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Zorluk: OrtaCapacitors

A circuit in a timing module contains two 30 μF30\text{ }\mu\text{F} capacitors connected in parallel. This parallel pair is then connected in series with a third 15 μF15\text{ }\mu\text{F} capacitor. What is the total equivalent capacitance of this network?

  1. A
    75 μF75\text{ }\mu\text{F}
  2. 12 μF12\text{ }\mu\text{F}Cevap
  3. C
    30 μF30\text{ }\mu\text{F}
  4. D
    0.012 μF0.012\text{ }\mu\text{F}

Cevap

The total equivalent capacitance of the network is 12 μF12\text{ }\mu\text{F}.
For capacitors in parallel, the total capacitance is the sum of the individual values: 30 μF+30 μF=60 μF30\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}. When this 60 μF60\text{ }\mu\text{F} equivalent combination is connected in series with a 15 μF15\text{ }\mu\text{F} capacitor, the product-over-sum rule is applied: (60×15)/(60+15)=900/75=12 μF(60 \times 15) / (60 + 15) = 900 / 75 = 12\text{ }\mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the two parallel capacitors (C1=30 μFC_1 = 30\text{ }\mu\text{F} and C2=30 μFC_2 = 30\text{ }\mu\text{F}).
Cp=C1+C2=30 μF+30 μF=60 μFC_p = C_1 + C_2 = 30\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}
Capacitors in parallel add their plate surface areas together, so their capacitances add directly.
2
Calculate the total capacitance when Cp=60 μFC_p = 60\text{ }\mu\text{F} is placed in series with C3=15 μFC_3 = 15\text{ }\mu\text{F}.
Ctotal=Cp×C3Cp+C3=60×1560+15=90075=12 μFC_{\text{total}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{60 \times 15}{60 + 15} = \frac{900}{75} = 12\text{ }\mu\text{F}
Capacitors in series combine inversely using the reciprocal formula, identical to the formula used for parallel resistors.

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Equivalent Capacitance in Series-Parallel Combinations
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