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Difficulty: MediumElectrical Power

A high-intensity auxiliary floodlight installed on a naval patrol craft has an internal operating resistance of 6 Ω6\text{ }\Omega and connects directly across a 24 V24\text{ V} direct-current (DC) electrical bus. How much electrical power does the floodlight consume during operation?

  1. A
    4 W4\text{ W}
  2. 96 W96\text{ W}Answer
  3. C
    144 W144\text{ W}
  4. D
    0.25 W0.25\text{ W}

Answer

96 W96\text{ W}
Electrical power is calculated using the formula P=V2RP = \frac{V^2}{R}. Substituting 24 V24\text{ V} and 6 Ω6\text{ }\Omega gives P=2426=5766=96 WP = \frac{24^2}{6} = \frac{576}{6} = 96\text{ W}.

Step-by-Step Solution

1
Identify the given circuit parameters
Voltage V=24 VV = 24\text{ V} and Resistance R=6 ΩR = 6\text{ }\Omega
These two values are provided directly in the problem statement.
2
Select the power formula relating voltage and resistance
P=V2RP = \frac{V^2}{R}
Combining Watt's Law (P=V×IP = V \times I) with Ohm's Law (I=VRI = \frac{V}{R}) allows direct calculation of power without first solving for current.
3
Substitute the values and calculate power
P=(24 V)26 Ω=5766=96 WP = \frac{(24\text{ V})^2}{6\text{ }\Omega} = \frac{576}{6} = 96\text{ W}
Squaring 2424 yields 576576, and dividing by 66 gives 96 W96\text{ W}.

Key Concept

Calculating electrical power from voltage and resistance using P=V2RP = \frac{V^2}{R}

Alternative Method

Calculate circuit current first using Ohm's Law: I=VR=24 V6 Ω=4 AI = \frac{V}{R} = \frac{24\text{ V}}{6\text{ }\Omega} = 4\text{ A}. Then apply P=V×I=24 V×4 A=96 WP = V \times I = 24\text{ V} \times 4\text{ A} = 96\text{ W}.
Estimated Time:45s
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Topics

Electrical Power

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