A high-intensity auxiliary floodlight installed on a naval patrol craft has an internal op… — Armed Services Vocational Aptitude Battery (ASVAB)

A high-intensity auxiliary floodlight installed on a naval patrol craft has an internal operating resistance of $6\text{ }\Omega$ and connects directly across a $24\text{ V}$ direct-current (DC) electrical bus. How much electrical power does the floodlight consume during operation?

A
$4\text{ W}$
$96\text{ W}$ Cevap
C
$144\text{ W}$
D
$0.25\text{ W}$

Cevap

96\text{ W}

Electrical power is calculated using the formula

P = \frac{V^2}{R}

. Substituting

24\text{ V}

and

6\text{ }\Omega

gives

P = \frac{24^2}{6} = \frac{576}{6} = 96\text{ W}

Adım Adım Çözüm

Identify the given circuit parameters

Voltage

V = 24\text{ V}

and Resistance

R = 6\text{ }\Omega

These two values are provided directly in the problem statement.

Select the power formula relating voltage and resistance

P = \frac{V^2}{R}

Combining Watt's Law (

P = V \times I

) with Ohm's Law (

I = \frac{V}{R}

) allows direct calculation of power without first solving for current.

Substitute the values and calculate power

P = \frac{(24\text{ V})^2}{6\text{ }\Omega} = \frac{576}{6} = 96\text{ W}

Squaring

24

yields

576

, and dividing by

6

gives

96\text{ W}

Anahtar Kavram

Calculating electrical power from voltage and resistance using

P = \frac{V^2}{R}

Alternatif Yöntem

Calculate circuit current first using Ohm's Law:

I = \frac{V}{R} = \frac{24\text{ V}}{6\text{ }\Omega} = 4\text{ A}

. Then apply

P = V \times I = 24\text{ V} \times 4\text{ A} = 96\text{ W}

Tahmini Süre:45s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Konular

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