Respuesta :

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