If a 0.047 ?F, 0.047 ?F and 0.47 ?F capacitor are connected in series across a 25 V source, the total capacitance equals ________ and each 0.047 ?F capacitor drops ________.
A) 0.564 ?F, 1.2 V
B) 0.564 ?F, 11.9 V
C) 0.022 ?F, 11.9 V
D) 0.022 ?F, 25 V

Question

Quizplus · Accepted Answer

The total capacitance (C_total) of capacitors in series is given by the reciprocal of the sum of the reciprocals of the individual capacitances: 1/C_total = 1/C1 + 1/C2 + 1/C3. Using the given values, 1/C_total = 1/0.047 + 1/0.047 + 1/0.47, which simplifies to C_total ≈ 0.022 µF. The voltage drop across each capacitor in a series circuit is proportional to its capacitance. The 0.047 µF capacitors have the same capacitance, so they will have the same voltage drop. Using the formula V = Q/C, where Q is the charge and is the same for all capacitors in series, and knowing the total voltage and total capacitance, we can find the voltage drop across the 0.047 µF capacitors to be approximately 11.9 V.

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