An electronics technician measures a power dissipation of 20\text{ W} across a 500\ \Omega… — Armed Services Vocational Aptitude Battery (ASVAB)

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

A
$40\text{ mA}$
$200\text{ mA}$ Cevap
C
$0.2\text{ mA}$
D
$100\text{ mA}$

Cevap

200\text{ mA}

According to Watt's Law, power dissipated in a resistive load is

P = I^2 R

. Rearranging to solve for current gives

I = \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\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

Adım Adım Çözüm

Identify known parameters and select the Watt's Law formula relating power, resistance, and current.

P = 20\text{ W}

R = 500\ \Omega

, and

P = I^2 R

The circuit power and resistance are given, and current is the unknown quantity.

Rearrange the formula to solve for current (

I

) and substitute the given values.

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}

Isolating

I

requires dividing power by resistance and taking the square root.

Convert the calculated current from amperes to milliamperes.

0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

The question asks for the final answer in milliamperes.

Anahtar Kavram

Watt's Law relates power, current, and resistance via

P = I^2 R

, which rearranges to

I = \sqrt{\frac{P}{R}}

Alternatif Yöntem

First compute the voltage drop across the resistor using

V = \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 = \frac{V}{R} = \frac{100\text{ V}}{500\ \Omega} = 0.2\text{ A} = 200\text{ mA}

Tahmini Süre:1m 0s

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Cevap

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Alternatif Yöntem

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