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------------------------------------------------ Three capacitors are connected to a battery with a potential difference of 12 V. Their capacitances are [tex]C_1 = 6 \, \text{F}[/tex], [tex]C_2 = 2 \, \text{F}[/tex], and [tex]C_3 = 4 \, \text{F}[/tex].

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.

To know more about "Capacitance" refer here:

brainly.com/question/29591088#

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