An electronics technician measures the operating parameters for four direct-current (DC) s… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician measures the operating parameters for four direct-current (DC) subsystems on an avionics test bench. Match each electrical subsystem described on the left with its correct operating power consumption on the right.

A relay solenoid with a resistance of $150\text{ }\Omega$ drawing a steady current of $0.4\text{ A}$ $24\text{ W}$
A radio transceiver operating at $28\text{ V}$ that draws a current of $2.5\text{ A}$ $70\text{ W}$
An instrument panel backlight operating from a $12\text{ V}$ line with an internal resistance of $24\text{ }\Omega$ $6\text{ W}$
A de-icing heating element with a resistance of $40\text{ }\Omega$ connected across a $120\text{ V}$ supply $360\text{ W}$

Cevap

The

150\text{ }\Omega

solenoid at

0.4\text{ A}

dissipates

24\text{ W}

; the

28\text{ V}

transceiver at

2.5\text{ A}

consumes

70\text{ W}

; the

12\text{ V}

24\text{ }\Omega

backlight consumes

6\text{ W}

; and the

120\text{ V}

40\text{ }\Omega

heating element dissipates

360\text{ W}

Each electrical load's power dissipation is determined by applying the appropriate variation of Watt's Law based on the two given quantities:

P = I^2 R

for the solenoid yielding

24\text{ W}

P = V \times I

for the transceiver yielding

70\text{ W}

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

for the instrument backlight yielding

6\text{ W}

, and

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

for the de-icing element yielding

360\text{ W}

Adım Adım Çözüm

Calculate power for the

150\text{ }\Omega

relay solenoid with

0.4\text{ A}

current.

P = I^2 R = (0.4)^2 \times 150 = 0.16 \times 150 = 24\text{ W}

When current and resistance are known, use Joule's law / Watt's law variation

P = I^2 R

Calculate power for the

28\text{ V}

transceiver drawing

2.5\text{ A}

P = V \times I = 28 \times 2.5 = 70\text{ W}

When voltage and current are known, use the fundamental power formula

P = V \times I

Calculate power for the

12\text{ V}

backlight with

24\text{ }\Omega

resistance.

P = \frac{V^2}{R} = \frac{12^2}{24} = \frac{144}{24} = 6\text{ W}

When voltage and resistance are known, use the power formula

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

Calculate power for the

120\text{ V}

heating element with

40\text{ }\Omega

resistance.

P = \frac{V^2}{R} = \frac{120^2}{40} = \frac{14400}{40} = 360\text{ W}

When voltage and resistance are known, use

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

Anahtar Kavram

Electrical power is calculated using the three primary variations of Watt's Law:

P = V \times I

P = I^2 R

, and

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

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru