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Difficulty: MediumElectrical Power and Watt's Law

A direct-current (DC) relay coil has an internal resistance of 80 Ω80\ \Omega and draws a steady current of 250 mA250\text{ mA}. How much electrical power is dissipated by the coil?

  1. A
    20 W20\text{ W}
  2. B
    5000 W5{}000\text{ W}
  3. 5 W5\text{ W}Answer
  4. D
    320 W320\text{ W}

Answer

The electrical power dissipated by the coil is 5 W5\text{ W}.
To find the power dissipated when current and resistance are known, apply Watt's Law in the form P=I2RP = I^2 R. First, convert 250 mA250\text{ mA} to 0.25 A0.25\text{ A}. Squaring the current gives (0.25 A)2=0.0625 A2(0.25\text{ A})^2 = 0.0625\text{ A}^2. Multiplying by the resistance yields 0.0625 A2×80 Ω=5 W0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}.

Step-by-Step Solution

1
Convert the current from milliamperes (mA) to amperes (A).
I=250 mA=0.25 AI = 250\text{ mA} = 0.25\text{ A}
Standard electrical formulas require current in base units of amperes.
2
Calculate the power dissipation using Watt's Law (P=I2RP = I^2 R).
P=(0.25 A)2×80 Ω=0.0625 A2×80 Ω=5 WP = (0.25\text{ A})^2 \times 80\ \Omega = 0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}
When current and resistance are known, power equals the square of current multiplied by resistance.

Key Concept

Calculating electrical power using current and resistance (P=I2RP = I^2 R) with metric prefix conversion.
Estimated Time:1m 0s
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Topics

Electrical Power and Watt's Law

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