A circuit in a timing module contains two 30\text{ }\mu\text{F} capacitors connected in pa… — Armed Services Vocational Aptitude Battery (ASVAB)

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

A
$75\text{ }\mu\text{F}$
$12\text{ }\mu\text{F}$ Answer
C
$30\text{ }\mu\text{F}$
D
$0.012\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

12\text{ }\mu\text{F}

For capacitors in parallel, the total capacitance is the sum of the individual values:

30\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}

. When this

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

equivalent combination is connected in series with a

15\text{ }\mu\text{F}

capacitor, the product-over-sum rule is applied:

(60 \times 15) / (60 + 15) = 900 / 75 = 12\text{ }\mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the two parallel capacitors (

C_1 = 30\text{ }\mu\text{F}

and

C_2 = 30\text{ }\mu\text{F}

C_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.

Calculate the total capacitance when

C_p = 60\text{ }\mu\text{F}

is placed in series with

C_3 = 15\text{ }\mu\text{F}

C_{\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.

Key Concept

Equivalent Capacitance in Series-Parallel Combinations

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