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Difficulty: MediumElectrical Power and Watt's Law

An electronics technician measures a power dissipation of 20 W20\text{ W} across a 500 Ω500\ \Omega test resistor in a direct-current circuit. What is the electrical current flowing through this resistor in milliamperes (mA\text{mA})?

  1. A
    40 mA40\text{ mA}
  2. 200 mA200\text{ mA}Answer
  3. C
    0.2 mA0.2\text{ mA}
  4. D
    100 mA100\text{ mA}

Answer

200 mA200\text{ mA}
According to Watt's Law, power dissipated in a resistive load is P=I2RP = I^2 R. Rearranging to solve for current gives I=PR=20 W500 Ω=0.04=0.2 AI = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}. Converting amperes to milliamperes gives 0.2 A×1000 mA/A=200 mA0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}.

Step-by-Step Solution

1
Identify known parameters and select the Watt's Law formula relating power, resistance, and current.
P=20 WP = 20\text{ W}, R=500 ΩR = 500\ \Omega, and P=I2RP = I^2 R
The circuit power and resistance are given, and current is the unknown quantity.
2
Rearrange the formula to solve for current (II) and substitute the given values.
I=PR=20 W500 Ω=0.04=0.2 AI = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}
Isolating II requires dividing power by resistance and taking the square root.
3
Convert the calculated current from amperes to milliamperes.
0.2 A×1000 mA/A=200 mA0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}
The question asks for the final answer in milliamperes.

Key Concept

Watt's Law relates power, current, and resistance via P=I2RP = I^2 R, which rearranges to I=PRI = \sqrt{\frac{P}{R}}.

Alternative Method

First compute the voltage drop across the resistor using V=P×R=20 W×500 Ω=10000=100 VV = \sqrt{P \times R} = \sqrt{20\text{ W} \times 500\ \Omega} = \sqrt{10{}000} = 100\text{ V}. Then apply Ohm's Law to find current: I=VR=100 V500 Ω=0.2 A=200 mAI = \frac{V}{R} = \frac{100\text{ V}}{500\ \Omega} = 0.2\text{ A} = 200\text{ mA}.
Estimated Time:1m 0s
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Topics

Electrical Power and Watt's Law

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