For 2 capacitors in series, the equivalent capacitance is:

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

For capacitors in parallel, the equivalent capacitance is:

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

Capacitor C_{5} and C_{6} are in parallel.

C_{56} = 0.65 + 13 = 13.65 μF

A combination of series and parallel connections of capacitors is shown in the figure. The sizes of these capacitors are given by the follow data:

C_{1} = 4.1 μF

C_{2} = 3.2 μF

C_{3} = 7.7 μF

C_{4} = 1.6 μF

C_{5} = 0.65 μF

C_{6} = 13 μF

Find the total capacitance of the combination of capacitors in microfarads.

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