The effective capacitance in the circuit is given by:
1/C = ¹/₂ + ¹/₃ + ¹/₆ = ⁶/₆
therefore C = 1μF or
1.0 x 10^-6F.
Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV
= 1.0 x 10^-6 x 12
= 12.0 x 10^-6 Coulomb.
∴ P.d across the 6μF,

V = Q/C =	12.0 x 10-6
	6.0 x 10-6F

= 2.0 volts