Equivalent capacitance for capacitors in parallel:

$\overline{){{\mathit{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathit{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathit{C}}}_{{\mathbf{2}}}{\mathbf{+}}{{\mathit{C}}}_{{\mathbf{3}}}}$

Equivalent capacitance for capacitors in series:

$\overline{)\frac{\mathbf{1}}{{\mathbf{C}}_{{\mathbf{eq}}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{C}}_{{\mathbf{1}}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{{\mathbf{2}}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{{\mathbf{3}}}}}$

C_{3} and C_{4} are in parallel:

C_{3,4} = (C_{3} + C_{4}) = (3.00 + 5.00) = 8.00 μF

Consider the combination of capacitors shown in the diagram, where *C*_{1} = 3.00 μF , *C*_{2} = 11.0 μF , *C*_{3} = 3.00 μF , and *C*_{4} = 5.00 μF . (Figure 1)

Find the equivalent capacitance *C*_{A} of the network of capacitors. Express your answer in microfarads.

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