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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
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