A technician is analyzing a power supply smoothing circuit. The circuit contains three cap… — Armed Services Vocational Aptitude Battery (ASVAB)

A technician is analyzing a power supply smoothing circuit. The circuit contains three capacitors arranged in a specific configuration: Capacitor $C_1$ ( $30 \mu\text{F}$ ) and Capacitor $C_2$ ( $15 \mu\text{F}$ ) are connected in series. This entire series branch is then connected in parallel with Capacitor $C_3$ ( $90 \mu\text{F}$ ). What is the total equivalent capacitance of this capacitor network?

$100 \mu\text{F}$ Answer
B
$30 \mu\text{F}$
C
$135 \mu\text{F}$
D
$9 \mu\text{F}$

Answer

The correct equivalent capacitance is

100 \mu\text{F}

To find the total capacitance, first evaluate the series branch. Unlike resistors, capacitors in series use the inverse formula:

C_{12} = (30 \times 15) / (30 + 15) = 10 \mu\text{F}

. This equivalent capacitor is in parallel with

C_3

. Parallel capacitors simply add their values together:

10 \mu\text{F} + 90 \mu\text{F} = 100 \mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the series branch containing

C_1

and

C_2

C_{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.

Calculate the total equivalent capacitance by combining the series branch (

C_{12}

) in parallel with

C_3

C_{\text{total}} = 10 + 90 = 100 \mu\text{F}

Capacitors in parallel add directly together, as the effective plate area is increased.

Key Concept

Calculating equivalent capacitance in mixed series and parallel circuits.

Estimated Time:1m 30s

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