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

C_eq= 1/(1/2.5+1/4+1/6+1/8))=1.062 µF ... Q=C_eqV= 1.062E-6*12=12.74E-6 C.

V_2.5 =2.5E-6/12.74E-6 = 5.10 volt

V₄ = 3.18 volt

V₆= 2.12 volt

V₈= 1.59 volt