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Difficulty: MediumElectrical Power

A direct-current (24 V24\text{ V} DC) cooling fan motor installed in an electronics enclosure has an internal resistance of 16 Ω16\text{ }\Omega. What is the total electrical power consumed by the fan motor during normal operation?

  1. A
    1.5 W1.5\text{ W}
  2. B
    24 W24\text{ W}
  3. 36 W36\text{ W}Answer
  4. D
    384 W384\text{ W}

Answer

The total electrical power consumed by the fan motor is 36 W36\text{ W}.
Electrical power (PP) delivered to a resistive load is given by P=V2RP = \frac{V^2}{R}. Substituting the given parameters yields P=(24 V)216 Ω=57616=36 WP = \frac{(24\text{ V})^2}{16\text{ }\Omega} = \frac{576}{16} = 36\text{ W}. Alternatively, by Ohm's Law, the current drawn is I=24 V16 Ω=1.5 AI = \frac{24\text{ V}}{16\text{ }\Omega} = 1.5\text{ A}, yielding P=V×I=24 V×1.5 A=36 WP = V \times I = 24\text{ V} \times 1.5\text{ A} = 36\text{ W}.

Step-by-Step Solution

1
Determine the current flowing through the motor using Ohm's Law (I=VRI = \frac{V}{R}).
I=24 V16 Ω=1.5 AI = \frac{24\text{ V}}{16\text{ }\Omega} = 1.5\text{ A}
Current is needed to calculate power using Watt's Law (P=V×IP = V \times I).
2
Calculate the electrical power using Watt's Law (P=V×IP = V \times I) or directly using P=V2RP = \frac{V^2}{R}.
P=24 V×1.5 A=36 WP = 24\text{ V} \times 1.5\text{ A} = 36\text{ W} (or P=24216=57616=36 WP = \frac{24^2}{16} = \frac{576}{16} = 36\text{ W})
Electrical power is the rate of energy consumption and is given by the product of voltage and current.

Key Concept

Electrical power dissipation in a resistive direct-current load calculated from voltage and resistance (P=V2R=V×IP = \frac{V^2}{R} = V \times I).
Estimated Time:1m 0s
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Topics

Electrical Power

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