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

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)