Which scenario requires the most electrical input power?

Which scenario requires the most electrical input power?

• A. A 1/5-horsepower, 24-volt motor which is 75 percent efficient
• B. A 1/4-horsepower, 24-volt motor which is 80 percent efficient
• C. A 1/3-horsepower, 120-volt motor which is 60 percent efficient
• D. A 1/2-horsepower, 24-volt motor which is 90 percent efficient

Answer: (A)

The amount of electrical input power a motor needs is the output power divided by its efficiency: P_in = P_out / η. The output power is the motor’s horsepower converted to watts, using 1 hp ≈ 746 W. - For the 1/5-hp, 24-V motor at 75% efficiency: P_out = 0.2 × 746 ≈ 149 W; P_in ≈ 149 / 0.75 ≈ 200 W. - For the 1/4-hp, 24-V motor at 80% efficiency: P_out = 0.25 × 746 ≈ 187 W; P_in ≈ 187 / 0.80 ≈ 233 W. - For the 1/3-hp, 120-V motor at 60% efficiency: P_out ≈ 0.333 × 746 ≈ 249 W; P_in ≈ 249 / 0.60 ≈ 414 W. - For the 1/2-hp, 24-V motor at 90% efficiency: P_out = 0.5 × 746 ≈ 373 W; P_in ≈ 373 / 0.90 ≈ 414 W. The highest input power occurs in the two higher-power scenarios, about 414 W each, so they require the most electrical input power (a tie between the 1/3-hp at 60% and the 1/2-hp at 90%).

The amount of electrical input power a motor needs is the output power divided by its efficiency: P_in = P_out / η. The output power is the motor’s horsepower converted to watts, using 1 hp ≈ 746 W.

• For the 1/5-hp, 24-V motor at 75% efficiency: P_out = 0.2 × 746 ≈ 149 W; P_in ≈ 149 / 0.75 ≈ 200 W.

• For the 1/4-hp, 24-V motor at 80% efficiency: P_out = 0.25 × 746 ≈ 187 W; P_in ≈ 187 / 0.80 ≈ 233 W.

• For the 1/3-hp, 120-V motor at 60% efficiency: P_out ≈ 0.333 × 746 ≈ 249 W; P_in ≈ 249 / 0.60 ≈ 414 W.

• For the 1/2-hp, 24-V motor at 90% efficiency: P_out = 0.5 × 746 ≈ 373 W; P_in ≈ 373 / 0.90 ≈ 414 W.

The highest input power occurs in the two higher-power scenarios, about 414 W each, so they require the most electrical input power (a tie between the 1/3-hp at 60% and the 1/2-hp at 90%).