For capacitors in series, the equivalent capacitance is:

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

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 in a capacitor:

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

We also need to know that, parallel capacitors have the same potential difference across them. Series capacitors carry the same amount of charge.

C_{1} and C_{2} are in series:

C_{12} = (1/C_{1} + 1/C_{2})^{-1} = (1/4.30 + 1/4.30)^{-1} = 2.15 μF

C_{12} and C_{3} are in parallel:

C_{123} = C_{12} + C_{3} = 2.15 + 4.30 = 6.45 μF

C_{123} and C_{4} are in sereis.

In the figure (Figure 1), each capacitor has 4.30 μF and V_{ab} = 32.0 V

Part A: Calculate the charge on capacitor *C*1.

Part B: Calculate the potential difference across capacitor *C*1

Part C: Calculate the charge on capacitor *C*2.

Part D: Calculate the potential difference across capacitor *C*2.

Part E: Calculate the charge on capacitor *C*3.

Part F: Calculate the potential difference across capacitor *C*3.

Part G: Calculate the charge on capacitor *C*4.

Part H: Calculate the potential difference across capacitor C4

Part I: Calculate the potential difference between points a and d

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