The capacitors have values C1 = 2.5 µF and C2 = 4.0 µF, C3 = 6.0 µF C4 = 8.0 µF and the potential difference across the battery is 12.0 V.

Question

The capacitors have values C1 = 2.5 µF and C2 = 4.0 µF, C3 = 6.0 µF C4 = 8.0 µF and the potential difference across the battery is 12.0 V. Assume that the capacitors are connected in series. Find the equivalent capacitance of the circuit and solve for the potential difference across each capacitors.

Answer 1

Let U = 12V be the voltage across the battery.

Let U1, U2, U3, U4 be the voltages across each of the capacitors (to be calculated).

Let Q (unknown) be the charge which left the battery and is then stored on each of the capacitors.

Let C be the equivalent capacitance of the circuit (to be calculated).

The key observation is that the same charge Q is on each plate of each capacitor.

(Of course, on one plate it is positive and on the other negative.)

By definition of capacitance,

U = Q/C (1)

And for each i

Ui = Q/Ci (2)

By Kirchoff's law

U1 + U2 + U3 + U4 = U

Substituting from (1) and (2)

Q/C1 + Q/C2 + Q/C3 + Q/C4 = Q/C

Dividing by Q

1/C1 + 1/C2 + 1/C3 + 1/C4 = 1/C

Substituting actual numbers

1/C = 1/2.5 + 1/4 + 1/6 + 1/8 = 113/120

C = 120/113 = 1.06 μF

Having calculated C, we can now calculate Q from (1), and each Ui from (2).

Q = UC

U1 = UC/C1 = 12×1.06/2.5 = 5.09V

and so on for the other three capacitors.