For capacitors in parallel:

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

For series:

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

**(A)**

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

C_{34} = 3.00 + 5.00 = 8.00 μF

C_{2} and C_{34} are in parallel.

C_{234} = 11.0 + 8.00 = 19.0 μF

A) 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.

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

B) Two capacitors of capacitance C_{5} = 6.00 μF and C_{6} = 3.00 μF are added to the network, as shown in the diagram. Find the equivalent capacitance C_{B} of the new network of capacitors.

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