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Zorluk: OrtaCapacitors

An industrial electronics technician is servicing a timing circuit on an automated assembly line. The circuit network contains two 12 μF12\text{ }\mu\text{F} capacitors connected in series, and this combination is connected in parallel with a third capacitor rated at 14 μF14\text{ }\mu\text{F}. What is the total equivalent capacitance of this circuit network in microfarads (μF\mu\text{F})?

Cevap: 20 μF

Cevap

20 μF20\text{ }\mu\text{F}
For the series branch, two identical 12 μF12\text{ }\mu\text{F} capacitors yield an equivalent capacitance of 6 μF6\text{ }\mu\text{F}. When placed in parallel with the 14 μF14\text{ }\mu\text{F} capacitor, the capacitances add directly (6 μF+14 μF6\text{ }\mu\text{F} + 14\text{ }\mu\text{F}), giving a total equivalent capacitance of 20 μF20\text{ }\mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the two 12 μF12\text{ }\mu\text{F} capacitors connected in series.
Cs=6 μFC_s = 6\text{ }\mu\text{F}
Capacitors in series combine reciprocally: 1Cs=1C1+1C2\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}, or Cs=Cn=122=6 μFC_s = \frac{C}{n} = \frac{12}{2} = 6\text{ }\mu\text{F}.
2
Add the capacitance of the parallel branch to the series equivalent.
Ctotal=20 μFC_{\text{total}} = 20\text{ }\mu\text{F}
Capacitors in parallel add directly together: Ctotal=Cs+C3=6 μF+14 μF=20 μFC_{\text{total}} = C_s + C_3 = 6\text{ }\mu\text{F} + 14\text{ }\mu\text{F} = 20\text{ }\mu\text{F}.

Anahtar Kavram

Series and parallel capacitor combinations and equivalent capacitance calculations
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