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Difficulty: EasyElectrical Power

A heating resistor with a resistance of 20 \Omega20\text{ \Omega} has a steady current of 3 A3\text{ A} flowing through it. What is the total electrical power dissipated by the resistor?

  1. A
    60 W60\text{ W}
  2. B
    120 W120\text{ W}
  3. 180 W180\text{ W}Answer
  4. D
    18 W18\text{ W}

Answer

180 W180\text{ W}
Electrical power dissipated by a resistor when current and resistance are known is determined by the formula P=I2×RP = I^2 \times R. Substituting the given values gives P=(3 A)2×20 \Omega=9×20=180 WP = (3\text{ A})^2 \times 20\text{ \Omega} = 9 \times 20 = 180\text{ W}.

Step-by-Step Solution

1
Identify the known circuit parameters and the appropriate power formula.
Known values: I=3 AI = 3\text{ A}, R=20 \OmegaR = 20\text{ \Omega}. Applicable formula: P=I2×RP = I^2 \times R.
When current and resistance are known, Watt's law in the form P=I2RP = I^2 R gives electrical power directly.
2
Square the current value.
I2=(3 A)2=9 A2I^2 = (3\text{ A})^2 = 9\text{ A}^2.
Power dissipation is proportional to the square of the current flowing through a resistor.
3
Multiply the squared current by the resistance.
P=9×20=180 WP = 9 \times 20 = 180\text{ W}.
Multiplying squared current by resistance yields total dissipated power in watts.

Key Concept

Electrical power dissipation in a resistive component using Joule's/Watt's law (P=I2RP = I^2 R).
Estimated Time:45s
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Topics

Electrical Power

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