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 charge stored in a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

Energy stored in a capacitor:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}}$

C_{eq} = (1/C_{1} + 1/C_{2})^{-1} = (1/2.00 + 1/7.40)^{-1} = 1.57 μF(10^{-6}F/μF) = 1.574 × 10^{-6} F

Suppose you have a 9.00-V battery, a 2.00-μF capacitor, and a 7.40-μF capacitor

Find the charge and energy stored if the capacitors are connected to the battery in series. .

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