A technician evaluates four direct-current (DC) electrical components operating under stea… — Armed Services Vocational Aptitude Battery (ASVAB)

A technician evaluates four direct-current (DC) electrical components operating under steady-state conditions. Match each component and operating condition on the left with its corresponding dissipated electrical power on the right.

An automotive indicator lamp operating at $12\text{ V}$ and drawing $750\text{ mA}$ of current $9\text{ W}$
A heating element with a resistance of $50\ \Omega$ carrying a steady current of $0.6\text{ A}$ $18\text{ W}$
An instrumentation sensor with an internal resistance of $600\ \Omega$ carrying $200\text{ mA}$ of current $24\text{ W}$
A communications relay coil with a resistance of $100\ \Omega$ connected across a $60\text{ V}$ DC source $36\text{ W}$

Answer

The automotive lamp (

12\text{ V}, 750\text{ mA}

) matches

9\text{ W}

; the heating element (

50\ \Omega, 0.6\text{ A}

) matches

18\text{ W}

; the instrumentation sensor (

600\ \Omega, 200\text{ mA}

) matches

24\text{ W}

; and the relay coil (

100\ \Omega, 60\text{ V}

) matches

36\text{ W}

Each electrical component's power dissipation is determined by selecting the appropriate algebraic form of Watt's Law based on the known parameters:

P = V \times I

for the indicator lamp (

12\text{ V} \times 0.75\text{ A} = 9\text{ W}

P = I^2 \times R

for the heating element (

(0.6\text{ A})^2 \times 50\ \Omega = 18\text{ W}

) and sensor (

(0.2\text{ A})^2 \times 600\ \Omega = 24\text{ W}

), and

P = \frac{V^2}{R}

for the relay coil (

(60\text{ V})^2 / 100\ \Omega = 36\text{ W}

Step-by-Step Solution

Calculate power for the automotive indicator lamp using voltage and current

P = 12\text{ V} \times 0.75\text{ A} = 9\text{ W}

Convert milliamperes to amperes (

750\text{ mA} = 0.75\text{ A}

) and apply the fundamental Watt's Law equation

P = V \times I

Calculate power for the heating element using current and resistance

P = (0.6\text{ A})^2 \times 50\ \Omega = 0.36 \times 50 = 18\text{ W}

Apply Watt's Law in the form

P = I^2 \times R

when current and resistance are known.

Calculate power for the instrumentation sensor using current and resistance

P = (0.2\text{ A})^2 \times 600\ \Omega = 0.04 \times 600 = 24\text{ W}

Convert

200\text{ mA}

0.2\text{ A}

and calculate power using

P = I^2 \times R

Calculate power for the communications relay coil using voltage and resistance

P = \frac{(60\text{ V})^2}{100\ \Omega} = \frac{3600}{100} = 36\text{ W}

Apply Watt's Law in the form

P = \frac{V^2}{R}

when potential difference and resistance are known.

Key Concept

Watt's Law and Electrical Power Dissipation

Estimated Time:1m 15s

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