An audio receiver circuit contains three capacitors arranged in a sub-network. Two 8\text{… — Armed Services Vocational Aptitude Battery (ASVAB)

An audio receiver circuit contains three capacitors arranged in a sub-network. Two $8\text{ }\mu\text{F}$ capacitors are connected in series with each other, and this series combination is connected in parallel with a third capacitor rated at $16\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this sub-network in microfarads ( $\mu\text{F}$ )?

Answer: 20 μF

Answer

20 μF

The equivalent capacitance of two

8\text{ }\mu\text{F}

capacitors in series is

4\text{ }\mu\text{F}

because series capacitors follow the formula

\frac{1}{C_s} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}

, resulting in

C_s = 4\text{ }\mu\text{F}

. When this

4\text{ }\mu\text{F}

equivalent section is wired in parallel with the

16\text{ }\mu\text{F}

capacitor, their values add directly (

4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Step-by-Step Solution

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

C_1 = 8\text{ }\mu\text{F}

and

C_2 = 8\text{ }\mu\text{F}

C_{\text{series}} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

, which yields

C_{\text{series}} = 4\text{ }\mu\text{F}

Add the series branch equivalent capacitance (

4\text{ }\mu\text{F}

) to the parallel branch capacitor (

C_3 = 16\text{ }\mu\text{F}

C_{\text{total}} = 20\text{ }\mu\text{F}

Capacitors connected in parallel add directly:

C_{\text{total}} = C_{\text{series}} + C_3 = 4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Key Concept

Series and Parallel Capacitance Calculation

Estimated Time:1m 0s

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