An avionics technician is assembling a filtering network using three capacitors connected … — Armed Services Vocational Aptitude Battery (ASVAB)

An avionics technician is assembling a filtering network using three capacitors connected in series. The capacitors have ratings of $30\text{ }\mu\text{F}$ , $20\text{ }\mu\text{F}$ , and $60\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this series network in microfarads ( $\mu\text{F}$ )?

Answer: 10 µF

Answer

The total equivalent capacitance of the series circuit is

10\text{ }\mu\text{F}

For capacitors connected in series, the total capacitance is found using the formula

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

. Substituting

30\text{ }\mu\text{F}

20\text{ }\mu\text{F}

, and

60\text{ }\mu\text{F}

gives

\frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10\text{ }\mu\text{F}}

. Inverting this fraction yields the correct total capacitance of

10\text{ }\mu\text{F}

, which is less than the smallest individual capacitor (

20\text{ }\mu\text{F}

Step-by-Step Solution

State the series capacitance relationship

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Capacitors in series combine inversely because the total dielectric separation increases, reducing total capacitance.

Calculate the sum of the reciprocals using a common denominator

\frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}

Finding a least common multiple of

60

allows direct fractional addition.

Invert the result to find

C_{\text{total}}

C_{\text{total}} = 10\text{ }\mu\text{F}

The final equivalent capacitance is the inverse of the sum of the reciprocals.

Key Concept

Series capacitance follows the reciprocal sum rule, resulting in a total capacitance smaller than any individual capacitor in the string.

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