Match each electrical load scenario on the left with its corresponding steady-state power … — Armed Services Vocational Aptitude Battery (ASVAB)

Match each electrical load scenario on the left with its corresponding steady-state power dissipation on the right.

A $12\text{ V}$ cooling fan motor drawing a steady current of $2.5\text{ A}$ $30\text{ W}$
A $0.5\text{ }\Omega$ current-limiting shunt resistor carrying a current of $8\text{ A}$ $32\text{ W}$
A $16\text{ }\Omega$ heating element connected across a $24\text{ V}$ direct current (DC) supply $36\text{ W}$
Two $50\text{ }\Omega$ load resistors connected in parallel across a $50\text{ V}$ direct current (DC) line $100\text{ W}$

The fan motor (

12\text{ V}, 2.5\text{ A}

) matches

30\text{ W}

; the shunt resistor (

0.5\text{ }\Omega, 8\text{ A}

) matches

32\text{ W}

; the heating element (

24\text{ V}, 16\text{ }\Omega

) matches

36\text{ W}

; and the parallel resistors (

50\text{ V}, 2 \times 50\text{ }\Omega

) match

100\text{ W}

Each load condition requires selecting the appropriate formula derived from Watt's Law and Ohm's Law:

P = VI

yields

30\text{ W}

for the motor;

P = I^2R

yields

32\text{ W}

for the current shunt;

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

yields

36\text{ W}

for the heating resistor; and parallel reduction with

P = \frac{V^2}{R_{eq}}

yields

100\text{ W}

for the dual load.

Calculate the power consumed by the cooling fan motor using

P = V \times I

P = 12\text{ V} \times 2.5\text{ A} = 30\text{ W}

When voltage and current are explicitly provided, the direct product gives the electrical power.

Calculate the power dissipated in the shunt resistor using

P = I^2 \times R

P = (8\text{ A})^2 \times 0.5\text{ }\Omega = 64 \times 0.5 = 32\text{ W}

When current and resistance are known, squaring the current and multiplying by resistance gives power dissipation.

Calculate the power dissipated in the heating element using

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

P = \frac{(24\text{ V})^2}{16\text{ }\Omega} = \frac{576}{16} = 36\text{ W}

When voltage across a component and its resistance are known, dividing the squared voltage by the resistance gives the dissipated power.

Determine the equivalent parallel resistance of the two

50\text{ }\Omega

resistors and calculate total circuit power.

R_{eq} = \frac{50\text{ }\Omega \times 50\text{ }\Omega}{50\text{ }\Omega + 50\text{ }\Omega} = 25\text{ }\Omega

;

P = \frac{(50\text{ V})^2}{25\text{ }\Omega} = \frac{2500}{25} = 100\text{ W}

Combining parallel branches determines the total circuit load before applying the power equation.

Calculation of electrical power dissipation across varied circuit configurations using Watt's Law formulas (

P = VI

P = I^2R

, and

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

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