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Difficulty: MediumElectrical Power

A direct-current starter relay coil in a heavy transport vehicle has an internal resistance of 15 Ω15\text{ }\Omega and draws a steady operating current of 6 A6\text{ A}. What is the total electrical power dissipated by the relay coil?

  1. A
    90 W90\text{ W}
  2. 540 W540\text{ W}Answer
  3. C
    180 W180\text{ W}
  4. D
    2.5 W2.5\text{ W}

Answer

540 W540\text{ W}
Electrical power dissipated by a resistor with a known current is found using the formula P=I2×RP = I^2 \times R. Squaring the current of 6 A6\text{ A} gives 36 A236\text{ A}^2, and multiplying by the coil resistance of 15 Ω15\text{ }\Omega results in 540 W540\text{ W}.

Step-by-Step Solution

1
Identify the given circuit parameters and required quantity.
Current I=6 AI = 6\text{ A}, Resistance R=15 ΩR = 15\text{ }\Omega, Unknown = Power (PP).
Selecting the correct formula depends on the given parameters.
2
Select the appropriate power formula relating power, current, and resistance.
P=I2×RP = I^2 \times R
When current and resistance are known, Joule's/Watt's Law formula P=I2RP = I^2 R directly determines electrical power.
3
Square the current value and multiply by resistance.
P=(6 A)2×15 Ω=36×15=540 WP = (6\text{ A})^2 \times 15\text{ }\Omega = 36 \times 15 = 540\text{ W}
Completing the arithmetic yields the power in watts.

Key Concept

Electrical power dissipation in a resistive component is directly proportional to resistance and the square of the current (P=I2RP = I^2 R).
Estimated Time:1m 0s
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Topics

Electrical Power

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