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Question

A 3.0 µF capacitor and a 4.0 µF capacitor are connected in series across a 46.0 V battery. A 13.0 µF capacitor is also connected directly across the battery terminals. Find the total charge that the battery delivers to the capacitors.

Answer 1

.....3mf...4mf........................The equivalent series capacitance
.........||.......||.........................of upper mesh is C1 = 12 / 7 mf
N..... ....| |.............P............. C1 and C2=13 mf are in parallel.
.........B:46V.........................So voltage across NP will be 46V
...........+.||.-............................V1=V2 & total Charge Q=Q1+Q2
...........13 mf.......................

Q1 = C1 * V1 = C2 * V =(12/7)*46 *10^-6 quolomb
Q2 = C2 * V2 = C2 * V =(13)*46 *10^-6 quolomb
Total charge supplied by battery (Q). It starts from + terminal, gets divided in to Q1 and Q2, depending on c1, c2, being in parallel. The pd accross PN is same. The series capacitors have same Q1 charge

Q=Q1+Q2=(12/7)*46 +(13)*46]*10^-6 =
= 6.768 *10^-4 quolomb

Answer 2

6.7685X10^-4 Q