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A 2.80-µF and a 4.50-µF capacitor are connected in series across a 30.0-V battery. A 6.95-µF capacitor is then connected in parallel across the 2.80-µF capacitor. Determine the voltage across the 6.95-µF capacitor.

Respuesta :

Answer:

9.47 V

Explanation:

Here, 6.95 μC and 2.8 μC are in parallel.

So, Cp = 6.95 + 2.8 = 9.75 μC

Now, Cp and 4.5 are in series.

C = 4.5 x 9.75 / (4.5 + 9.75) = 3.08 μC

Let q be the charge.

q = C x V = 3.08 x 30 = 92.4 μC

Voltage across 4.5 μF, V' = 92.4 / 4.5 = 20.53 V

Voltage across 6.95 μF, V'' = V - V' = 30 - 20.53 = 9.47 V

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