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Zorluk: OrtaCapacitors

A pulse-generator circuit contains two 50 μF50\text{ }\mu\text{F} capacitors connected in parallel. This parallel pair is then connected in series with a third capacitor rated at 25 μF25\text{ }\mu\text{F}. What is the total equivalent capacitance of the circuit in microfarads (μF\mu\text{F})?

Cevap: 20 uF

Cevap

The total equivalent capacitance of the circuit is 20 μF20\text{ }\mu\text{F}.
The correct answer is 20 μF20\text{ }\mu\text{F}. When two 50 μF50\text{ }\mu\text{F} capacitors are wired in parallel, their total capacitance is 50 μF+50 μF=100 μF50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}. When this 100 μF100\text{ }\mu\text{F} equivalent capacitance is connected in series with the 25 μF25\text{ }\mu\text{F} capacitor, the product-over-sum formula yields 100×25100+25=2500125=20 μF\frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the two parallel capacitors (C1C_1 and C2C_2).
Cp=50 μF+50 μF=100 μFC_p = 50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}
Capacitors in parallel add their plate surface areas directly, so their individual capacitance values are summed (Cp=C1+C2C_p = C_1 + C_2).
2
Calculate the total series capacitance combining the 100 μF100\text{ }\mu\text{F} parallel equivalent with the 25 μF25\text{ }\mu\text{F} third capacitor (C3C_3).
Ctotal=100×25100+25=2500125=20 μFC_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}
Capacitors in series combine reciprocally according to the product-over-sum rule for two components: Ctotal=Cp⋅C3Cp+C3C_{\text{total}} = \frac{C_p \cdot C_3}{C_p + C_3}.

Anahtar Kavram

Capacitors combine opposite to resistors: parallel capacitors add directly (Cp=C1+C2C_p = C_1 + C_2), whereas series capacitors combine using reciprocal formulas (Cs=C1⋅C2C1+C2C_s = \frac{C_1 \cdot C_2}{C_1 + C_2}).

Alternatif Yöntem

Use the standard reciprocal formula for the series stage: 1Ctotal=1Cp+1C3=1100+125=1100+4100=5100=120\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{100} + \frac{1}{25} = \frac{1}{100} + \frac{4}{100} = \frac{5}{100} = \frac{1}{20}, hence Ctotal=20 μFC_{\text{total}} = 20\text{ }\mu\text{F}.
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