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------------------------------------------------ What is the ampacity of a single-phase resistive load with a nameplate rating of [tex]38.4 \, \text{kW}[/tex], [tex]240 \, \text{V}[/tex]?

A. 160A
B. 240A
C. 384A
D. 580A

Answer :

Final answer:

The ampacity for a single-phase resistive load with a nameplate rating of 38.4 kW and 240 V is 160 A. The correct answer is option a.

Explanation:

The ampacity of a single-phase resistive load can be calculated using the formula P = VI, where P is the power (in watts), V is the voltage (in volts), and I is the current (in amperes).

Given a nameplate rating of 38.4 kW and 240 V, we can rearrange the formula to solve for I (current): I = P / V.

Using the provided information, we calculate the current as follows:

  • Power (P) = 38,400 W (since 38.4 kW is equal to 38,400 W)
  • Voltage (V) = 240 V
  • Current (I) = P / V = 38,400 W / 240 V = 160 A

Therefore, the ampacity for the given load is 160 A.

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