An avionics technician constructs a filtering network using three capacitors. Two capacito… — Armed Services Vocational Aptitude Battery (ASVAB)

An avionics technician constructs a filtering network using three capacitors. Two capacitors, rated at $0.05\text{ }\mu\text{F}$ and $50\text{ nF}$ , are connected in series with each other. This series combination is then connected in parallel with a third capacitor rated at $15\text{ nF}$ . What is the total equivalent capacitance of the network in nanofarads ( $\text{nF}$ )?

Cevap: 40 nF

Cevap

The total equivalent capacitance of the network is

40\text{ nF}

First, convert

0.05\text{ }\mu\text{F}

to nanofarads:

0.05\text{ }\mu\text{F} \times 1000\text{ nF}/\mu\text{F} = 50\text{ nF}

. Next, the two series capacitors (

50\text{ nF}

and

50\text{ nF}

) combine via the reciprocal formula to give

25\text{ nF}

. Finally, adding the parallel capacitor (

15\text{ nF}

) yields

25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Adım Adım Çözüm

Convert the microfarad rating to nanofarads to ensure uniform units across all calculations.

0.05\text{ }\mu\text{F} = 0.05 \times 10^3\text{ nF} = 50\text{ nF}

Capacitor combination formulas require all terms to share the same metric prefix.

Calculate the equivalent capacitance of the two capacitors connected in series (

C_1 = 50\text{ nF}

and

C_2 = 50\text{ nF}

C_{\text{series}} = \frac{1}{\frac{1}{50\text{ nF}} + \frac{1}{50\text{ nF}}} = 25\text{ nF}

Capacitors in series combine reciprocally, meaning two identical series capacitors yield half of an individual capacitor's value.

Add the parallel branch capacitance (

C_3 = 15\text{ nF}

) to the series combination equivalent.

C_{\text{total}} = C_{\text{series}} + C_3 = 25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Capacitors in parallel add directly because their plate areas effectively combine to store more charge at a given voltage.

Anahtar Kavram

Series and parallel equivalent capacitance rules with metric prefix conversion

Alternatif Yöntem

Convert all components into microfarads first:

C_1 = 0.05\text{ }\mu\text{F}

C_2 = 0.05\text{ }\mu\text{F}

, and

C_3 = 0.015\text{ }\mu\text{F}

. The series pair gives

\frac{0.05}{2} = 0.025\text{ }\mu\text{F}

. Adding the parallel capacitor gives

0.025\text{ }\mu\text{F} + 0.015\text{ }\mu\text{F} = 0.040\text{ }\mu\text{F}

. Finally, converting

0.040\text{ }\mu\text{F}

back to nanofarads gives

0.040 \times 1000 = 40\text{ nF}

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Konular

Bu konudan daha fazla soru