Question

my custom study session: question 2 of 58
unit rating
one motor (75 percent efficient) 1/5 hp
three position lights 20 watts each
one heating element 5 amp
one anticollision light 3 amp
(note: 1 horsepower = 746 watts)
a 402 watts.
b 385 watts.
c 450 watts.
correct choice: c
asa question number 8016
category: all
reference: faa - h - 8083 - 30
aktr code: am.ia.k12
the efficiency rating, 149/x = 75/100; x = 199. therefore, the 1/5 horsepo
motor that is 75 percent efficient requires 199 watts. the three position l
require a total of 60 watts. the heating element requires 120 watts. the
anticollision light requires 72 watts. the total power the generator must p
is 451 watts.

Explanation:

Step1: Calculate motor power

Given \(1\) horsepower \(= 746\) watts, so \(\frac{1}{5}\) horsepower \(=\frac{1}{5}\times746 = 149.2\) watts. Let the power input to the motor be \(X\). Using the efficiency formula \(\text{efficiency}=\frac{\text{power output}}{\text{power input}}\), \(\frac{149.2}{X}=\frac{75}{100}\), solving for \(X\) gives \(X=\frac{149.2\times100}{75}\approx199\) watts.

Step2: Calculate position - lights power

Three - position lights with \(20\) watts each: \(3\times20 = 60\) watts.

Step3: Calculate heating - element power

Assume voltage \(V = 24\) (common in such set - ups, if not given, we can also use \(V = 120\) for standard electrical systems). Using \(P=VI\), for \(I = 5\) amp, \(P=120\times5=60\) (if \(V = 12\)) or \(P = 120\times5 = 600\) (incorrect, likely \(V=24\), \(P=24\times5 = 120\) watts).

Step4: Calculate anticollision - light power

Using \(P = VI\), with \(I = 3\) amp and \(V = 24\), \(P=24\times3=72\) watts.

Step5: Calculate total power

Total power \(P_{total}=199 + 60+120 + 72=451\approx450\) watts (rounding differences may occur).

Answer:

C. 450 watts.