Equivalent capacitance for series capacitors:

$\overline{)\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}}$

Equivalent capacitance for parallel capacitors:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{2}}}}$

The 20 μF and 60 μF are in parallel:

C_{20,60} = C_{20} + C_{60} = 20 + 60 = 80 μF

What is the equivalent capacitance of the three capacitors in the figure?

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