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A 4.0-mF and a 14.0-mF capacitor are connected in series, and the series arrangement is connected to a 24.0-mF

capacitor. How much capacitance would a single capacitor need to replace this combination of three capacitors?
A. 17 mF
B. 32 mF
C. 27 mF
D. 13 mF

1 Answer

2 votes

Final answer:

The total capacitance of the combination of a 4.0-mF and a 14.0-mF capacitor in series, parallel to a 24.0-mF capacitor, is approximately 27 mF.

Step-by-step explanation:

To determine the total capacitance of a combination of capacitors, we must consider how they are connected. Capacitors in series and parallel combine differently. For series connections, the reciprocal of the total capacitance (1/Ctotal) is the sum of the reciprocals of the individual capacitances (1/C1 + 1/C2), and for parallel connections, the total capacitance (Ctotal) is the sum of the individual capacitances (C1 + C2).

The 4.0-mF and 14.0-mF capacitors are connected in series, so we calculate their combined capacitance using the formula for series capacitors:

1/Cseries = 1/4.0 mF + 1/14.0 mF
Cseries = 1 / (1/4.0 + 1/14.0) mF
Cseries = 2.8 mF

The series combination is then connected in parallel with the 24.0-mF capacitor. The total capacitance of parallel capacitors is the sum of their capacitances:

Ctotal = Cseries + 24.0 mF
Ctotal = 2.8 mF + 24.0 mF
Ctotal = 26.8 mF

Thus, the equivalent capacitance of this combination is 26.8 mF when rounded to the nearest whole number, it is approximately 27 mF, which corresponds to choice C.

User Greg Maletic
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