What is the maximum current that a 32 KW, 480-volt generator can deliver to a load?

What is the maximum current that a 32 KW, 480-volt generator can deliver to a load?

• A. 25.0
• B. 30.5
• C. 38.5
• D. 40.0

Answer: (C)

To determine the maximum current that a 32 kW, 480-volt generator can deliver to a load, you can use the formula for three-phase power, which is given by: \[ P = \sqrt{3} \times V \times I \] Where: - \( P \) is the power in watts (W), - \( V \) is the voltage in volts (V), - \( I \) is the current in amperes (A). First, convert kilowatts to watts: 32 kW = 32,000 W. Next, since the generator is operating at a voltage of 480 volts, you can rearrange the formula to solve for current \( I \): \[ I = \frac{P}{\sqrt{3} \times V} \] Now, substitute the values into the formula: \[ I = \frac{32,000}{\sqrt{3} \times 480} \] Calculating \( \sqrt{3} \) gives approximately 1.732. Now insert that into the equation: \[ I = \frac{32,000}{1.732 \times 480} \] \[ I = \frac{32,000}{829.

To determine the maximum current that a 32 kW, 480-volt generator can deliver to a load, you can use the formula for three-phase power, which is given by:

[ P = \sqrt{3} \times V \times I ]

Where:

• ( P ) is the power in watts (W),

• ( V ) is the voltage in volts (V),

• ( I ) is the current in amperes (A).

First, convert kilowatts to watts:

32 kW = 32,000 W.

Next, since the generator is operating at a voltage of 480 volts, you can rearrange the formula to solve for current ( I ):

[ I = \frac{P}{\sqrt{3} \times V} ]

Now, substitute the values into the formula:

[ I = \frac{32,000}{\sqrt{3} \times 480} ]

Calculating ( \sqrt{3} ) gives approximately 1.732. Now insert that into the equation:

[ I = \frac{32,000}{1.732 \times 480} ]

[ I = \frac{32,000}{829.