (in C) on each capacitor. 2.52 pF capacitor C N C 6.00 pF capacitor 7.94 pF capacitor C C 2.00 uF capacitor CNC μα (c) Find the potential difference (in V) on each capacitor. 2.52 pF capacitor 6.00 uF capacitor 7.94 uF capacitor 2.00 uF capacitor (d) Find the total energy (in m3) stored by the group. Cm

Answer 1

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a) C & GuF Capacitor are in series so - 1 + 1 = L td - é a 6MF 2.52uF GUF G = 1.775 uF & similarly Cg & Zuf are also inseris so 1-0 g x 7.94 v66 = 1- 598 u F 7.94+2 So equivalent capacitance ca & G are in Cega ei + Cg' = 1.598+1.775= 3.37 uF

int Q=CV charge on top fra branch & Co V Q = 1.775 X600x 90 = 159.75 ŭe l charge is same in series so change on 2.52 uF = 159.7 Suc! change on GMF = 15975 MC sinulanly Qb = Ca V=1.598x90 MC=143-82uc so change on 7.94 MF = 1143.82 uel & charge on 2 MF = [143.82 me! Potential diff verence on each capacitor For 2.52 uF V=159.75 -163.4V 2.52 For GuF V = 159.75 = 26. 62v] For 709 4uF V = 143.82 = [18.11V / For 2 UE r = 143.82 - (71.918) Toth energy E= { leqriz _ *3-37 X1536x96 7.96 2 ay 2 2 E=113.65m ]