Please solve the Problem.

The sum of the capacitance is calculated as follows;
1/Ct = 1/C + 1/3
1/Ct = 1/10μf + 1/3μf
1/Ct = (3μf + 10μf)/30μf²
1/Ct = 13μf/30μf²
Ct = 30μf²/13μf
Ct = 2.31μf
The total capacitance in parallel arrangement is calculated as follows;
Ct = 2.31μf + 6μf = 8.31μf
1/Ct = 1/8.31μf + 1/20μf
1/Ct = 0.1703
Ct = 5.87μf
Maximum voltage is delivered in 20μf, q = CV
charge for 20μf:
q = (5.87 x 16)μC
q = 93.92 μC
Ct = 2.31μf + 6μf = 8.31uf
charge for 6μf:
q = (6/8.31) x 93.92μC
q = 67.8μC
q = 93.92μC - 67.8μC = 26.12 μC
charge for C = charge 3μf = 26.12 μC
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