An electronics technician is testing an optical pulse circuit and measures that a capacito… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician is testing an optical pulse circuit and measures that a capacitor stores an electrostatic charge of $450\text{ }\mu\text{C}$ when connected across a potential difference of $15\text{ V}$ . What is the capacitance of this component?

A
$6{}750\text{ }\mu\text{F}$
B
$30\text{ }\mu\text{H}$
$30\text{ }\mu\text{F}$ Answer
D
$0.03\text{ }\mu\text{F}$

Answer

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

Capacitance is defined by the fundamental relationship

C = \frac{Q}{V}

. Dividing the stored charge of

450\text{ }\mu\text{C}

by the applied potential difference of

15\text{ V}

yields

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

Step-by-Step Solution

Identify the relationship between charge, capacitance, and voltage

The governing equation is

Q = C \times V

, which rearranges to

C = \frac{Q}{V}

Capacitance is defined as the amount of electric charge stored per unit of electrical potential difference.

Substitute the given values into the formula and solve for capacitance

C = \frac{450\text{ }\mu\text{C}}{15\text{ V}} = 30\text{ }\mu\text{F}

Dividing the charge in microcoulombs by the potential difference in volts directly yields the capacitance in microfarads.

Key Concept

The capacitance (

C

) of a component is the ratio of electric charge (

Q

) stored on its plates to the applied voltage (

V

), measured in Farads (F).

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Key Concept

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