In the figure below each capacitance C1 is 69 mu f and each

In the figure below each capacitance C_1 is 6.9 mu f, and each capacitance C_2 is 4.36 mu F. Compute the equivalent capacitance of the network between points a and b. Compute the charge on each of the three capacitors nearest a and b when v_ab = 480 v. With 480 V across a and b, compute V_cd.

Solution

A. you have already calculated Ceq = 2.3 uf

B. from a to b we have three 6.9 uf capacitor in series, so voltage will be equally distributed between these three.

V1 = V2 = V3 = V/3 = 480/3 = 160 V

Q1 = V1*C1 = 6.9*10^-12*160 = 1.104*10^-3 C

Q2 = V2*C2 = 4.3*10^-12*160 = 0.688*10^-3 C

C. on the leftmost C2 voltage is 160 V

Now again this voltage will be equally distributed between next three capacitors.

So Vcd = 160/3 = 53.33 V

Let me know if you have any doubt.