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An avionics communication filter assembly contains a capacitor network designed for signal conditioning. The network consists of two 40 μF40\text{ }\mu\text{F} capacitors connected in series. This series combination is connected in parallel with an 80 μF80\text{ }\mu\text{F} capacitor. Finally, this entire parallel arrangement is connected in series with a single 25 μF25\text{ }\mu\text{F} capacitor. What is the total equivalent capacitance of the network, in microfarads (μF\mu\text{F})?

Cevap: 20 μF

Cevap

The total equivalent capacitance of the network is 20 μF20\text{ }\mu\text{F}.
The correct equivalent capacitance is 20 μF20\text{ }\mu\text{F}. In capacitor networks, series components combine reciprocally (1Cs=1C1+1C2\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}) and parallel components add directly (Cp=C1+C2C_p = C_1 + C_2). Resolving the initial series branch yields 20 μF20\text{ }\mu\text{F}, which in parallel with 80 μF80\text{ }\mu\text{F} produces 100 μF100\text{ }\mu\text{F}. Finally, combining 100 μF100\text{ }\mu\text{F} in series with 25 μF25\text{ }\mu\text{F} results in 100×25100+25=20 μF\frac{100 \times 25}{100 + 25} = 20\text{ }\mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the initial series branch consisting of two 40 μF40\text{ }\mu\text{F} capacitors.
Cs1=20 μFC_{s1} = 20\text{ }\mu\text{F}
Capacitors in series combine reciprocally: 1Cs1=140 μF+140 μF=240 μF=120 μF\frac{1}{C_{s1}} = \frac{1}{40\text{ }\mu\text{F}} + \frac{1}{40\text{ }\mu\text{F}} = \frac{2}{40\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}.
2
Calculate the equivalent capacitance of the parallel combination formed by Cs1C_{s1} and the 80 μF80\text{ }\mu\text{F} capacitor.
Cp=100 μFC_p = 100\text{ }\mu\text{F}
Capacitors in parallel add directly: Cp=Cs1+C3=20 μF+80 μF=100 μFC_p = C_{s1} + C_3 = 20\text{ }\mu\text{F} + 80\text{ }\mu\text{F} = 100\text{ }\mu\text{F}.
3
Calculate the total circuit capacitance by placing the parallel equivalent CpC_p in series with the 25 μF25\text{ }\mu\text{F} capacitor.
Ctotal=20 μFC_{\text{total}} = 20\text{ }\mu\text{F}
Applying the series formula for the final combination: 1Ctotal=1Cp+1C4=1100 μF+125 μF=1+4100 μF=5100 μF=120 μF\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_4} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{25\text{ }\mu\text{F}} = \frac{1+4}{100\text{ }\mu\text{F}} = \frac{5}{100\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}.

Anahtar Kavram

Calculation of total equivalent capacitance in compound series-parallel networks using reciprocal series addition and direct parallel addition rules.

Alternatif Yöntem

Use the product-over-sum shortcut for each two-component series reduction: Cs1=40×4040+40=160080=20 μFC_{s1} = \frac{40 \times 40}{40 + 40} = \frac{1600}{80} = 20\text{ }\mu\text{F}. Next, add the parallel component directly: Cp=20+80=100 μFC_p = 20 + 80 = 100\text{ }\mu\text{F}. Finally, apply product-over-sum for the last series stage: Ctotal=100×25100+25=2500125=20 μFC_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}.
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