For a 90 KVA, 240-volt motor, how much current will each circuit conductor carry?

For a 90 KVA, 240-volt motor, how much current will each circuit conductor carry?

• A. 180
• B. 200
• C. 217
• D. 250

Answer: (C)

To determine the current that each circuit conductor will carry for a 90 KVA, 240-volt motor, you can use the formula: \[ \text{Current (I)} = \frac{\text{Power (P) in KVA} \times 1000}{\text{Voltage (V)}} \] In this case, the power is 90 KVA and the voltage is 240 volts. Substituting these values into the formula gives: \[ I = \frac{90 \times 1000}{240} \] Calculating this: \[ I = \frac{90000}{240} \] \[ I = 375 \text{ amperes} \] Since the motor is typically three-phase, each conductor would carry one-third of the total current. Therefore, divide the total current by 3 to find the current per conductor: \[ \text{Current per conductor} = \frac{375}{3} = 125 \text{ amperes} \] However, if the question specifically indicates the rating of a single conductor for calculations involving a three-phase system, the answer is approached differently. Instead of dividing 375 by three, we should consider the need for the full load of

To determine the current that each circuit conductor will carry for a 90 KVA, 240-volt motor, you can use the formula:

[ \text{Current (I)} = \frac{\text{Power (P) in KVA} \times 1000}{\text{Voltage (V)}} ]

In this case, the power is 90 KVA and the voltage is 240 volts. Substituting these values into the formula gives:

[ I = \frac{90 \times 1000}{240} ]

Calculating this:

[ I = \frac{90000}{240} ]

[ I = 375 \text{ amperes} ]

Since the motor is typically three-phase, each conductor would carry one-third of the total current. Therefore, divide the total current by 3 to find the current per conductor:

[ \text{Current per conductor} = \frac{375}{3} = 125 \text{ amperes} ]

However, if the question specifically indicates the rating of a single conductor for calculations involving a three-phase system, the answer is approached differently.

Instead of dividing 375 by three, we should consider the need for the full load of