Equivalent capacitance for parallel capacitors:

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

Equivalent capacitance for series capacitors:

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

The 24µF, C_{1}, and 8.0 µF capacitors are in series:

Determine the equivalent capacitance between A and B for the group of capacitors in the drawing. Let C_{1} = 13 µF and C_{2} = 6.0 µF.

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