In a single-phase system, what is the total current drawn by a load if the load’s kVA rating is 17.2?

In a single-phase system, what is the total current drawn by a load if the load’s kVA rating is 17.2?

• A. 82.7 amps
• B. 48.6 amps
• C. 59.2 amps
• D. 166 amps

Answer: (A)

To determine the total current drawn by a load in a single-phase system from its kVA rating, you can use the formula: \[ \text{Current (I)} = \frac{\text{kVA} \times 1000}{\text{Voltage (V)}} \] In this case, if we assume a common voltage level (e.g., 120V or 240V), we can proceed to calculate the current based on a kVA rating of 17.2. For example, at a standard voltage of 240 volts: \[ I = \frac{17.2 \times 1000}{240} \] \[ I = \frac{17200}{240} \] \[ I = 71.67 \text{ amps} \] However, if you multiply that kVA by a different standard voltage, say 480V, you would get: \[ I = \frac{17.2 \times 1000}{480} \] \[ I = \frac{17200}{480} \] \[ I = 35.83 \text{ amps} \] For option A suggesting 82.7 amps,

To determine the total current drawn by a load in a single-phase system from its kVA rating, you can use the formula:

[

\text{Current (I)} = \frac{\text{kVA} \times 1000}{\text{Voltage (V)}}

]

In this case, if we assume a common voltage level (e.g., 120V or 240V), we can proceed to calculate the current based on a kVA rating of 17.2.

For example, at a standard voltage of 240 volts:

[

I = \frac{17.2 \times 1000}{240}

]

[

I = \frac{17200}{240}

]

[

I = 71.67 \text{ amps}

]

However, if you multiply that kVA by a different standard voltage, say 480V, you would get:

[

I = \frac{17.2 \times 1000}{480}

]

[

I = \frac{17200}{480}

]

[

I = 35.83 \text{ amps}

]

For option A suggesting 82.7 amps,