Answer :
When capacitors are connected in parallel, the total capacitance (C_total) is the sum of the individual capacitances:
C_total = C1 + C2 + C3
C_total = 6F + 2F + 4F
C_total = 12F
So, the total capacitance when these capacitors are connected in parallel is 12F.
When capacitors are connected in series, the inverse of the total capacitance (1/C_total) is the sum of the inverses of the individual capacitances:
1/C_total = 1/C1 + 1/C2 + 1/C3
1/C_total = 1/6F + 1/2F + 1/4F
1/C_total = (2/12 + 6/12 + 3/12)F
1/C_total = 11/12F
C_total = 12F/11
So, the total capacitance when these capacitors are connected in series is 12F/11.
The potential difference across each capacitor in a parallel connection is the same as the potential difference of the battery, which is 12V.
The potential difference across each capacitor in a series connection is divided among the capacitors according to their capacitance. To calculate the potential difference across each capacitor, we can use the formula:
V_capacitor = (C_total / C_individual) * V_battery
For C1:
V1 = (12F/11 / 6F) * 12V = 2.1818V
For C2:
V2 = (12F/11 / 2F) * 12V = 10.909V
For C3:
V3 = (12F/11 / 4F) * 12V = 5.4545V
So, the potential difference across each capacitor when they are connected in series is approximately V1 = 2.1818V, V2 = 10.909V, and V3 = 5.4545V.
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