In Fig. P24.59, each capacitance C1 is 6.9 µF, and each capacitance C2 is 4.6 µF. (a) Compute the equivalent capacitance of the network between points a and b . (b) Compute the charge on each of the three capacitors nearest a and b when Vab = 420 V. (c) With 420 V across a and b, compute vcd.

Solution 63P Step 1 : Introduction : Here, we need to find the equivalent capacitance between a and b In the second part, we need to find charge on each of the three capacitors nearest to a and b and c and d Data given C1= 6.9 F C1= 4.6 F V = 420 V ab Considering the given figure