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:

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

The charge stored by a capacitor:

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

From the figure, C_{1} and C_{2} are in parallel.

Therefore, C_{12} = C_{1} + C_{2} = 4 + 12 = 16 μF

C_{12} and C_{3} are in series.

What are the charge on and the potential difference across each capacitor in the figure ?

Part A. Q_{1}

Part B. V_{1}

Part C. Q_{2}

Part D. V_{2}

Part E. Q_{3}

Part F. V_{3}

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