Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:
[tex]C=C_1+C_2+C_3[/tex]In this case:
C1 = 30µF
C2 = 20µF
C3 = 12µF
Replace the previous values into the formula for C and simplify:
[tex]C=30\mu F+20\mu F+12\mu F=62\mu F[/tex]Hence, the total capacitance is 62µF
