What fuse size is required for a 100 HP DC motor at 220V with 90% efficiency?

Study for the New York City DOB Master Electrician Exam. Use flashcards and multiple choice questions with hints and explanations. Boost your exam readiness!

Multiple Choice

What fuse size is required for a 100 HP DC motor at 220V with 90% efficiency?

Explanation:
To determine the required fuse size for a 100 HP DC motor operating at 220V with a 90% efficiency, we first need to convert the horsepower to watts, as electrical calculations typically use watts or kilowatts. 1. Calculate the output power in watts: - 1 HP is approximately 746 watts. Therefore, 100 HP equals: \[ 100 \, \text{HP} \times 746 \, \text{W/HP} = 74600 \, \text{W} \] 2. Since the motor has an efficiency of 90%, the input power can be calculated using the formula: \[ \text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} \] Thus, we have: \[ \text{Input Power} = \frac{74600 \, \text{W}}{0.90} = 82888.89 \, \text{W} \] 3. To find the current drawn by the motor, we use the formula: \[ I = \frac{P}{V} \] Where \(P\)

To determine the required fuse size for a 100 HP DC motor operating at 220V with a 90% efficiency, we first need to convert the horsepower to watts, as electrical calculations typically use watts or kilowatts.

  1. Calculate the output power in watts:
  • 1 HP is approximately 746 watts. Therefore, 100 HP equals:

[

100 , \text{HP} \times 746 , \text{W/HP} = 74600 , \text{W}

]

  1. Since the motor has an efficiency of 90%, the input power can be calculated using the formula:

[

\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}}

]

Thus, we have:

[

\text{Input Power} = \frac{74600 , \text{W}}{0.90} = 82888.89 , \text{W}

]

  1. To find the current drawn by the motor, we use the formula:

[

I = \frac{P}{V}

]

Where (P)

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy