Soru

Zorluk: Çok zorCapacitors

A technician is analyzing a power supply smoothing circuit. The circuit contains three capacitors arranged in a specific configuration: Capacitor C1C_1 (30μF30 \mu\text{F}) and Capacitor C2C_2 (15μF15 \mu\text{F}) are connected in series. This entire series branch is then connected in parallel with Capacitor C3C_3 (90μF90 \mu\text{F}). What is the total equivalent capacitance of this capacitor network?

  1. 100μF100 \mu\text{F}Cevap
  2. B
    30μF30 \mu\text{F}
  3. C
    135μF135 \mu\text{F}
  4. D
    9μF9 \mu\text{F}

Cevap

The correct equivalent capacitance is 100μF100 \mu\text{F}.
To find the total capacitance, first evaluate the series branch. Unlike resistors, capacitors in series use the inverse formula: C12=(30×15)/(30+15)=10μFC_{12} = (30 \times 15) / (30 + 15) = 10 \mu\text{F}. This equivalent capacitor is in parallel with C3C_3. Parallel capacitors simply add their values together: 10μF+90μF=100μF10 \mu\text{F} + 90 \mu\text{F} = 100 \mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the series branch containing C1C_1 and C2C_2.
C12=30×1530+15=45045=10μFC_{12} = \frac{30 \times 15}{30 + 15} = \frac{450}{45} = 10 \mu\text{F}
Capacitors in series add inversely, which is the exact opposite of resistors. The product-over-sum formula applies to two capacitors in series.
2
Calculate the total equivalent capacitance by combining the series branch (C12C_{12}) in parallel with C3C_3.
Ctotal=10+90=100μFC_{\text{total}} = 10 + 90 = 100 \mu\text{F}
Capacitors in parallel add directly together, as the effective plate area is increased.

Anahtar Kavram

Calculating equivalent capacitance in mixed series and parallel circuits.
Tahmini Süre:1m 30s
Bu soruyu puanla

Konular