High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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.